Two blocks with masses M 1 and M 2 hang one under the other. For this problem, t

Question

Two blocks with masses M1 and M2 hang one under the other. For this problem, take the positive direction to be upward, and use g for the magnitude of the acceleration due to gravity

Part A and B assume blocks are at rest

1, the tension in the upper rope.

For Parts C and D the blocks are now accelerating upward (due to the tension in the strings) with acceleration of magnitude a.

Part C) Find

a

Two blocks with masses M1 and M2 hang one under the other. For this problem, take the positive direction to be upward, and use g for the magnitude of the acceleration due to gravity Find T2, the tension in the lower rope. Express your answer in terms of

Explanation / Answer

Part A)

Drawing the FBD of block M2 there are only 2 forces T2 upwards and M2*g downwards hence for equilibrium

T2 = M2*g

Part B)

on M1 there are 3 forces T1 upwards and T2 and M1*g downwards

Hence T1 = T2 + M1*g = (M2+M1)*g

Now they are accelerating upward with an acceleration a

Now we can simply see that the form of the equation won't change . If we analyze with respect to the system the new acceleration can be obtained as g+a..

Hence both the tensions can be calculated by replacing g by g+a

Part C)

T2 - M2*g = M2 *a

T2 = M2*(g+a)

Part D)

T1 = (M1 +M2)*(g+a)

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