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Atwood Machine, Pulleys

An Atwood machine uses a cable drawn over a pulley to connect two or more masses. One of the masses acts as a counterbalance or counterweight to reduce acceleration because of gravity. Elevators in multi-level buildings are examples of Atwood machines. The counterweight in an elevator is typically the mass of the elevator plus about half of the mass of the allowable load.

Atwood Machine and Pulley Problems

(See bottom of page for answers.)

A simple pulley problem is illustrated, with two masses hanging from a frictionless pulley.

1. In an Atwood's machine, the larger mass is 1.8 kg and the smaller mass is 1.2 kg.
a. Ignoring friction, what is the acceleration of the masses?
b. What is the tension in the string?

2. A 10.0 kg mass, m1, on a frictionless table is accelerated by a 5.0 kg mass, m2, hanging over the edge of the table. What is the acceleration of the mass along the table?

3. Two identical blocks are tied together with a string which passes over a pulley at the crest of the inclined planes, one of which makes an angle ø1 = 28° to the horizontal, the other makes the complementary angle ø2 = 62°. If there is no friction anywhere, with what acceleration do the blocks move?

A pulley is shown with a string connecting two masses on a slope.

4.

A 2.00-kg and a 6.00-kg block are connected by a light string over a frictionless pulley. The two blocks are allowed to move on a fixed s wedge inclined 30º. µ = 0.18 Determine the acceleration of the two blocks and the tension in the string.

5. Two blocks are moving down a ramp that is inclined at 30º. Box 1(of mass 1.55 kg) is above box 2(of 3.1 kg) and they are connected by a rod having negligible mass. Find the tension of the rod if the coefficient of kinetic friction for box 1 is .226 and box 2 is .113.

Pulley problems sometimes involve masses on surfaces that have friction.  The string in this diagram connects a mass on a horizontal surface and a mass on an inclined plane.

6.

Two blocks, one 0.8 kg and the other 2.0 kg are connected by a massless string over a frictionless pulley. The coefficient of kinetic friction is 0.14, and the downward ramp angle is 60 degrees.

a) Determine the acceleration of the blocks.

b) Calculate the tension of the string.

A 5.0kg mass is accelerated from rest at the bottom of the 4.0 m long ramp by a falling 20.0kg mass suspended over a frictionless pulley.

7.

A 5.0kg mass is accelerated from rest at the bottom of the 4.0 m long ramp by a falling 20.0kg mass suspended over a frictionless pulley. The ramp is inclined 30º ramp from the horizontal, and the coefficient of kinetic friction = .26.
a. Determine the acceleration of the 5.0 kg mass along the ramp.
b. Determine the tension in the rope during the acceleration on the 5.0 kg mass along the ramp.

A string through a pulley attaches a mass on a table to a mass hanging over the edge.

8.

A box of mass m1 sits on a table. It is connected, by a rope drawn through a pulley, to a box m2 = 2.10 kg that is hanging off the side of the table. The coefficient of static friction between mass m1 and the table is 0.400, whereas the coefficient of kinetic friction is 0.295.
a. What minimum value of m1 will keep the boxes from starting to move?
b. What value of m1 will keep the boxes moving at constant speed?

A string tied to a mass on a ramp goes through a pulley to a mass sitting on top of the first mass.

9.

A 20 kg block, m1, is sliding on a 10 kg block, m2. The blocks are on a 20º slope and are connected by a light string looped over a pulley. All surfaces are frictionless. Find the acceleration of each block and the tension in the sting that connects the blocks.

A string through a pulley attaches a mass on a table to a mass hanging over the edge.

10.

A 20 kg weight on a table is attached by a string to another 20 kg weight that is hanging over the edge of the table. Assuming that the coefficient of friction is 0.2, what is the acceleration and what is the tension in the string?

Answers to Atwood Machine and Pulley Problems

Selected solutions are printed below.
For solutions to all the problems on this page click here.

Free body diagrams for attached masses show the forces on the individual masses.

1.
a. The net force on each mass is given in the two equations (g = 9.8 m/s/s):
1.8a = 1.8g -T
1.2a = T - 1.2g
Since a and T are the same for both masses, add the two equations:
3.0a = .6g
Therefore,
a = .2g = 1.96 m/s/s

b. Substituting for a in the first equation:
1.8(1.96) = 1.8(9.8) - T
T = 14.1 N

For solutions to all the problems on this page click here.

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