A 65kg gymnast wedges himself between two closely spaced vertical walls by pressing his hands and feet against the walls. What is the magnitude of the friction force on each hand and foot? Assume they are all equal.

Short Answer

Expert verified

The friction force on each hand and foot has a value of 160N.

Step by step solution

01

Magnitude : 

The greatest extent of size and direction of an item is defined as magnitude.

02

Explanation : 

Mass of a gymnast = 65kg.

Gymnast's free body diagram,

For the equilibrium requirement, Newton's second law is used,

Fy=0

In order to resolve the components of friction forces and the gymnast's weight in the y-direction,

-mg+4Ff=0

Ff= Friction force on each hand and foot.

m= mass of the gymnast.

g= acceleration due to gravity.

Substitute 65kg for m, localid="1647704475058" 9.81m/s2for gto find Ff.

-65kg(9.81m/s2)+4Ff=0Ff=637.65kg·m/s24160N

Hence, the friction force on each hand and foot has a value of160N.

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