Newton's laws problems — pulleys and slopes — Physics

Before this

This leads to

Friction — the model and its limits · coming soon
Resolving forces · coming soon
Conservation of momentum

Two objects, one string

Hang two masses on a string over a pulley: 8 kg on one side and 5 kg on the other. Let go. The 8 kg mass falls and the 5 kg mass rises. This arrangement is called an Atwood machine.

Two questions sound simple. How fast does the motion speed up? And how hard is the string pulling? Most students can draw the forces on one object. Here there are two objects, joined by a string, and each one affects the other. That is the new difficulty, and it lives in the string. Once you know exactly what the string does, the rest is the method from the last two lectures, used twice.

The last lecture promised this: each object gets its own free-body diagram, and the string's tension appears in both. Before any equations, make a prediction.

Predict, then reveal — the tension in the string

An 8 kg mass (weight 78.4 N) and a 5 kg mass (weight 49 N) hang over a pulley and start to move. What is the tension in the string while they move?

The tension in the string of an Atwood machine with 8 kg and 5 kg masses. Many students choose one of the two weights, but the tension lies between them.

The tension is about 60.3 N. It must be more than 49 N, because the 5 kg mass accelerates upwards. It must be less than 78.4 N, because the 8 kg mass accelerates downwards.

What the string does

Every problem in this lecture uses the same three assumptions. The string is light (its mass is so small it can be ignored). The string is inextensible (it does not stretch). The pulley is smooth (no friction at its axle). Each assumption gives one fact, and it is worth knowing why.

The tension is the same all along the string. Think of a short piece of light string. The string on each side pulls it, in opposite directions. If those two pulls were different, there would be a resultant force on a piece with almost no mass. By $a = F \div m$ , its acceleration would be enormous. That does not happen, so the two pulls must be equal. A smooth pulley changes the direction of the string, but not the size of its pull.
Both objects have the same size of acceleration. The string does not stretch. So when one mass moves down 1 cm, the other must move up 1 cm. They move the same distance in the same time, so they have the same speed and the same size of acceleration at every moment, in different directions.

That is it. That is the whole idea. The string links the two objects through one tension $T$ and one acceleration $a$ . Two unknowns need two equations, and each object's diagram gives one.

Connected objects: the method

1. Decide which way the system moves. For each object, take that direction as positive.

2. Draw a free-body diagram for each object separately. The same $T$ appears in both.

3. Write $F = ma$ for each object, along its direction of motion.

4. Add the equations to remove $T$ . Solve for $a$ , then substitute back to find $T$ .

5. Check the answer with the whole-system view and with common sense.

The Atwood machine, solved

An Atwood machine with 8 kg and 5 kg masses. Each mass has two forces: the tension T upwards and its weight downwards. The tension is the same on both sides.

Take $g = 9.8$ N/kg. The 8 kg mass moves down, so take down as positive for it. The 5 kg mass moves up, so take up as positive for it.

8 kg, down: 5 kg, up: 78.4 - T = 8 a T - 49 = 5 a

Add the two equations. The $T$ and $- T$ cancel, which is exactly why the method chooses the positive directions this way:

29.4 a = 13 a = \frac{29.4}{13} \approx 2.26 m/s^{2}

Then from the second equation, $T = 49 + 5 a$ . Keep $a$ unrounded: $T = 49 + 5 \times \frac{29.4}{13} \approx 60.3$ N. Check with the first equation: $78.4 - 60.3 = 18.1$ , and $8 \times 2.26 \approx 18.1$ . They agree.

Now look at the answer as a whole system. Treat both masses and the string as one object of mass 13 kg. The tension is now inside the object, so its two pulls cancel, just as third-law pairs did in the third-law lecture. The only forces that drive the motion are the two weights, pulling opposite ways round the pulley: $78.4 - 49 = 29.4$ N. That force accelerates 13 kg, so $a = 29.4 \div 13$ . The same answer, in one line. In general:

a T = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = \frac{2 m _{1} m _{2} g}{m _{1} + m _{2}}

Check these formulas at the extremes. If the masses are equal, $a = 0$ and $T = m g$ : nothing moves, and the string holds each weight exactly. If the lighter mass is almost zero, $a$ approaches $g$ and $T$ approaches zero: the heavy mass simply falls freely. A formula that passes both checks is very likely right.

Why Atwood built it

George Atwood, a mathematician at Cambridge, described this machine in 1784 in A Treatise on the Rectilinear Motion and Rotation of Bodies. His aim was to demonstrate the laws of uniformly accelerated motion under gravity (MacTutor History of Mathematics, University of St Andrews). Free fall is too fast to watch and time by hand: a stone falls 1 m in under half a second. With two nearly equal masses, the same gravity drives a much slower motion, which can be timed with a clock. Try it below.

Simulation — an Atwood machine

Choose the two masses. Press Play, or Step 0.1 s at a time. Compare the time to fall 1 m with free fall. Then make the masses nearly equal.

Mass A: 8.0 kg

Mass B: 5.0 kg

The acceleration, tension and time to fall 1 m from rest for three pairs of masses (g = 9.8 m/s²). Free fall takes about 0.45 s.

Masses	a (m/s²)	T (N)	Time to fall 1 m
8 kg, 5 kg	≈ 2.26	≈ 60.3	≈ 0.94 s
5.2 kg, 4.8 kg	≈ 0.39	≈ 48.9	≈ 2.26 s
10 kg, 0.5 kg	≈ 8.87	≈ 9.3	≈ 0.47 s

The closer the two masses, the smaller the acceleration and the longer the fall, which is what makes the motion easy to time.

With 5.2 kg and 4.8 kg, the acceleration is only $9.8 \times 0.4 \div 10 \approx 0.39$ m/s², and the fall takes about 2.26 s instead of 0.45 s. Measure that time, and you can work backwards to $g$ . The machine slows gravity down without changing it.

Friction on a sliding object

The free-body lecture drew friction as an arrow along the surface, without saying how big it is. Here is the model used at this level. When an object slides, the friction on it is:

F = μ N

Here $N$ is the normal force and $μ$ (the Greek letter "mu") is the coefficient of friction, a number for the two surfaces. Rubber on dry road has a large $μ$ ; ice on steel has a small one. The formula makes sense: press the surfaces together harder (larger $N$ ), and they grip harder. Before an object slides, friction can be anything from zero up to $μ N$ , just enough to stop the motion. $F = μ N$ is the most it can be, and it is reached when sliding starts. Friction always acts against the sliding, or against the way the object would slide without it.

A block on a slope, pulled by a hanging mass

A 4 kg block on a 30° slope, joined over a smooth pulley to a 3 kg hanging mass. The forces on the block along the slope are the tension T (up the slope), the part of its weight mg sin 30° (down the slope) and friction (against the motion). The normal force N is at right angles to the slope.

Worked Example 1 — Slope, friction and a hanging mass

A 4 kg block rests on a slope at 30°. The coefficient of friction is 0.2. A light string runs up the slope, over a smooth pulley, to a hanging 3 kg mass. Find the acceleration and the tension.

Step 1: which way will it move? Friction's direction depends on the motion, so decide this first. The hanging mass pulls with its weight, 29.4 N. The block's weight pulls back along the slope with $4 \times 9.8 \times sin 3 0^{\circ} = 19.6$ N. Friction can resist with at most $μ N = 0.2 \times 4 \times 9.8 \times cos 3 0^{\circ} \approx 6.79$ N. Since $29.4 > 19.6 + 6.79 \approx 26.4$ , the system moves: the hanging mass falls and the block slides up the slope. So friction on the block acts down the slope.

Step 2: each object. Block, up the slope positive: $T - 19.6 - 6.79 = 4 a$ . Hanging mass, down positive: $29.4 - T = 3 a$ .

Step 3: add. $29.4 - 26.39 \approx 3.01 = 7 a$ , so $a \approx 0.43$ m/s². Then $T = 29.4 - 3 a \approx 28.1$ N.

Check. Only about 3 N of driving force moves 7 kg, so a small acceleration makes sense. And the tension is a little less than the hanging weight, as it must be for that mass to accelerate downwards.

Stop and think

In the same set-up, the 3 kg mass is replaced by a 2 kg mass. Does anything move?

Show the answer

No. The hanging weight is now $2 \times 9.8 = 19.6$ N, exactly equal to the block's pull down the slope, $m g sin 3 0^{\circ} = 19.6$ N. The two balance, so friction does not even need to act: it is zero. The block would stay still even on a smooth slope. A student who writes $F = 6.79$ N down the slope and solves will get a negative acceleration, a sign that the assumed motion is wrong. Always decide whether, and which way, the system moves before you use $F = μ N$ .

Worked Example 2 — A block on a smooth table

A 4 kg block lies on a smooth horizontal table. A string runs from it over a pulley at the table's edge to a hanging 1 kg mass. Find the acceleration and the tension.

Block, towards the pulley positive: $T = 4 a$ . Hanging mass, down positive: $9.8 - T = 1 a$ . Add: $9.8 = 5 a$ , so $a = 1.96$ m/s². Then $T = 4 \times 1.96 = 7.84$ N.

Whole-system check: the only driving force is the 9.8 N weight, and it moves 5 kg: $9.8 \div 5 = 1.96$ m/s². The table's normal force balances the block's weight, so it plays no part along the motion.

Three classic mistakes

Trap 1 — Tension equals the hanging weight

The tension equals the hanging weight only when nothing accelerates. If the hanging mass speeds up downwards, the tension is less than its weight; in Worked Example 1, it is 28.1 N, not 29.4 N.

Trap 2 — Friction drawn before the motion is known

Friction opposes the motion, so you must know the motion first. Compare the driving force with the largest possible resistance before you write $F = μ N$ . If the driving force is not larger, the system stays still and friction is smaller than $μ N$ .

Trap 3 — Using the whole weight on the slope

On a slope, only $m g sin θ$ acts along the slope, and the normal force is $m g cos θ$ , so friction is $μ m g cos θ$ . Using $m g$ in either place makes the forces along the slope far too large.

Try it yourself

Take $g = 9.8$ N/kg. Light, inextensible strings and smooth pulleys throughout.

An Atwood machine has masses of 3 kg and 2 kg. Find the acceleration and the tension.

Hint 1
Write one equation for each mass, positive in its own direction of motion.

Hint 2
$29.4 - T = 3 a$ and $T - 19.6 = 2 a$ .

Full answer
Adding: $9.8 = 5 a$ , so $a = 1.96$ m/s². Then $T = 19.6 + 2 \times 1.96 = 23.52$ N. Check with the formula: $T = 2 \times 3 \times 2 \times 9.8 \div 5 = 23.52$ N. The tension lies between the two weights, 19.6 N and 29.4 N.
A 6 kg block on a rough horizontal table ( $μ = 0.25$ ) is joined over a pulley at the edge to a hanging 4 kg mass. Find the acceleration and the tension.

Hint 1
On a horizontal table, $N = 6 \times 9.8$ . Find the friction, then check that the system moves.

Hint 2
Friction $= 0.25 \times 58.8 = 14.7$ N, less than the hanging weight of 39.2 N.

Full answer
Block: $T - 14.7 = 6 a$ . Hanging mass: $39.2 - T = 4 a$ . Adding: $24.5 = 10 a$ , so $a = 2.45$ m/s². Then $T = 39.2 - 4 \times 2.45 = 29.4$ N. Check the block: $29.4 - 14.7 = 14.7 = 6 \times 2.45$ .
A 5 kg block on a smooth slope at 20° is joined over a pulley at the top to a hanging 1 kg mass. Which way does the system move? Find the acceleration and the tension.

Hint 1
Compare the block's pull down the slope, $5 \times 9.8 \times sin 2 0^{\circ}$ , with the hanging weight.

Hint 2
$49 sin 2 0^{\circ} \approx 16.76$ N, more than 9.8 N. So the block slides down and lifts the hanging mass.

Full answer
Block, down the slope positive: $16.76 - T = 5 a$ . Hanging mass, up positive: $T - 9.8 = 1 a$ . Adding: $6.96 \approx 6 a$ , so $a \approx 1.16$ m/s². Then $T \approx 9.8 + 1.16 \approx 10.96$ N. The tension is a little more than 9.8 N, because the hanging mass accelerates upwards.
A student uses an Atwood machine with masses of 0.52 kg and 0.48 kg. From rest, the heavier mass falls 1.00 m in 2.26 s. Use these measurements to find $g$ . Then explain why a timing error of 0.05 s matters much less here than in a free-fall experiment over 1 m.

Hint 1
From rest, $s = \frac{1}{2} a t^{2}$ , so $a = 2 s \div t^{2}$ .

Hint 2
Then $g = a \times (m_{1} + m_{2}) \div (m_{1} - m_{2}) = a \times 1.00 \div 0.04$ .

Full answer
$a = 2 \times 1.00 \div 2.2 6^{2} \approx 0.392$ m/s². Then $g \approx 0.392 \div 0.04 \approx 9.79$ m/s², close to 9.8. In free fall, 1 m takes only about 0.45 s, so an error of 0.05 s is about 11% of the time. Here it is about 2% of 2.26 s. Because $g$ depends on $1 \div t^{2}$ , a smaller percentage error in time gives a much smaller error in $g$ . This is the advantage Atwood's design gives.

The bigger idea

This lecture completes Newton's laws, and the method is now the same for every problem: choose an object, draw its forces, apply $F = ma$ along each direction, and let shared quantities, such as one tension or one acceleration, link the objects together. Lifts, cranes, trains of carriages and tow ropes are all the same problem in different clothes.

Notice also the two views that kept agreeing: one equation for each object, or one equation for the whole system, where internal forces cancel. When internal forces cancel, the motion of the whole system depends only on outside forces. That idea leads to one of the great conservation laws of physics, momentum, in a later topic.

Key concepts in this lesson

A light, inextensible string over a smooth pulley has the same tension all along it, and the objects it joins have the same size of acceleration.
Connected objects are solved by writing F = ma for each object, positive in its own direction of motion, then adding the equations to remove the tension.
For an Atwood machine, a = (m₁ − m₂)g ÷ (m₁ + m₂), and the tension lies between the two weights.
A sliding object has friction F = μN against its motion, and friction before sliding can be anything up to μN.
Decide whether and which way a system moves before using F = μN, and check answers with the whole-system view.