MHB Boy Exerts Force on 6kg Chair to Prevent Slipping

[Shah 72]

A chair of mass 6kg is at rest on a rough horizontal floor with coefficient of friction 0.35. It is pulled horizontally by a force of 25N. A boy pushes down on the chair so that the chair is on the point of slipping but remains at rest. Find the force that the boy exerts on the chair.

 

[HOI]

The friction force, opposing the motion, is the coefficient of friction times the downward force In this case the downward force is the weight of the box, 6g N (g is the acceleration due to gravity, 9.81 meters per second squared), plus the downward force applied by the boy, Calling that force, F, that is a total downward force, 6g+ F Newtons so the friction force is 0.35(6g+ F) Newtons. Since the box " is on the point of slipping but remains at rest" that must be equal to the applied force 25 Newtons.

Solve 0.35(6g+ F)= 25 for F.

 

[Shah 72]

The friction force, opposing the motion, is the coefficient of friction times the downward force In this case the downward force is the weight of the box, 6g N (g is the acceleration due to gravity, 9.81 meters per second squared), plus the downward force applied by the boy, Calling that force, F, that is a total downward force, 6g+ F Newtons so the friction force is 0.35(6g+ F) Newtons. Since the box " is on the point of slipping but remains at rest" that must be equal to the applied force 25 Newtons.

Solve 0.35(6g+ F)= 25 for F.

Thank you so much!