Question

A 55 kg gymnast wedges himself between two closely spaced vertical walls by pressing his hands and feet against the walls. Part a what is the magnitude of the friction force on each hand and foot? Assume they are all equal.

Answer 1

solution:

$Let frictional forces due to both hands and feets be "Ff" each(since its given that they all are equal), acting in upward direction( in opposite direction of supposed motion).\\ Then since there is no motion of gymnast thus net frictional force due to both hands and feets will be exactly balanced by the weight of the gymnast,\\ i.e\\ 4f_{f}=weight =mg\\ f_{f}=\frac{55x9.8}{4}\\ =134.75N$