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# Mass - Pulley problem

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Two pulley system supporting two masses; one sliding on a rough surface and the other hanging. The system is released from rest. To determine the linear acceleration of the sliding mass.

Block A weighs 25 lbs and it slides on a surface with a coefficient of friction 0.30. Block B weighs 7 lbs. The cable does not slip on either pulley. Both pulleys are massless and frictionless. The system is released from rest. Determine the linear acceleration of mass A.

(See attachment for fig.)

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Problem: Block A weighs 25 lbs and it slides on a surface with a coefficient of friction 0.30. Block B weighs 7 lbs. The cable does not slip on either pulley. Both pulleys are massless and frictionless. The system is released from rest. Determine the linear acceleration of mass A.

Solution:
Please refer to the attachment.

(Please see attached file for diagram)

All the forces acting on the system are shown in the fig.. Weights of the blocks act vertically downwards. Normal reaction N acts vertically upwards. T is the tension in the string. Frictional force acts backwards (opposite to the direction of motion).

Normal reaction N = weight of block A = 25 lb

Limiting frictional force f = μN = 0.3 x 25 = 7.5 lb

If block A moves to left by a distance x, block B moves down by a distance x/2 (because the length x of the string is equally distributed on string sections on two sides of pulley 2). Hence, speed and acceleration of block B are ½ of the speed and acceleration of bock A.

Let linear acceleration of block A be a. Then linear acceleration of block B is a/2.

Net force acting on block A = T - f = T - 7.5

Acceleration of block A = a = Force/Mass

Mass of block A = 25/32.2 = 0.78 slug

Hence, a = (T - 7.5)/0.78 .........(1)

Net force acting on block B = 7 - 2T

Mass of block B = 7/32.2 = 0.22 slug

Hence, acceleration of block B a/2 = (7 - 2T)/0.22

Or a = (7 - 2T)/0.11 ..............(2)

From (1) and (2) we get: (T - 7.5)/0.78 = (7 - 2T)/0.11

Or T - 7.5 = 7.1(7 - 2T) = 49.7 - 14.2T

Or 15.2T = 57.2 or T = 3.76 lb

Substituting in (1) we get: a = - 4.8 ft/s2

We get a -ve value for the acceleration of block A, which implies it accelerates towards right. As that s not possible, block A (and block B) remains at rest.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

© BrainMass Inc. brainmass.com October 5, 2022, 6:18 pm ad1c9bdddf>
https://brainmass.com/physics/acceleration/mass-pulley-problem-251961