A man of mass m stands on a crate of mass M. He pulls on a light rope passing over a smooth light pulley. The other end of the rope is attached to the crate. For the system to be in equilibrium, the force exerted by the man on the rope will be
A
(M+m)g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
12(M+m)g
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
Mg
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
mg
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B12(M+m)g From the free body diagram of man and crate system:
For vertical equilbrium 2T=(M+m)g ∴T=(M+m)g2