C. Forces · Forces

Connected Objects, Tension and Pulleys

C2.2 C2.4 C3.3About 75 minutes

Learning goals

I can find the common acceleration of connected objects by treating them as one system.
I can find the tension in a rope or string by drawing the free-body diagram of one object.
I can explain how an ideal pulley changes the direction of a tension force.
I can solve problems with toboggan trains, a cart pulled by a hanging mass, and an Atwood machine.

Words to know

system
The object or group of objects you choose to study. Forces from outside are external; forces between parts of the system are internal.

external force
A force on the system from something outside it, such as gravity or friction from the ground.

internal force
A force between two parts of the system, such as the tension in a rope joining two sleds. Internal forces cancel in pairs for the whole system (third law).

ideal string and pulley
A string with no mass that does not stretch, and a pulley with no mass or friction. The tension is the same everywhere along the string.

Atwood machine
Two masses hanging on either side of a pulley, joined by a string.

Two toboggans, one rope

A parent pulls two toboggans along a level trail: a young child rides in the front toboggan, and a second toboggan is tied behind it. The rope from the parent feels the pull of both toboggans. The short rope between them only has to pull the back one. Which rope is under more tension, and by how much?

When objects are tied together, they move together, with the same acceleration. That single fact, plus Newton's second law, solves trains, tow trucks, elevators and pulley systems.

Two-step method for connected objects

Step 1, the whole system: Treat all the connected objects as one object, with mass m(total). Include only external forces (internal tensions cancel). Use Fnet = m(total) × a to find the common acceleration.

Step 2, one object: Draw the FBD of one object, ideally the one with the fewest forces. Use Fnet = ma for that object alone to find the tension.

Check your answer with the other object.

Worked example

Example 1: The toboggan train

A parent pulls a 12 kg toboggan (with its rider) which is tied by a rope to an 8.0 kg toboggan behind it. The parent's horizontal pull is 40 N and μk = 0.10 for both on the snow. Find (a) the acceleration and (b) the tension in the rope between the toboggans.

Forward is +.
(a) System, m(total) = 20 kg. Friction on both: μk m(total) g = 0.10 × 20 × 9.8 = 19.6 N. Fnet = 40 − 19.6 = 20.4 N. a = 20.4 ÷ 20 = 1.02 m/s².
(b) Rear toboggan alone (8.0 kg): forward tension FT, backward friction 0.10 × 8.0 × 9.8 = 7.84 N.
FT − 7.84 = 8.0 × 1.02, so FT = 8.16 + 7.84 = 16.0 N.
Check with the front toboggan: 40 − FT − 11.76 = 12 × 1.02 → 40 − 16.0 − 11.76 = 12.24 ✓.

(a) 1.02 m/s² forward. (b) 16 N, much less than the parent's 40 N pull.

Treat both toboggans as one 20 kg system to find the acceleration. The tension in rope 2 is an internal force: it pulls the rear toboggan forward and the front toboggan back by the same amount, so it cancels for the whole system.

Pulleys

An ideal pulley changes the direction of a string without changing its tension. For a cart on a table pulled by a hanging mass, both objects move the same distance along the string at the same time, so they have the same size of acceleration: the cart moves right while the mass moves down.

The trick: choose "positive" to follow the string. For the cart, + is toward the pulley; for the hanging mass, + is down. Then the whole system acts like one object pulled along a line, and you can use the two-step method.

Worked example

Example 2: Cart and hanging mass

A 2.0 kg cart on a level, frictionless track is tied to a 0.50 kg mass hanging over a pulley. Find (a) the acceleration and (b) the tension.

Along the string, the only external force that drives the system is the hanging mass's weight: 0.50 × 9.8 = 4.9 N. (The cart's weight is balanced by the normal force.)
(a) m(total) = 2.5 kg. a = 4.9 ÷ 2.5 = 1.96 m/s².
(b) Cart alone: only the tension acts horizontally. FT = ma = 2.0 × 1.96 = 3.92 N.
Check with the hanging mass (down +): 4.9 − 3.92 = 0.98 = 0.50 × 1.96 ✓. The tension is less than the hanging weight, because the hanging mass is accelerating down.

(a) About 2.0 m/s². (b) About 3.9 N.

The pulley only changes the direction of the string. Along the string, the system is pulled forward by the hanging mass's weight (4.9 N) and must accelerate 2.5 kg in total.

Worked example

Example 3: Atwood machine

A 3.0 kg mass and a 2.0 kg mass hang on either side of an ideal pulley. Find the acceleration and the tension.

The 3.0 kg side goes down. Take + as down for the 3.0 kg mass and up for the 2.0 kg mass (following the string).
System: driving force = 3.0g − 2.0g = 29.4 − 19.6 = 9.8 N. m(total) = 5.0 kg. a = 9.8 ÷ 5.0 = 1.96 m/s².
2.0 kg mass (up +): FT − 19.6 = 2.0 × 1.96, so FT = 19.6 + 3.92 = 23.5 N.
Check, 3.0 kg (down +): 29.4 − 23.5 = 5.9 ≈ 3.0 × 1.96 = 5.88 ✓.

a ≈ 2.0 m/s²; FT ≈ 23.5 N.

Just for fun

The Atwood machine is named after George Atwood, who described it in 1784. By hanging two nearly equal masses, he made gravity's effect slow enough to time by eye, like an 18th-century slow-motion replay of free fall.

Hands-on activity

Measure tension with a luggage scale

You need: two cardboard boxes or laundry baskets with some books in them, string, a luggage scale or spring scale (or two rubber bands to compare stretches), and a smooth floor.

Tie box B behind box A with a short string, and tie a pull string to box A.
Put the luggage scale in the front string and pull steadily so the boxes speed up gently. Read the scale.
Move the scale to the string between the boxes and repeat with the same kind of pull. Which reading is larger? Why?
Swap the boxes (heavier one in the back). Predict how the middle reading changes, then test.
Explain your results using the two-step method: the front string accelerates both boxes; the middle string accelerates only the back box.

Pull slowly; keep toes away from the boxes.

Your turn

Two carts (3.0 kg and 1.0 kg) are tied together on a frictionless floor. A 12 N force pulls the 3.0 kg cart. Find the acceleration.

In the question above, find the tension in the string between the carts.

What would the tension be if the 12 N force pulled the 1.0 kg cart instead (with the 3.0 kg cart behind)?

What does an ideal pulley do to the tension in a string?

A 4.0 kg cart on a frictionless table is pulled by a 1.0 kg hanging mass. Find the acceleration.

In an Atwood machine with 5.0 kg and 3.0 kg masses, find the acceleration.

More practice: warm-up

Do connected objects (with a tight rope) have the same acceleration?

Is the tension between two sleds an internal or external force for the two-sled system?

Which forces do you include when finding the acceleration of the whole system?

Can a string push?

Two equal masses hang on either side of a pulley at rest. What is the acceleration?

More practice: on your own

A 2.0 kg cart on a table with μk = 0.10 is pulled by a 0.50 kg hanging mass. Find the acceleration and tension.

In an Atwood machine, a 4.0 kg mass and a 6.0 kg mass hang over a pulley. Find the tension.

A truck tows a 1200 kg car with a cable. The car accelerates at 0.80 m/s², and rolling friction and air resistance on the car total 300 N. Find the cable tension.

A snowmobile pulls three identical 50 kg sleds in a line with 600 N, on snow with μk = 0.20. Find the acceleration.

For the sled train above, find the tension in the rope between the 2nd and 3rd sleds.

Challenge

A 1.0 × 10³ kg elevator carries a 70 kg passenger. The cable tension is 1.2 × 10⁴ N. Find the acceleration of the elevator and the normal force on the passenger.

In Example 2 (cart and hanging mass), explain why the tension (3.9 N) is less than the hanging weight (4.9 N). What would the tension be if the cart were held still?

Two masses hang over an ideal pulley. One is 2.0 kg. The system accelerates at 2.45 m/s², with the other mass moving down. Find the other mass.

Check your understanding

Quiz

Two blocks (2.0 kg and 3.0 kg) tied together are pulled across a frictionless floor by a 10 N force. Their acceleration is:

For the blocks above, the 10 N force pulls the 3.0 kg block. The tension in the rope between them is:

An ideal pulley:

When finding the acceleration of a whole connected system, you should include:

An Atwood machine has 6.0 kg and 4.0 kg masses. Its acceleration is about:

A hanging mass pulls a cart across a table and the system speeds up. The string tension is:

For parents and teachers

Big idea: Applying Newton's laws and free-body diagrams to systems of connected objects (C2.2), including friction (C2.4), and the third-law idea that internal forces cancel (C3.3).

How to help: Ask your teen to explain in words before calculating: "What is the system? What drives it? What resists it? Which rope pulls more?"

Common difficulties:

Using only one object's mass when finding the system's acceleration.
Assuming the tension equals the hanging weight when the system accelerates.
Getting signs tangled around a pulley. Choosing + to follow the string fixes this.
Including internal forces in the whole-system equation.