Figure shows a block (mass ) on a smooth horizontal surface, connected by a thin cord that passes over a pulley to a second block which hangs vertically. (a) Draw a free-body diagram for each block, showing the force of gravity on each, the force (tension) exerted by the cord, and any normal force. (b) Apply Newton’s second law to find formulas for the acceleration of the system and for the tension in the cord. Ignore friction and the masses of the pulley and cord.

Question

Figure shows a block (mass  ) on a smooth horizontal surface, connected by a thin cord that passes over a pulley to a second block   which hangs vertically. (a) Draw a free-body diagram for each block, showing the force of gravity on each, the force (tension) exerted by the cord, and any normal force. (b) Apply Newton’s second law to find formulas for the acceleration of the system and for the tension in the cord. Ignore friction and the masses of the pulley and cord.

Damon · Accepted Answer

m1 up top
m2 falling
tension T
distance m2 moves DOWN is x is also distance m1 moves horizontal (string does not stretch)
now for m1
force down on table = m1 g
force up from table = m1 g
no vertical acceleration
only force horizontal is T
so horizontal acceleration = a1 = T/m1
now for m2
no horizontal force or acceleration
force down = m2 g
force up = T
total force down = m2 g - T
acceleration a2 = (m2 g -T)/ m2
now if the string does not stretch then a2 = a1
T/m1 = (m2 g - T)/m2
T (m2/m1) = m2 g - T
T (1 + m2/g) = m2 g
now if by any wild chance (to check answer) m2 = m1 = m
2 T = m g
T = mg/2 LOL, yes accelerates half as fast as one would alone :)