A boy of mass 4 kg is standing on a piece of wood having mass 5 kg. If the coefficient of friction between the wood and the floor is 0.5, the maximum force that boy can exert on the rope so that the piece of wood does not move from its place is ______ N. (Round off to the Nearest Integer) [Take g=10ms−2]
Answer: 30
Solution
The rope the boy holds pulls back on him with the same force F, directed upward along the rope, so he presses on the plank with 4g−F and the floor carries
N=(4+5)g−F=90−F
The far end of the rope is tied to the plank and drags it horizontally with F, which the friction at the floor has to balance. The plank stays where it is as long as
F≤μN=0.5(90−F)
1.5F≤45⇒F≤30
The greatest force the boy can exert on the rope is 30 N.
JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Numerical answer
