In the figure here three ballot boxes are connected by cords, one of which wraps

Question

In the figure here three ballot boxes are connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. The masses are mA = 32 kg, mB = 41 kg, and mC = 12 kg. When the assembly is released from rest, (a) what is the tension in the cord connecting B and C, and (b) how far does A move in the first 0.250 s (assuming it does not reach the pulley)?


The picture shows box A on a horizontal, tied to B and C via a rope. , which hangs them vertical from the edge of a cliff on a pulley. B is above C, with a rope between.

Explanation / Answer

mass of the box A is mA = 32 kg mass of the box B is mB = 41 kg mass of the box C is mC = 12 kg a) for tension in the cord: apply Newton's law of motion for mass mA ,    T = mAa      ........ (1) apply Newton's law of motion for mass mC + mB ,    (mC+mB)g - T  = (mC+mB)a      ........ (2) apply Newton's law of motion for mass mC ,    mCg -T' = mCa      ........ (3) adding equations (1) and (3), we get    acceleration a = {(mC+mB)g} / {(mA+mB+mC)}                          = {(41 kg + 12 kg)(9.8 m/s2)} / {(32 kg+41 kg+12 kg)}                          = 6.11 m/s2 substitute this value in equation (3), we get         mCg -T' = mCa            (12 kg)(9.8 m/s2) -T' = (12 kg)(6.11 m/s2)       tension T' = 44.28 N b) from kinematic equation's              distance d = ut + (1/2) at2                              = (0 m/s)(0.25 s) + (1/2)(6.11 m/s2)(0.25 s)2                              = 0.190 m      mass of the box B is mB = 41 kg mass of the box C is mC = 12 kg a) for tension in the cord: apply Newton's law of motion for mass mA ,    T = mAa      ........ (1) apply Newton's law of motion for mass mC + mB ,    (mC+mB)g - T  = (mC+mB)a      ........ (2) apply Newton's law of motion for mass mC ,    mCg -T' = mCa      ........ (3) adding equations (1) and (3), we get    acceleration a = {(mC+mB)g} / {(mA+mB+mC)}                          = {(41 kg + 12 kg)(9.8 m/s2)} / {(32 kg+41 kg+12 kg)}                          = 6.11 m/s2 substitute this value in equation (3), we get         mCg -T' = mCa            (12 kg)(9.8 m/s2) -T' = (12 kg)(6.11 m/s2)       tension T' = 44.28 N b) from kinematic equation's              distance d = ut + (1/2) at2                              = (0 m/s)(0.25 s) + (1/2)(6.11 m/s2)(0.25 s)2                              = 0.190 m      apply Newton's law of motion for mass mC + mB ,    (mC+mB)g - T  = (mC+mB)a      ........ (2) apply Newton's law of motion for mass mC ,    mCg -T' = mCa      ........ (3) adding equations (1) and (3), we get    acceleration a = {(mC+mB)g} / {(mA+mB+mC)}                          = {(41 kg + 12 kg)(9.8 m/s2)} / {(32 kg+41 kg+12 kg)}                          = 6.11 m/s2 substitute this value in equation (3), we get         mCg -T' = mCa            (12 kg)(9.8 m/s2) -T' = (12 kg)(6.11 m/s2)       tension T' = 44.28 N b) from kinematic equation's              distance d = ut + (1/2) at2                              = (0 m/s)(0.25 s) + (1/2)(6.11 m/s2)(0.25 s)2                              = 0.190 m                              = 0.190 m     

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