Physics · Secondary School · Statics

Statics

The bridge that carries lorries, the picture hanging crooked on the wall, the wheel brace that only works with a pipe slipped over the handle and the pile of books that collapses at the wrong moment all obey the same two rules. Statics studies bodies in equilibrium, and we will see that they need a zero resultant force and, when they can turn, a zero resultant moment as well. We start from a point hanging by strings, move through torque, levers and the centre of mass, and arrive at bridges and pulleys. In the calculations of this lesson we use \(g = 10\ \text{m/s}^2\).

1Equilibrium of a point
2Moment of a force
3Extended body
4Levers
5Centre of mass
6Beam on two supports
7Pulleys
✓Challenges

STEP 1

Why is the washing line never straight?

When the size of a body and its rotation do not matter for the problem, we can treat it as a particle. It is in equilibrium when the resultant of the forces acting on it is zero, and since forces are vectors, the sum has to be zero horizontally and vertically, separately.

To add forces at an angle, we resolve each one into two perpendicular components, using the sine and cosine of the angle. In a picture hanging from two identical strings, the horizontal components of the tensions cancel, and the vertical ones, together, hold up the weight.

Out of this comes a result that tends to surprise. The wider the strings open, the smaller the vertical component of each tension, and the tension has to grow to keep holding the same weight. With the strings horizontal no tension would be enough, and that is why a washing line with clothes on it sags a little in the middle, however tightly we pull it.

\(\vec F_R = \vec 0\)\(\sum F_x = 0\ \text{ and }\ \sum F_y = 0\)\(F_x = F\cos\theta\)\(F_y = F\sin\theta\)\(T = \dfrac{W}{2\cos\theta}\) (symmetric strings)With symmetric strings, \(\theta\) is the angle of each string to the vertical. We assume ideal strings, massless and inextensible, and \(g = 10\ \text{m/s}^2\). In the triangle of forces we draw \(\vec W\), \(\vec T_2\) and \(\vec T_1\) one after the other, and the figure closes because the resultant is zero.

weight W tension T₁ (left string) tension T₂ (right string)

Let's discuss

Start with the strings almost vertical and open them little by little. At what angle does each tension become equal to the weight of the picture?
Take the opening up to 85° and compare the tensions with the weight. Why can a very taut washing line snap with only a few clothes on it?
Create an asymmetry and see which string now pulls harder. Our first intuition tends to bet on the longer string, and it is worth checking whether it gets it right.
Double the mass of the picture without touching the angles. Does the triangle of forces change shape, or only size?

Opening of the strings

Asymmetry

Mass of the picture

Weight W

Tension T₁

Tension T₂

Larger tension

Resultant

Exercises for step 1 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A 2 kg flowerpot hangs from the ceiling by a single vertical string and stays still. What is the tension in the string? Use \(g = 10\ \text{m/s}^2\).

Show solution
At rest, the pot is in equilibrium, and the two forces acting on it, the weight downwards and the tension upwards, have to cancel.
The weight is \(W = m\,g = 2 \cdot 10\), and the tension has the same magnitude.
\(T = 20\ \text{N}\)
basic
When we hang wet clothes on a washing line, nobody manages to leave the line perfectly straight, however hard they pull it. Why does this happen?

Show solution
At the point where the clothes are pegged, the tensions on the two sides of the line need a vertical component that balances the weight of the clothes.
With the line straight and horizontal, that component would be zero, and no tension, however large, would hold the clothes up.
The line always sags a little, because only when it is at an angle does it have a vertical component to hold the weight.
basic
A 4 kg picture hangs from two identical vertical strings, fixed at the two ends of the frame. What is the tension in each string? Use \(g = 10\ \text{m/s}^2\).

Show solution
With the strings vertical and symmetric, each one holds half of the weight \(W = 4 \cdot 10 = 40\ \text{N}\).
\(T = 20\ \text{N}\) in each string
basic
A 10 N force pulls a crate at 60° to the floor. What are the horizontal and vertical components of the force? Use \(\cos 60^\circ = 0.5\) and \(\sin 60^\circ \approx 0.87\).

Show solution
The horizontal component is the projection of the force on the floor, \(F_x = F\cos 60^\circ = 10 \cdot 0.5\), and the vertical one is \(F_y = F\sin 60^\circ = 10 \cdot 0.87\).
\(F_x = 5\ \text{N}\) and \(F_y \approx 8.7\ \text{N}\)
intermediate
A 3 kg picture hangs on a nail from two symmetric strings, each at 60° to the vertical. What is the tension in each string? Use \(g = 10\ \text{m/s}^2\). Remember that \(\cos 60^\circ = 0.5\).

Show solution
The horizontal components of the two tensions cancel, and the vertical ones add up to the weight, \(2\,T\cos\theta = W\).
With \(W = 30\ \text{N}\), we have \(T = \dfrac{W}{2\cos\theta} = \dfrac{30}{2 \cdot 0.5}\).
\(T = 30\ \text{N}\), the whole weight in each string, even though there are two
intermediate
A 2.4 kg lamp hangs from two symmetric cables, each at 37° to the vertical. What is the tension in each cable? Use \(g = 10\ \text{m/s}^2\). Use \(\cos 37^\circ = 0.8\).

Show solution
In the symmetric case, \(2\,T\cos\theta = W\), with \(W = 24\ \text{N}\).
Hence \(T = \dfrac{24}{2 \cdot 0.8} = \dfrac{24}{1.6}\).
\(T = 15\ \text{N}\)
intermediate
A 5 kg sack hangs from a rope. Gabriel pushes the sack with a horizontal force until the rope makes 37° with the vertical, and the sack stays still. What is Gabriel's force, and what is the tension in the rope? Use \(g = 10\ \text{m/s}^2\). Use \(\sin 37^\circ = 0.6\) and \(\cos 37^\circ = 0.8\).

Show solution
Vertically, only the rope holds the weight, \(T\cos 37^\circ = 50\), and so \(T = \dfrac{50}{0.8} = 62.5\ \text{N}\).
Horizontally, Gabriel's force balances the horizontal component of the tension, \(F = T\sin 37^\circ = 62.5 \cdot 0.6\).
\(F = 37.5\ \text{N}\) and \(T = 62.5\ \text{N}\)
intermediate
A 6 kg traffic light is attached to a horizontal cable, tied to a post, and to a second cable that rises to the top of another post at 37° to the horizontal. What are the tensions in the two cables? Use \(g = 10\ \text{m/s}^2\). Use \(\sin 37^\circ = 0.6\) and \(\cos 37^\circ = 0.8\).

Show solution
Only the sloping cable has a vertical component, and it holds the whole weight, \(T_2\sin 37^\circ = 60\), which gives \(T_2 = \dfrac{60}{0.6} = 100\ \text{N}\).
The horizontal cable balances the horizontal component of the other, \(T_1 = T_2\cos 37^\circ\) \(= 100 \cdot 0.8\).
\(T_1 = 80\ \text{N}\) in the horizontal cable and \(T_2 = 100\ \text{N}\) in the sloping one
challenge
A 2 kg wet T-shirt is pegged right in the middle of a washing line, and each half of the line makes an angle \(\alpha\) with the horizontal. The line snaps above 200 N. What is the smallest angle \(\alpha\) it can take? And what would the tension be if the angle were 5°? Use \(g = 10\ \text{m/s}^2\). Use \(\sin 5^\circ \approx 0.087\).

Show solution
With the angle measured from the horizontal, the vertical components add up to the weight, \(2\,T\sin\alpha = W\), and \(T = \dfrac{W}{2\sin\alpha}\).
At the limit, \(\sin\alpha = \dfrac{20}{2 \cdot 200} = 0.05\), which corresponds to \(\alpha \approx 2.9^\circ\).
With \(\alpha = 5^\circ\), \(T = \dfrac{20}{2 \cdot 0.087} \approx 115\ \text{N}\), more than five times the weight of the T-shirt.
The angle cannot be smaller than about 2.9°, and at 5° the tension is some 115 N.
challenge
A 5 kg picture hangs from two strings fixed to different nails. String A makes 37° with the horizontal and string B, 53°, so that the two strings are perpendicular to each other. Work out the tensions. Use \(g = 10\ \text{m/s}^2\). Use \(\sin 37^\circ = \cos 53^\circ = 0.6\) and \(\cos 37^\circ = \sin 53^\circ = 0.8\).

Show solution
Horizontally, \(T_A\cos 37^\circ = T_B\cos 53^\circ\), or \(0.8\,T_A = 0.6\,T_B\).
Vertically, \(T_A\sin 37^\circ + T_B\sin 53^\circ = 50\), or \(0.6\,T_A + 0.8\,T_B = 50\).
Since the strings are perpendicular, the triangle of forces is right-angled, with the weight on the hypotenuse and the tensions on the other two sides, \(T_A = 50 \cdot 0.6\) and \(T_B = 50 \cdot 0.8\) (substituting, both equations hold).
\(T_A = 30\ \text{N}\) and \(T_B = 40\ \text{N}\); the string steeper to the horizontal, B, pulls harder

STEP 2

What makes a door turn

A force applied to a body that can turn about an axis produces a moment, also called a torque. It depends on the force and on the distance between the axis and the force's line of action, which we call the lever arm \(d\).

With the force perpendicular to the bar, the arm is simply the distance from the axis to the point of application. At an angle, only the perpendicular component makes the bar turn, and the moment gains the factor \(\sin\theta\). If we pull the brace along its own length, nothing turns, because the line of action passes through the axis.

That is why the door handle sits far from the hinges, and anyone who pushes the door close to them needs much more force to open it. On a wheel brace, a pipe slipped over the handle lengthens the arm and can undo a nut that the hand alone could not shift.

To add moments, we give each one the sign of the sense in which it turns the body, and here we take anticlockwise as positive.

\(M = F\,d\)\(d = r\sin\theta\)\(M = F\,r\sin\theta\)unit: \(\text{N}\cdot\text{m}\)\(r\) is the distance from the axis to the point of application and \(\theta\), the angle between the force and the bar. In the simulation, the nut only comes loose with at least 60 N·m anticlockwise, and the door, which we assume has a spring, only opens with 9 N·m.

Let's discuss

With the force perpendicular, drag the point of application from the end of the brace to near the nut. What force would be needed there to reach the 60 N·m?
Drag the tip of the arrow until the force lies along the brace. What happens to the arm \(d\), and to the moment?
Turn the force to the other side of the brace. The moment changes sign, and the nut is now tightened instead of loosened.
Switch to the door and compare pushing at the handle with pushing near the hinge, keeping the same force.

Force

Distance from the axis

Angle to the bar

Force

Distance r

Arm d

Moment

Sense

Exercises for step 2 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
To undo a wheel nut, Rafael applies 50 N perpendicular to the end of a 40 cm brace. What is the moment of the force about the nut?

Show solution
With the force perpendicular to the brace, the arm is its full length, \(d = 0.40\ \text{m}\), and \(M = F\,d = 50 \cdot 0.40\).
\(M = 20\ \text{N}\cdot\text{m}\)
basic
Why are door handles on the side opposite the hinges, and not close to them?

Show solution
The door turns about the hinges, and the moment of the hand's force is \(M = F\,d\), with \(d\) measured from them.
Far from the hinges the arm is large, and a small force already produces the moment needed to turn the door.
With a longer arm, we need less force for the same moment.
basic
Lívia pulls a wheel brace with 30 N, 20 cm from the nut, but along the brace itself, as if she wanted to stretch it. What moment does she produce?

Show solution
The line of action of the force passes through the axis of the nut, and the lever arm, which is the distance from the axis to that line, is zero.
In terms of the formula, the angle between the force and the brace is 0°, and \(\sin 0^\circ = 0\).
\(M = 0\), and the nut does not turn, however hard she pulls.
basic
A nut needs a moment of 60 N·m to come loose. With a 30 cm brace, what perpendicular force do we need to apply at the end?

Show solution
From \(M = F\,d\), we make the force the subject, \(F = \dfrac{M}{d} = \dfrac{60}{0.30}\).
\(F = 200\ \text{N}\)
intermediate
On the same 40 cm brace, Rafael now applies the 50 N at the end, but at 30° to the brace. What is the moment, and what is the effective lever arm? Use \(\sin 30^\circ = 0.5\).

Show solution
Only the perpendicular component of the force makes it turn, and \(M = F\,d\sin\theta\) \(= 50 \cdot 0.40 \cdot 0.5\) \(= 10\ \text{N}\cdot\text{m}\).
The same result comes from the effective arm, the distance from the axis to the line of action, \(d\sin\theta = 0.40 \cdot 0.5\).
An effective arm of \(0.20\ \text{m}\), and the moment drops to half the perpendicular case.
intermediate
A heavy door needs a moment of 16 N·m to start opening. Compare the perpendicular force needed when we push 10 cm from the hinges with the force when we push at the handle, 80 cm from them.

Show solution
From \(F = M/d\), near the hinge we have \(F = \dfrac{16}{0.10} = 160\ \text{N}\).
At the handle, \(F = \dfrac{16}{0.80} = 20\ \text{N}\).
160 N near the hinge and 20 N at the handle, eight times less.
intermediate
A bar can turn about an axis at its centre. A 20 N force, 0.5 m from the axis, tends to turn it anticlockwise, and another, of 30 N, 0.2 m on the other side, tends to turn it clockwise. Both are perpendicular to the bar. What is the resultant moment? Take anticlockwise as positive.

Show solution
The first moment is \(M_1 = +20 \cdot 0.5 = +10\ \text{N}\cdot\text{m}\), and the second, \(M_2 = -30 \cdot 0.2 = -6\ \text{N}\cdot\text{m}\).
Adding with the signs, \(M = 10 - 6\).
\(M = +4\ \text{N}\cdot\text{m}\), and the bar tends to turn anticlockwise, even though the larger force pulls the other way.
intermediate
A wheel nut needs 120 N·m to come loose, and Tomás can push the brace with at most 400 N, perpendicularly. The car's brace is 25 cm long. Can he do it? What is the shortest brace that solves the problem?

Show solution
With the car's brace, the largest moment is \(M = 400 \cdot 0.25 = 100\ \text{N}\cdot\text{m}\), less than the 120 needed.
The minimum length comes from \(d = \dfrac{M}{F} = \dfrac{120}{400}\).
He cannot with the 25 cm brace; he needs at least 0.30 m, which a pipe slipped over the handle provides.
challenge
Unable to undo a nut by hand, Joaquim, who has a mass of 60 kg, stands on the end of a 40 cm wheel brace. The brace is tilted 30° above the horizontal. What is the moment of his weight about the nut? And if the brace were horizontal? Use \(g = 10\ \text{m/s}^2\). Use \(\sin 60^\circ \approx 0.87\).

Show solution
The weight, \(W = 600\ \text{N}\), is vertical, and the brace makes 30° with the horizontal, that is, 60° with the vertical.
The moment is \(M = W\,d\sin 60^\circ\) \(= 600 \cdot 0.40 \cdot 0.87 \approx 209\ \text{N}\cdot\text{m}\).
With the brace horizontal, the angle would be 90° and \(M = 600 \cdot 0.40 = 240\ \text{N}\cdot\text{m}\).
About 209 N·m tilted and 240 N·m horizontal, the most efficient position.
challenge
With her arm stretched out horizontally, Helena holds a 5 kg dumbbell 0.6 m from her shoulder. The arm has a mass of 4 kg, with its centre of mass 0.3 m from the shoulder. What is the total moment about the shoulder? And if she lowers the straight arm until it makes 60° with the horizontal? Use \(g = 10\ \text{m/s}^2\). Use \(\cos 60^\circ = 0.5\).

Show solution
The two weights turn the arm in the same sense.
\(M = 50 \cdot 0.6 + 40 \cdot 0.3\) \(= 30 + 12 = 42\ \text{N}\cdot\text{m}\)
With the arm tilted, each lever arm is the horizontal distance to the shoulder, which is multiplied by \(\cos 60^\circ\), and \(M = 42 \cdot 0.5\).
42 N·m horizontal and 21 N·m with the arm at 60°, which is why holding a weight with the arm stretched out in front is so tiring.

STEP 3

How to balance a seesaw

In an extended body, size matters, because the forces act at different points and can make it turn. To stay in equilibrium it has to meet two conditions at once, a zero resultant force, which stops it moving along, and a zero resultant moment, which stops it turning.

On a seesaw, each child produces a moment about the pivot, one clockwise and the other anticlockwise. The plank is balanced when the two are equal, and that is how a light child manages to balance a heavy one, by sitting further from the pivot.

The upward force from the pivot, the normal force, balances the sum of all the weights. In a body in equilibrium, the sum of the moments is zero about any point, and so we can choose the point that makes the calculation simplest, which is usually the one where a force we do not yet know acts.

\(\sum \vec F = \vec 0\)\(\sum M = 0\)\(W_1\,d_1 = W_2\,d_2\)\(N = W_1 + W_2 + W_{\text{plank}}\)In the simulation, the plank is 4 m long, has a mass of 10 kg and rests on its middle, so its weight produces no moment. We assume the pivot is at the height of the plank, and the end that goes down touches the ground when the tilt reaches about 14°.

Let's discuss

Unbalance the seesaw by changing Caio's mass, wait for the plank to tilt and click 'Balance'. Our intuition may expect it to go back to horizontal, and it is worth checking whether it does.
Drag Caio to the end and try to guess, without any sums, where Ana needs to sit. Then check with \(W_1\,d_1 = W_2\,d_2\).
When one end touches the ground, the normal force at the pivot decreases. Where does the rest of the weight seem to have gone?
Double Ana's mass. How far from the pivot does she need to sit now to keep the balance?

Ana's mass

Ana's distance from the pivot

Caio's mass

Caio's distance from the pivot

Ana's moment

Caio's moment

Resultant

Normal at the pivot

Tilt

Exercises for step 3 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
Ana, 30 kg, sits 2 m from the pivot of a seesaw. Where should Caio, 40 kg, sit on the other side for the seesaw to be in equilibrium? Ignore the mass of the plank, or assume it rests on its middle.

Show solution
The moments of the two weights about the pivot have to be equal, \(m_1\,g\,d_1 = m_2\,g\,d_2\), and \(g\) cancels.
With the numbers, \(30 \cdot 2 = 40 \cdot d_2\), and \(d_2 = \dfrac{60}{40}\).
Caio should sit 1.5 m from the pivot.
basic
What are the two conditions for an extended body to be in equilibrium? Give an example in which only one of them is met.

Show solution
The resultant of the forces must be zero, \(\sum \vec F = 0\), so that the body does not move along, and so must the sum of the moments, \(\sum M = 0\), so that it does not turn.
On a car's steering wheel, the two hands can exert equal and opposite forces on the two sides. The resultant is zero, but the moments add up and the wheel turns.
\(\sum \vec F = 0\) and \(\sum M = 0\); a pair of opposite forces on the steering wheel meets only the first.
basic
A 10 kg seesaw, resting on its middle, has two 25 kg children at its ends and is in equilibrium. What force does the pivot exert on the plank? Use \(g = 10\ \text{m/s}^2\).

Show solution
The plank neither rises nor falls, and the force from the pivot balances all the weights, \(N = 250 + 250 + 100\).
\(N = 600\ \text{N}\)
basic
A light ruler rests on its middle. A 6 N weight sits 30 cm from the pivot, on the left. What weight, placed 45 cm from the pivot on the right, keeps the ruler in equilibrium?

Show solution
The moments on the two sides are equal, \(6 \cdot 30 = W \cdot 45\), and \(W = \dfrac{180}{45}\).
\(W = 4\ \text{N}\)
intermediate
On a seesaw resting on its middle, Bruno (50 kg) sits 1.2 m from the pivot, on the left, and Luísa (30 kg), 1.5 m away, on the right. Where should Teo, 20 kg, sit to balance the seesaw?

Show solution
Bruno produces \(50 \cdot 1.2 = 60\) (in kg·m, since \(g\) cancels) and Luísa, \(30 \cdot 1.5 = 45\). Luísa's side is 15 short.
Teo should sit on the right, at a distance such that \(20\,x = 15\).
0.75 m from the pivot, on Luísa's side.
intermediate
A light seesaw is in equilibrium with a 300 N child 1 m from the pivot on one side, and a 200 N child 1.5 m away on the other. Work out the force from the pivot and check that the sum of the moments is also zero about the point where the first child sits.

Show solution
The force from the pivot balances the weights, \(N = 300 + 200 = 500\ \text{N}\).
About the first child, her weight has zero arm. The normal force is 1 m away and turns the plank one way, with \(500 \cdot 1 = 500\ \text{N}\cdot\text{m}\), and the second child is \(1 + 1.5 = 2.5\ \text{m}\) away and turns it the other way, with \(200 \cdot 2.5 = 500\ \text{N}\cdot\text{m}\).
\(N = 500\ \text{N}\), and the moments also cancel about that point, as they must about any point.
intermediate
A uniform plank, 3 m long and 20 kg, rests on a support 1 m from one of its ends. What mass should we put on the shorter end to balance it? What is the force from the support then? Use \(g = 10\ \text{m/s}^2\).

Show solution
The weight of the plank acts at its centre, 1.5 m from the end, that is, 0.5 m from the support, on the longer side, and turns the plank with \(200 \cdot 0.5 = 100\ \text{N}\cdot\text{m}\).
The mass on the short end, 1 m from the support, has to balance that moment, \(m \cdot 10 \cdot 1 = 100\), and so \(m = 10\ \text{kg}\).
10 kg on the short end, with the support exerting \(N = 200 + 100 = 300\ \text{N}\).
intermediate
Two children are balanced on a seesaw resting on its middle, with the plank horizontal. Someone tilts the plank and lets go. Does it go back to horizontal, go down on one side, or stay where it was left? Assume the pivot is at the height of the plank.

Show solution
With the plank tilted at an angle \(\theta\), the arm of each weight becomes the horizontal distance to the pivot, \(d\cos\theta\).
Both arms are multiplied by the same factor, and the moments, which were equal, stay equal.
It stays where it was left: the equilibrium is neutral.
challenge
A uniform seesaw, 4 m long and 30 kg, rests on its middle. Clara, 40 kg, sits at one end. Where should Márcio, 50 kg, sit to balance the seesaw? If Márcio moves to the other end, what extra mass do we need to place 1 m from the pivot, on Clara's side?

Show solution
The weight of the plank acts at the pivot and produces no moment. Clara's end is 2 m away, and \(40 \cdot 2 = 50\,x\), which gives \(x = 1.6\ \text{m}\).
With Márcio at the end, his side produces \(50 \cdot 2 = 100\), and Clara's, \(40 \cdot 2 = 80\). We are 20 short, and the extra mass at 1 m needs \(m \cdot 1 = 20\).
Márcio sits 1.6 m from the pivot; at the end, we need 20 kg more on Clara's side.
challenge
A uniform 1 m rule of 100 g lies on a table, with 30 cm sticking out over the edge. What is the largest mass we can hang on the outer end without the rule tipping over? Use \(g = 10\ \text{m/s}^2\).

Show solution
When it is about to tip, the rule rests only on the edge of the table, and it is about that edge that we compare the moments.
The weight of the rule, \(1\ \text{N}\), acts at its middle, 50 cm from each end, that is, 20 cm in from the edge. The hanging mass is 30 cm out.
At the limit, \(1 \cdot 0.20 = m \cdot 10 \cdot 0.30\), and \(m = \dfrac{0.2}{3}\ \text{kg}\).
About 67 g; with more than that, the rule tips over.

STEP 4

Levers, from pliers to the biceps

A lever is a bar that turns about a pivot, called the fulcrum. Two forces act on it, the effort, which we apply, and the load, which we want to overcome, and in equilibrium the moments of the two about the fulcrum are equal.

We call the distances from each force to the fulcrum the effort arm \(b_e\) and the load arm \(b_l\). The ratio \(b_e/b_l\) is the mechanical advantage, which tells us how many times smaller the effort is than the load.

We classify levers by what sits in the middle. In a first-class lever, the fulcrum lies between the forces, as in scissors, pliers and the seesaw. In a second-class lever, the load is in the middle, as in the wheelbarrow and the nutcracker, and the mechanical advantage is always greater than 1.

In a third-class lever, the effort is in the middle, as in tweezers and in our forearm, where the biceps pulls a few centimetres from the elbow. The muscle has to exert a force several times larger than the weight in the hand, and in return the hand sweeps a wide arc while the muscle shortens only a little.

\(F_e\,b_e = F_l\,b_l\)\(\text{MA} = \dfrac{b_e}{b_l} = \dfrac{F_l}{F_e}\)We ignore the weight of the bar and assume the forces are perpendicular to it. In the simulation, the bar is 1 m long, positions are measured from the left end, and the effort shown is the one that keeps the bar in equilibrium.

Let's discuss

On the first-class lever, move the effort away from the fulcrum. Does the force seem to fall in the same proportion as the mechanical advantage grows?
Choose the second-class lever and bring the load close to the wheel, as in a well-loaded wheelbarrow.
Switch to the third-class lever and look for a position where the effort is smaller than the load. Why does this seem impossible with this kind of lever?
Drag the effort on top of the fulcrum and see what happens to the force needed.

Fulcrum position

Load position

Effort position

Weight of the load

Class

Effort arm

Load arm

Effort

Mechanical advantage

Exercises for step 4 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
Classify the levers in scissors, a wheelbarrow and eyebrow tweezers, saying what sits in the middle of each.

Show solution
In scissors, the pivot (fulcrum) lies between the hand and the paper, and the lever is first-class.
In the wheelbarrow, the load lies between the wheel (fulcrum) and the hands, and the lever is second-class.
In tweezers, the fingers squeeze between the hinge (fulcrum) and the tips that hold the hair, and the lever is third-class.
Scissors first-class, wheelbarrow second-class and tweezers third-class.
basic
With a crowbar, Mateus wants to lift a stone that needs 600 N. The tip is 5 cm from the fulcrum and his hand, 60 cm. What force does he need to apply, and what is the mechanical advantage?

Show solution
By the lever rule, \(F_e\,b_e = F_l\,b_l\), and \(F_e = \dfrac{600 \cdot 5}{60} = 50\ \text{N}\).
The mechanical advantage is the ratio between the arms, \(\dfrac{b_e}{b_l} = \dfrac{60}{5}\).
50 N, with a mechanical advantage of 12.
basic
A wheelbarrow carries a 600 N load whose centre of mass is 0.4 m from the wheel axle. The hands hold the handles 1.2 m from the axle. What vertical force do the hands exert? Ignore the weight of the wheelbarrow.

Show solution
The fulcrum is the wheel, and the moments about it are equal, \(F \cdot 1.2 = 600 \cdot 0.4\).
Hence \(F = \dfrac{240}{1.2}\).
\(F = 200\ \text{N}\), a third of the weight of the load.
basic
In third-class levers, such as tweezers and our forearm, the effort is always greater than the load. Why would anyone use a lever like that?

Show solution
In a third-class lever, the effort lies between the fulcrum and the load, and the effort arm is always the shorter one, which gives a mechanical advantage of less than 1.
In return, the end of the lever sweeps a bigger arc than the point where we apply the force, and the movement comes out wider and faster. With tweezers, we gain delicacy and precision.
We lose force and gain range, speed or precision.
intermediate
The biceps is attached to the forearm 4 cm from the elbow. With the forearm horizontal, the hand holds a 3 kg ball 32 cm from the elbow. What force does the biceps need to exert? Ignore the weight of the forearm. Use \(g = 10\ \text{m/s}^2\).

Show solution
The elbow is the fulcrum, and the biceps pulls between it and the ball, which makes the forearm a third-class lever.
Equating the moments, \(F_b \cdot 4 = 30 \cdot 32\), and \(F_b = \dfrac{960}{4}\).
\(F_b = 240\ \text{N}\), eight times the weight of the ball.
intermediate
In a nutcracker, the nut sits 3 cm from the hinge and the hand squeezes 15 cm from it. If the shell cracks at 200 N, what force does the hand need to exert? What class of lever is this?

Show solution
The nut lies between the hinge and the hand, and the lever is second-class.
By the lever rule, \(F_e \cdot 15 = 200 \cdot 3\), and \(F_e = \dfrac{600}{15}\).
\(F_e = 40\ \text{N}\), on a second-class lever.
intermediate
A pair of pliers has 12 cm handles, measured from the pivot, and cuts wire 2 cm from the pivot. If the hand squeezes with 100 N, what is the force on the wire?

Show solution
The pivot lies between the hand and the wire, on a first-class lever, and \(100 \cdot 12 = F_l \cdot 2\).
\(F_l = 600\ \text{N}\), six times the force of the hand.
intermediate
Isadora squeezes the handles of a pair of scissors with 40 N, 8 cm from the pivot. What is the cutting force on a piece of cardboard placed 2 cm from the pivot? And at the tip of the blades, also 8 cm from the pivot?

Show solution
Near the pivot, \(40 \cdot 8 = F \cdot 2\), and the force is \(160\ \text{N}\).
At the tip, the arms are equal, and the cutting force is the same as the hand's, \(40\ \text{N}\).
160 N near the pivot and 40 N at the tip; anyone who cuts thick cardboard learns to push it right into the scissors.
challenge
A 12 kg wheelbarrow has its centre of mass 0.5 m from the wheel axle. It carries 60 kg of sand, with centre of mass 0.4 m from the axle, and the hands hold it at 1.5 m. What vertical force do the hands exert, and how much does the wheel support? Use \(g = 10\ \text{m/s}^2\).

Show solution
The moments about the wheel balance.
\(F \cdot 1.5 = 120 \cdot 0.5 + 600 \cdot 0.4\) \(= 60 + 240\)
Hence \(F = \dfrac{300}{1.5} = 200\ \text{N}\).
The wheel supports the rest of the total weight, \(720 - 200\).
The hands exert 200 N and the wheel supports 520 N.
challenge
Renan's forearm has a mass of 1.5 kg, with centre of mass 15 cm from the elbow, and his hand holds a 4 kg ball 35 cm from the elbow. The biceps pulls vertically 5 cm from the elbow. What force does the biceps exert, and what force does the elbow joint exert on the forearm? Use \(g = 10\ \text{m/s}^2\).

Show solution
About the elbow,
\(F_b \cdot 0.05 = 15 \cdot 0.15 + 40 \cdot 0.35\) \(= 2.25 + 14\)
Hence \(F_b = \dfrac{16.25}{0.05} = 325\ \text{N}\).
The sum of the vertical forces is also zero. The biceps pulls up with 325 N, the weights add up to 55 N downwards, and the elbow has to push the forearm down with the difference, \(325 - 55\).
The biceps exerts 325 N and the elbow, 270 N downwards.

STEP 5

How far can the pile go?

The centre of mass is the point that behaves as if all the mass of the body were concentrated in it. For a set of parts, it is the average of the positions weighted by the masses, and so it lies closer to the heavier parts.

Near the ground, where gravity is practically uniform, we can treat the weight of the body as a single force, applied at the centre of mass. A body resting on a surface does not topple as long as the vertical line through the centre of mass falls inside the base of support, and it topples when the line moves outside.

We call the equilibrium stable when the body, after a small push, returns to its initial position, like a ball at the bottom of a bowl. It is unstable when the body moves further and further away, like a ball on top of an upturned bowl, and neutral when the body stays where we leave it, like a ball on a flat table.

Lowering the centre of mass and widening the base tend to make a body more stable, which is what we do when we spread our feet on a moving bus. The Leaning Tower of Pisa, tilted for centuries, is still standing because the vertical through its centre of mass still passes inside the base.

\(x_{cm} = \dfrac{\sum m_i\,x_i}{\sum m_i}\)topples if the vertical through the CM leaves the base\(\tan\theta_{\text{topple}} = \dfrac{b}{h}\)The last formula holds for a block of base \(b\) and height \(h\) on a ramp, if friction stops it sliding. In the simulation, the books are identical, 20 cm long, and the centre of mass of each one is at its middle; the pile is checked from the bottom up, and each group of books needs its centre of mass over the book that supports it.

Let's discuss

Choose the top book and push it out slowly. It falls when its end goes past what fraction of the book below?
Now move the bottom book and follow the centre of mass of them all. What seems to decide whether the whole pile falls at once?
Before clicking 'Ideal pile' with 4 books, make a guess about the next question. Can the top book get completely past the edge of the table?
Build the pile starting from the bottom book and then starting from the top one. Which of the two strategies seems to get further?

End of the chosen book

Number of books

CM of all (edge = 0)

Overhang of the top book

Off the table

State

Exercises for step 5 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A home-made dumbbell has a light 50 cm bar with 2 kg at one end and 3 kg at the other. How far from the 2 kg end is the centre of mass?

Show solution
Putting the origin at the 2 kg end, \(x_{cm} = \dfrac{2 \cdot 0 + 3 \cdot 50}{2 + 3} = \dfrac{150}{5}\).
\(x_{cm} = 30\ \text{cm}\), closer to the larger mass.
basic
A small ball can stay still at the bottom of a bowl, on top of an upturned bowl and on a flat table. Classify the equilibrium in each case.

Show solution
At the bottom of the bowl, a little push makes the ball roll up, and it comes back, so the equilibrium is stable.
On top of the upturned bowl, any push moves it further and further away, and the equilibrium is unstable.
On the flat table, it stays still wherever we leave it, and the equilibrium is neutral.
Stable, unstable and neutral.
basic
A uniform 30 cm ruler of 40 g has a 20 g rubber fixed to one end. Where is the centre of mass of the set, measured from the other end?

Show solution
The centre of mass of the ruler is at its middle, 15 cm, and the rubber is at 30 cm.
\(x_{cm} = \dfrac{40 \cdot 15 + 20 \cdot 30}{40 + 20}\) \(= \dfrac{1200}{60}\)
\(x_{cm} = 20\ \text{cm}\)
basic
The Leaning Tower of Pisa has been tilted for centuries and does not fall. What has to remain true for it to stay standing?

Show solution
A body resting on a surface does not topple as long as the vertical through its centre of mass falls inside the base of support.
For the tower, that vertical still passes inside the base, and the weight tends to bring it back. If the tilt grew until the vertical through the centre of mass left the base, the weight would start turning it outwards.
The vertical through the centre of mass has to stay inside the base.
intermediate
Two identical 20 cm books are stacked on a table. We slide the top one 8 cm out relative to the bottom one. Does it fall? What is the largest possible shift?

Show solution
The centre of mass of the top book is at its middle, which is \(10 + 8 = 18\ \text{cm}\) in front of the back edge of the bottom book, still before the front edge, at 20 cm.
It only falls when its centre goes past the edge of the bottom book, that is, with more than half its length sticking out.
It does not fall; the maximum is 10 cm, half the book.
intermediate
A block with a 30 cm base and 60 cm height sits on a ramp whose slope we can increase, and friction is enough to stop it sliding. From what slope does it topple?

Show solution
The block topples when the vertical through the centre of mass passes through the lower corner of the base. At that limit, \(\tan\theta = \dfrac{b}{h}\), with \(b\) the width of the base and \(h\) the height.
Hence \(\tan\theta = \dfrac{30}{60} = 0.5\).
From the angle whose tangent is 0.5, about 27°.
intermediate
Three small spheres lie in a plane: 1 kg at (0, 0), 1 kg at (4, 0) and 2 kg at (0, 4), with the coordinates in centimetres. Where is the centre of mass?

Show solution
We take the weighted average along each axis, \(x_{cm} = \dfrac{1 \cdot 0 + 1 \cdot 4 + 2 \cdot 0}{4}\) and \(y_{cm} = \dfrac{1 \cdot 0 + 1 \cdot 0 + 2 \cdot 4}{4}\).
\((x_{cm}, y_{cm}) = (1, 2)\ \text{cm}\)
intermediate
Felipe, 70 kg, puts on a 10 kg rucksack whose centre of mass is 20 cm behind the centre of mass of his body. How far does the centre of mass of the set shift, and what does he do to avoid falling backwards?

Show solution
With the origin at the centre of mass of the body and the axis pointing forwards, \(x_{cm} = \dfrac{70 \cdot 0 + 10 \cdot (-20)}{80}\), that is, \(-2.5\ \text{cm}\).
The centre of mass moves 2.5 cm back, and to keep it over his feet Felipe leans forwards.
2.5 cm backwards, and Felipe compensates by leaning his body forwards.
challenge
We have four identical 24 cm books and want to stack them at the edge of a table with the top one as far out as possible. Given that the top book can stick out up to half its length over the second, that the top two together can stick out a quarter over the third, that the top three can stick out a sixth over the fourth and that all four together can stick out an eighth over the table, how far beyond the edge is the top book?

Show solution
Each overhang puts the centre of mass of the upper block exactly over the edge of the support below.
Adding up, the total overhang is
\(\dfrac{24}{2} + \dfrac{24}{4} + \dfrac{24}{6} + \dfrac{24}{8}\) \(= 12 + 6 + 4 + 3\).
The total is more than the length of one book, and the top one is entirely off the table.
25 cm beyond the edge, more than the whole book; in practice, we leave a little margin so the pile is not right at the limit.
challenge
A block with a 20 cm base and 50 cm height sits on a ramp, and the coefficient of static friction between the two is 0.6. Increasing the slope little by little, does the block topple before it slides, or slide before it topples?

Show solution
The block starts to slide when \(\tan\theta = \mu = 0.6\), about 31°.
It topples when \(\tan\theta = \dfrac{b}{h} = \dfrac{20}{50} = 0.4\), about 22°.
The toppling condition is reached first, with the ramp less steep.
It topples first, at around 22°, without ever sliding.

STEP 6

How much does each pier hold?

A beam resting on two points, such as a bridge over two piers or the builder's plank on two trestles, is perhaps the most common example of an extended body in equilibrium. The forces from the supports, \(N_A\) and \(N_B\), are the reactions, and the two conditions of equilibrium are enough to work them out.

The sum of the reactions is the total weight. To separate one from the other, we take moments about one of the supports, which removes its reaction from the calculation and leaves an equation with a single unknown.

\(N_A + N_B = W_{\text{bridge}} + W_t\)\(N_B\,L = W_{\text{bridge}}\,\dfrac{L}{2} + W_t\,x\)\(N_B = \dfrac{W_{\text{bridge}}}{2} + W_t\,\dfrac{x}{L}\)\(x\) is the distance from the lorry to pier A and \(L\), the span. We treat the lorry as a single load, applied at its centre of mass, and the bridge as uniform, with its weight at the middle. In the simulation, \(L = 20\ \text{m}\) and the bridge has a mass of 20 t; with \(g = 10\ \text{m/s}^2\), each tonne weighs 10 kN.

weight of the lorry reaction at pier A reaction at pier B

Following the lorry as it crosses, we see each reaction change in a straight line with its position. On top of pier A, the lorry presses only on A, and in the middle of the bridge its weight is shared equally between the two piers.

The same calculation works for the builder's plank. If one of the trestles is far from the end, the plank overhangs, and a builder who walks past the trestle can make it tip, because the reaction at the other support would have to pull the plank down, something a trestle cannot do.

Let's discuss

Click 'Cross' and watch the two arrows at the piers. Which one grows while the other shrinks, and do both seem to change at the same rate?
Stop the lorry in the middle of the bridge. Are the reactions equal, as the symmetry suggests?
Switch off the weight of the bridge and drag the lorry on top of A. How much does pier B hold?
Double the mass of the lorry and watch what happens to the slope of the lines on the graph.

Mass of the lorry

Distance from pier A

Weight of the lorry

Weight of the bridge

N_A

N_B

N_A + N_B

Exercises for step 6 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A light 4 m plank rests on two trestles, one at each end. An 80 kg builder stands exactly in the middle. How much does each trestle hold? Use \(g = 10\ \text{m/s}^2\).

Show solution
By symmetry, the two trestles share the weight, \(N_A = N_B = \dfrac{800}{2}\).
\(N_A = N_B = 400\ \text{N}\)
basic
A 200 kN bridge rests on two piers, one at each end, and there is no vehicle on it. How much does each pier hold?

Show solution
The bridge is uniform, and its weight acts at the middle, the same distance from both piers, which share the weight equally.
100 kN on each pier
basic
A lorry stops exactly on top of pier A of a bridge. How much of the lorry's weight goes to pier B?

Show solution
About pier A, the weight of the lorry has zero arm and produces no moment.
The reaction at B only needs to balance the weight of the bridge, and the whole lorry is left to A.
None; pier B holds only its half of the weight of the bridge.
basic
A light 3 m board rests on its ends A and B. A 30 kg tin of paint is 1 m from A. How much does each support hold? Use \(g = 10\ \text{m/s}^2\).

Show solution
About A, the moment of the tin, \(300 \cdot 1\), is balanced by the reaction at B, 3 m away, \(N_B \cdot 3 = 300\), and \(N_B = 100\ \text{N}\).
The sum of the reactions is the weight, and \(N_A = 300 - 100\).
\(N_A = 200\ \text{N}\) and \(N_B = 100\ \text{N}\); the support closer to the tin holds more.
intermediate
A bridge 20 m long and weighing 200 kN rests on piers A and B, at its ends. A 100 kN lorry is 5 m from A. Work out the two reactions.

Show solution
About A, the bridge (at the middle, 10 m away) and the lorry (5 m away) are balanced by \(N_B\), 20 m away, \(N_B \cdot 20 = 200 \cdot 10 + 100 \cdot 5\), and \(N_B = \dfrac{2500}{20} = 125\ \text{kN}\).
The reactions add up to the total weight, \(N_A = 300 - 125\).
\(N_A = 175\ \text{kN}\) and \(N_B = 125\ \text{kN}\)
intermediate
On the same bridge (20 m, 200 kN, 100 kN lorry), show that \(N_B\) grows in a straight line with the distance \(x\) from the lorry to pier A. Where is the lorry when \(N_B = 180\ \text{kN}\)?

Show solution
About A, \(N_B \cdot 20 = 200 \cdot 10 + 100\,x\), and so \(N_B = 100 + 5\,x\) (in kN, with \(x\) in metres), a linear function.
For \(N_B = 180\), \(5\,x = 80\).
\(x = 16\ \text{m}\), 4 m from pier B.
intermediate
A 4 m, 20 kg board, resting on its ends A and B, carries an 80 kg builder 1 m from A. How much does each support hold? Use \(g = 10\ \text{m/s}^2\).

Show solution
About A, the weight of the board acts at the middle (2 m) and that of the builder at 1 m, \(N_B \cdot 4 = 200 \cdot 2 + 800 \cdot 1\) \(= 1200\), and so \(N_B = 300\ \text{N}\).
The total weight is 1000 N, and \(N_A = 1000 - 300\).
\(N_A = 700\ \text{N}\) and \(N_B = 300\ \text{N}\)
intermediate
A light 6 m beam, resting on its ends A and B, carries 600 N 2 m from A and 300 N 4 m from A. Work out the reactions.

Show solution
About A, \(N_B \cdot 6 = 600 \cdot 2 + 300 \cdot 4\) \(= 2400\), and \(N_B = 400\ \text{N}\).
The sum of the reactions is \(900\ \text{N}\), and \(N_A = 900 - 400\).
\(N_A = 500\ \text{N}\) and \(N_B = 400\ \text{N}\)
challenge
A uniform board, 4 m long and 20 kg, rests on two trestles, A at the left end and B 3 m from it, so that 1 m overhangs beyond B. How far past B can a 60 kg builder walk before the board starts to tip? Use \(g = 10\ \text{m/s}^2\).

Show solution
When the board is about to tip, it stops touching A, and everything rests on B. So we compare the moments about B.
The weight of the board acts at the middle, 1 m before B, and turns the board towards A with \(200 \cdot 1 = 200\ \text{N}\cdot\text{m}\). The builder, at a distance \(s\) past B, turns it the other way with \(600\,s\).
At the limit, \(600\,s = 200\), and \(s = \dfrac{1}{3}\ \text{m}\).
Only about 33 cm past B; this is why builders nail the board to the trestle.
challenge
The two piers of a 200 kN bridge can take up to 250 kN each. What is the heaviest lorry that can cross the whole bridge?

Show solution
The reaction at a pier is greatest when the lorry is on top of it, and then \(N = 100 + W\) (in kN), with 100 kN coming from the bridge.
The condition \(100 + W \le 250\) gives \(W \le 150\ \text{kN}\).
Up to 150 kN, or about 15 t; the critical moment is arriving over each pier, and not the middle of the bridge.

STEP 7

Lifting loads with half the force

A fixed pulley turns while attached to the ceiling and only changes the direction of the rope. The force is the same on both sides, and what we gain is being able to pull downwards, helped by our own weight, instead of lifting the load with our arms.

With a movable pulley, the load hangs from the pulley itself, held up by two lengths of rope, and each pulls with half the weight. When we combine \(n\) movable pulleys in a pulley system, in which each pulley holds up the one below, the force halves with every pulley.

This trade does not create energy. For the load to rise 1 m with a movable pulley, the two lengths of rope shorten by 1 m each, and the hand pulls 2 m of rope. In the pulley system, the rope pulled grows as \(2^n\), and the work done by the hand, force times rope, equals the weight times the height. A real pulley system tends to need a little more force than that, because pulleys have friction and mass.

\(F = W\) (fixed)\(F = \dfrac{W}{2^n}\)\(s = 2^n\,h\)\(F\,s = W\,h\)We assume massless ropes and frictionless pulleys, and \(g = 10\ \text{m/s}^2\). In the simulation, the pulleys have no mass, unless you switch on the 1 kg per pulley option.

Let's discuss

With only the fixed pulley, lift the load and compare the rope pulled with the rise.
Switch to 3 movable pulleys and lift 1 m. How much rope went through the hand, and what is the work on each side?
Switch on the 1 kg pulleys and choose a light load, of 10 kg. Does the 3-pulley system still divide the force by 8, or does the weight of the pulleys tend to count?
A 50 kg person, pulling the rope downwards, cannot exert more force than their own weight. What load could they lift with 3 movable pulleys?

Mass of the load

Weight of the load

Force at the hand

W / F

Rope to rise 1 m

The load rose

Work: hand / load

Exercises for step 7 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A 20 kg bucket is lifted by a rope that passes over a fixed pulley at the top of a building under construction. What force does the worker exert, and what is the pulley for? Ignore friction. Use \(g = 10\ \text{m/s}^2\).

Show solution
The fixed pulley only changes the direction of the rope, and the tension is the same on both sides, equal to the weight of the bucket, \(F = 20 \cdot 10\).
\(F = 200\ \text{N}\); the pulley lets the worker pull downwards, using their own weight, instead of pulling upwards.
basic
A 400 N load is attached to a light movable pulley, with one end of the rope tied to the ceiling. What force do we need to exert on the other end to hold the load?

Show solution
The movable pulley hangs from two lengths of rope, which share the weight, \(2\,F = 400\).
\(F = 200\ \text{N}\)
basic
A pulley system has three movable pulleys combined, each dividing the force by two. What force lifts an 800 N load? Ignore the weight of the pulleys.

Show solution
With \(n\) movable pulleys, \(F = \dfrac{W}{2^n} = \dfrac{800}{2^3} = \dfrac{800}{8}\).
\(F = 100\ \text{N}\)
basic
With a movable pulley, we exert half the force. Why is this not a way of getting energy for free?

Show solution
For the load to rise 1 m, the two lengths of rope holding it up have to shorten by 1 m each, and the hand pulls 2 m of rope.
The work done by the hand, half the force times twice the distance, is the same as that of the force lifting the load.
The force halves and the rope pulled doubles, and the work stays the same.
intermediate
With a pulley system of two movable pulleys, what force lifts a 60 kg sack, and how much rope do we need to pull for it to rise 3 m? Use \(g = 10\ \text{m/s}^2\).

Show solution
The force is \(F = \dfrac{600}{2^2} = 150\ \text{N}\).
The rope pulled is \(2^2\) times the rise of the load, \(4 \cdot 3\).
150 N, pulling 12 m of rope.
intermediate
How many movable pulleys do we need, in a pulley system like the one in the simulation, to lift 1600 N exerting only 200 N?

Show solution
We need \(2^n = \dfrac{1600}{200} = 8\).
\(n = 3\) movable pulleys
intermediate
A pulley system of three movable pulleys lifts 800 N to a height of 2 m. Work out the force, the rope pulled and the work done by the hand, and compare it with the work needed to lift the load.

Show solution
The force is \(F = \dfrac{800}{8} = 100\ \text{N}\), and the rope pulled is \(8 \cdot 2 = 16\ \text{m}\).
The work done by the hand is \(100 \cdot 16 = 1600\ \text{J}\), and lifting the load takes \(800 \cdot 2 = 1600\ \text{J}\).
100 N, 16 m of rope and 1600 J in both cases.
intermediate
A movable pulley weighs 20 N and holds a 380 N load. What force do we need to exert?

Show solution
The pulley rises together with the load, and the two lengths of rope hold up both, \(2\,F = 380 + 20\).
\(F = 200\ \text{N}\)
challenge
Paulo has a mass of 70 kg and pulls the rope downwards, standing on the ground. What is the largest mass he can hold with a single fixed pulley? And with a pulley system of one movable pulley? And with two? Use \(g = 10\ \text{m/s}^2\).

Show solution
Pulling downwards, Paulo cannot exert more force than his own weight, 700 N, without being lifted off the ground.
With the fixed pulley, the load can weigh up to 700 N, or 70 kg. With one movable pulley, \(2 \cdot 700 = 1400\ \text{N}\), and with two, \(4 \cdot 700 = 2800\ \text{N}\).
70 kg, 140 kg and 280 kg.
challenge
A pulley system has two movable pulleys of 1 kg each and lifts a 300 N load. What force do we need to exert? Compare with the case of massless pulleys. Use \(g = 10\ \text{m/s}^2\).

Show solution
The bottom pulley holds up the load and its own weight, and its rope has tension \(T_1 = \dfrac{300 + 10}{2} = 155\ \text{N}\).
The second pulley holds up that tension and its own weight, \(F = \dfrac{155 + 10}{2}\).
Without the mass of the pulleys, it would be \(\dfrac{300}{4} = 75\ \text{N}\).
\(F = 82.5\ \text{N}\), against 75 N in the ideal case.

WRAP-UP

Challenges

Equilibrium of a point

A 6 kg chandelier hangs from two identical chains, each at 60° to the vertical. What is the tension in each chain? (\(g = 10\ \text{m/s}^2\), \(\cos 60^\circ = 0.5\))

Show solution

In the symmetric case, the vertical components add up to the weight, and \(T = \dfrac{W}{2\cos\theta} = \dfrac{60}{2 \cdot 0.5} = 60\ \text{N}\).
Each chain pulls with the whole weight of the chandelier, because both are wide open.

Moment of a force

Pedro applies 200 N perpendicular to the end of a 30 cm brace, and the nut only comes loose with 90 N·m. Can he do it? What would be the minimum length of brace for the same force?

Show solution

The moment is \(M = 200 \cdot 0.30 = 60\ \text{N}\cdot\text{m}\), below the 90 needed, and the nut does not come loose.
With the same force, the brace would need \(d = \dfrac{90}{200} = 0.45\ \text{m}\).

Extended body

A father of 80 kg and his 20 kg daughter play on a 4 m seesaw resting on its middle. If the daughter sits at the end, where should the father sit to balance the seesaw?

Show solution

The daughter is 2 m from the pivot, and the moments are equal, \(20 \cdot 2 = 80\,x\).
Hence \(x = 0.5\ \text{m}\), very close to the pivot, on the other side.

Levers

A wheelbarrow carries 1000 N of soil, with centre of mass 0.3 m from the wheel axle, and the hands hold the handles 1.5 m from the axle. What force do the hands exert, and what is the mechanical advantage? Ignore the weight of the wheelbarrow.

Show solution

About the wheel, \(F \cdot 1.5 = 1000 \cdot 0.3\), and \(F = 200\ \text{N}\).
The mechanical advantage is the ratio between the arms, \(\dfrac{1.5}{0.3} = 5\).

Centre of mass

With two identical 20 cm books, stacked at the edge of a table, how far beyond the edge can the top book reach, at most?

Show solution

The top book sticks out up to half its length over the bottom one, 10 cm. The two together stick out a quarter over the table, 5 cm, which puts the centre of mass of the pair exactly over the edge.
The maximum overhang is \(10 + 5 = 15\ \text{cm}\).

Beam on two supports

A bridge 12 m long and weighing 60 kN rests on its ends. A 120 kN lorry stops 3 m from pier A. How much does each pier hold?

Show solution

About A, \(N_B \cdot 12 = 60 \cdot 6 + 120 \cdot 3 = 720\), and \(N_B = 60\ \text{kN}\).
The reactions add up to 180 kN, and \(N_A = 120\ \text{kN}\).

Pulleys

A pulley system of two movable pulleys lifts a 50 kg bag of cement to a height of 4 m. Work out the force, the rope pulled and the work done by the hand. (\(g = 10\ \text{m/s}^2\))

Show solution

The force is \(F = \dfrac{500}{2^2} = 125\ \text{N}\), and the rope pulled, \(4 \cdot 4 = 16\ \text{m}\).
The work done by the hand is \(125 \cdot 16 = 2000\ \text{J}\), the same as \(500 \cdot 4\) to lift the bag.

Thinking, no sums

Anyone carrying a heavy bucket in one hand leans their body to the other side. Why is it usually more comfortable to carry two buckets, one in each hand?

Show solution

With a bucket on one side only, the centre of mass of the set moves off the line of the body, and to keep it over our feet we lean our trunk to the opposite side.
With a bucket in each hand, the centre of mass stays in the middle, over our feet, and we can walk upright.

ENEM-style questions

The ENEM is Brazil's national secondary-school exam, which most students sit to get into university, and we wrote the five questions below in its format, with an everyday context, a question and five options, each one returning to a different step of the lesson. Further down we list real exam questions on statics, which may well be the best practice once these are done.

Equilibrium of a point · Step 1
Marcos wants to hang a hammock on the veranda, fixed to two hooks at the same height. He has a mass of 70 kg and, lying in the middle of the hammock, makes the two ropes form the same angle \(\alpha\) with the horizontal. The manufacturer guarantees that each hook takes up to 600 N. We will use \(g = 10\ \text{m/s}^2\), ignore the mass of the hammock and assume that Marcos's weight is shared equally between the two ropes.

Sine of some angles
Angle α sin α

15° 0.26

30° 0.50

45° 0.71

60° 0.87

Of the angles in the table, which is the smallest that keeps the tension in each rope within the limit of the hooks?
1. 15°
2. 30°
3. 45°
4. 60°
5. Any of them, because each rope holds half the weight, 350 N, at any angle.
Show solution

Answer: C.
Only the vertical components of the tensions hold up the weight, and \(2\,T\sin\alpha = 700\), that is, \(T = \dfrac{350}{\sin\alpha}\).
With 30°, \(T = \dfrac{350}{0.50} = 700\ \text{N}\), above the limit. With 45°, \(T = \dfrac{350}{0.71} \approx 493\ \text{N}\), within it.
Option A appears when we swap the sine for the cosine, and B when we forget to compare the tension with the limit. D is safe, but it is not the smallest angle, and E forgets that only the vertical component of the tension holds the weight.
In practice, the rope stretches a little when Marcos lies down, and the angle tends to decrease, which calls for some margin when putting it up.
Moment of a force · Step 2
The fire door on the staircase of a block of flats has a spring that keeps it closed, and starting to open it takes a moment of 24 N·m about the hinges. The handle is 80 cm from the hinges. Rita, the building manager, wants to know what force a person needs to exert, pushing perpendicular to the door, at the handle and at a point 20 cm from the hinges.

What are these two forces, respectively?
1. 19.2 N and 4.8 N
2. 30 N and 7.5 N
3. 30 N and 30 N
4. 30 N and 120 N
5. 120 N and 30 N
Show solution

Answer: D.
With the force perpendicular to the door, \(M = F\,d\), and the force needed is \(F = M/d\).
At the handle, \(F = \dfrac{24}{0.8} = 30\ \text{N}\). Near the hinges, \(F = \dfrac{24}{0.2} = 120\ \text{N}\).
Option A multiplies the moment by the distance instead of dividing, and B assumes that the force decreases along with the distance. C ignores the lever arm, and E swaps the order asked for.
Levers · Step 4
In a physiotherapy session, Sônia holds a 2 kg dumbbell with her forearm horizontal. The dumbbell is 30 cm from the elbow, and the biceps tendon is attached to the forearm 4 cm from the elbow, pulling it upwards. We will use \(g = 10\ \text{m/s}^2\) and ignore the weight of the forearm itself.

What force does the biceps exert, and what class of lever is the forearm?
1. 2.7 N, on a first-class lever.
2. 20 N, on a third-class lever.
3. 150 N, on a second-class lever.
4. 150 N, on a third-class lever.
5. 600 N, on a third-class lever.
Show solution

Answer: D.
The elbow is the fulcrum, and the moments about it are equal, \(F_b \cdot 4 = 20 \cdot 30\), which gives \(F_b = \dfrac{600}{4} = 150\ \text{N}\).
The biceps pulls between the fulcrum and the dumbbell, and so the lever is third-class, with the effort greater than the load.
Option A inverts the ratio between the arms, and B ignores the lever. C gets the force right and the class wrong, and E forgets to divide by the arm of the biceps.
Beam on two supports · Step 6
Engineers installed sensors on the two piers of a small 20 m bridge to follow the passage of a lorry. The graph shows the force measured at pier A as a function of the distance \(x\) between the lorry and that pier, while it crosses the bridge on its own. We will treat the lorry as a load concentrated at its centre of mass and the bridge as uniform, resting on its ends.

What is the weight of the lorry, and how much does pier B hold when it is 8 m from A?
1. 100 kN and 160 kN
2. 150 kN and 160 kN
3. 150 kN and 190 kN
4. 250 kN and 160 kN
5. 350 kN and 190 kN
Show solution

Answer: B.
With the lorry over B (\(x = 20\ \text{m}\)), pier A holds only its half of the weight of the bridge, and the graph gives \(\dfrac{W_{\text{bridge}}}{2} = 100\ \text{kN}\).
With the lorry over A (\(x = 0\)), \(N_A = 100 + W_t = 250\ \text{kN}\), and the lorry weighs \(W_t = 150\ \text{kN}\). The total weight is \(200 + 150 = 350\ \text{kN}\).
The line falls by 7.5 kN per metre, and at \(x = 8\ \text{m}\) we have \(N_A = 250 - 7.5 \cdot 8 = 190\ \text{kN}\), which leaves \(N_B = 350 - 190 = 160\ \text{kN}\).
Option A takes the final value on the graph as the weight of the lorry, and D the initial value, forgetting the bridge. C reads \(N_A\) instead of \(N_B\), and E confuses the total weight with that of the lorry.
Pulleys · Step 7
On a small building site, with no winch, the site foreman Wagner sets up a pulley system with two movable pulleys and one fixed pulley, like the one in the simulation in step 7, to take 50 kg bags of cement up to the slab, 6 m above the ground. We will assume light ropes, massless and frictionless pulleys, and \(g = 10\ \text{m/s}^2\).

What force does the worker exert on the rope, and how much rope does he need to pull to take a bag up to the slab?
1. 250 N and 12 m
2. 125 N and 6 m
3. 125 N and 24 m
4. 500 N and 6 m
5. 125 N and 1.5 m
Show solution

Answer: C.
Each movable pulley divides the force by two, and \(F = \dfrac{500}{2^2} = 125\ \text{N}\).
The rope pulled grows in the same proportion, \(s = 2^2 \cdot 6 = 24\ \text{m}\), and the work checks out, \(125 \cdot 24 = 500 \cdot 6 = 3000\ \text{J}\).
Option A counts only one movable pulley, and B forgets that the rope pulled increases. D holds for the fixed pulley on its own, and E divides the displacement instead of multiplying it, as if the pulley system also saved rope.

Sine of some angles
Angle α	sin α
15°	0.26
30°	0.50
45°	0.71
60°	0.87

Real ENEM questions on this topic

The official papers and answer keys are on the INEP website, where we can look each question up by year, day, booklet colour and number.

ENEM 2012, Day 1, blue booklet, question 55. A door held to its frame by two hinges stays still, and the question asks for the drawing of the forces the hinges exert on it, which requires cancelling both the resultant force and the moment at the same time.
ENEM 2015, Day 1, blue booklet, question 82. A teacher balances a uniform bar on a triangular support with a bag of rice hanging from one end, and the mass of the bar comes from equating the moments.
ENEM 2016, Day 1, blue booklet, question 82. Archimedes' compound pulley appears pulling a ship, and we need to find out how many movable pulleys are enough for a small force to overcome friction.
ENEM 2018, Day 2, blue booklet, question 104. Among objects such as tweezers, pliers, a nutcracker and a wheelbarrow, the question asks in which of them the force we exert is greater than the load, which leads us to the third-class lever.
ENEM 2024, Day 2, blue booklet, question 96. When a car wheel is balanced, the centre of mass of the set lies off the geometric centre, and the question asks for the point where the counterweight should be fixed to bring it back.

ENEM papers and answer keys on the INEP website (in Portuguese) →

Cheat sheet

Equilibrium of a point\(\sum F_x = 0\)
\(\sum F_y = 0\)

Components\(F_x = F\cos\theta\)
\(F_y = F\sin\theta\)

Two symmetric strings\(T = \dfrac{W}{2\cos\theta}\)

Moment of a force\(M = F\,r\sin\theta = F\,d\)

Extended body\(\sum \vec F = \vec 0\) and \(\sum M = 0\)

Lever and seesaw\(F_e\,b_e = F_l\,b_l\)

Mechanical advantage\(\text{MA} = b_e/b_l\)

Centre of mass\(x_{cm} = \dfrac{\sum m_i\,x_i}{\sum m_i}\)

Toppling on a ramp\(\tan\theta = b/h\)

Beam on two supports\(N_B = \dfrac{W_{\text{beam}}}{2} + W\,\dfrac{x}{L}\)

Pulley system with n movable pulleys\(F = W/2^n\)

Rope pulled\(s = 2^n\,h\)