A 20N block rests on a table, connected by a massless rope to a 12N block hanging off the table. Assume the pulley is light and frictionless.

Question

What is the tension in the rope?
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I calculated the masses to be 2.04kg and 1.22kg respectively. From there, the free-body diagram for the 12N block (block 2) suggested to me that using Newton's 2nd Law would yield:

Fnet = ma
T + W2 = (m1 + m2)g
T = (m1 + m2)g - W2
T = (3.265)(9.81)-12
T = 20N

But this is not correct. I believe the correct answer is 7.50N but I do not understand how to calculate that. What am I missing? Thanks

Answer 1

What is missing, is that the block on a table is moving, and the table is frictionless. So the 12N block is pulling the larger block

at the smaller block
tension=m(g-a)
at the large block
tension=Ma
Ma=mg-ma
a(M+m)=mg
a=mg/(M+m)
now you can solve for tension in either equation

Answer 2

For the smaller block, wouldn't the sum of forces be:

T - W = ma
T - mg = ma
T = ma + mg = m(a+g)

Why is it m(g-a) instead?

Answer 3

nope, if if it were falling free fall, a=g
tension=m(g-a)=zero Think that out.

Answer 4

Ok, thanks