# PhysicsMechanics- friction

#### Shah 72

##### Member
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.

#### Country Boy

##### Well-known member
MHB Math Helper
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.

Last edited:

#### Shah 72

##### Member
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!!!