Two blocks on a frictionless table, A and B, are connected by a massless string.

Question

Two blocks on a frictionless table, A and B, are connected by a massless string. When block A is pulled with a certain force, dragging block B, the tension in the string is 28N . When block B is pulled by the same force, dragging block A, the tension is 12N. What is the ratio (ma/mb) of the blocks

Two blocks on a frictionless table, A and B, are connected by a massless string. When block A is pulled with a certain force, dragging block B, the tension in the string is 28N . When block B is pulled by the same force, dragging block A, the tension is 12N. What is the ratio (ma/mb) of the blocks masses?

Explanation / Answer

In all cases the acceleration of the blocks is the same a = F/(mA + mB)
When block A is pulled we have a = (F - 30)/mA

Equating F we have a*(mA + mB) = a*mA + 30 or mB = 30/a

Similarly in the second case we have a = (F - 16)/mB or a*(mA + mB) = a*mB + 16

or mA = 16/a

Therefore mA/mB = 16/30

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