# Relative acceleration of bodies on opposite sides of a pulley [closed]

This is a problem that I encountered while solving problems related to constrained motion:

With all the data given in the above question, I believe that the monkey just holds on to the rope and does not climb it up or slip down. So if that is the case , and assuming that the string is inextensible, shouldn’t the monkey and the block have the same accelerations? However, when I saw the solution to the above problem, the monkey and the block have different accelerations. How is that possible?

First of all the monkey is trying to climb up the rope. For that he is applying a force of 80 N downward on the rope. As per newton's third law of motion the rope exerts a force of 80 N upwards on the monkey.

If we take the up direction as '$$+$$' and down direction as '$$-$$', then force on monkey is $$Mg=-100$$ N, and rope force = $$+80$$ N. By $$F=ma$$ we get acceleration (monkey) = $$-2$$ ms$$^{-2}$$.

Now as the string is massless we get tension in string = $$80$$ N. This $$T$$ will lift the $$5$$ kg block. Applying $$F=ma$$ on $$5$$ kg block (forces are $$T$$ and $$Mg$$) we get a (rope) = $$+6$$ ms$$^{-2}$$.

Now, the acceleration of the Monkey relative to rope = $$A$$(monkey) — $$A$$(rope) $$= -2-6 = -8$$ (downward direction).

• Here, acc. of rope and 5 kg block is same. – Brijesh Shah Nov 26 '17 at 3:48
• Ohhhh so it means that if the monkey didnt climb up , he wouldnt exert a force of $80N$? – Aditi Nov 26 '17 at 4:14
• Yes if the monkey did not climb up then he would not exert a 80 N force on the rope. And in such a case the acc. of the rope and monkey will be same. – Brijesh Shah Nov 26 '17 at 4:36
• Ohhhh thank you very much ! I was really confused , but now it’s clear to me. – Aditi Nov 26 '17 at 6:22
• If the monkey is accelerating downward at 2 and the rope on his side of the pulley is accelerating downward at 6 , then his acceleration relative to the rope is upward at 4 (m/sec^2). – R.W. Bird Aug 31 '19 at 17:51