Forces acting on a box

[fro]

Problem: I am pushing a box holding it stationary against a vertical wall. The box weighs 8N and I am pushing with a steady 40N horizontal force (into the wall). The coefficient of static friction between the wall and box is 0.2 and the coefficient of kinetic friction between the wall and box in 0.15.

Requirement: List all the forces in action and their magnitude.

I know that 40N is one of them (not sure what to call it), the normal force (which should also be 40N?), and gravitational force of 80N. I am confused as to how the two coefficients of friction would affect this problem. Any ideas?

[OlderDan]

Problem: I am pushing a box holding it stationary against a vertical wall. The box weighs 8N and I am pushing with a steady 40N horizontal force (into the wall). The coefficient of static friction between the wall and box is 0.2 and the coefficient of kinetic friction between the wall and box in 0.15.

Requirement: List all the forces in action and their magnitude.

I know that 40N is one of them (not sure what to call it), the normal force (which should also be 40N?), and gravitational force of 80N. I am confused as to how the two coefficients of friction would affect this problem. Any ideas?
You appear to have a typo on the 8N/80N weight of the box. One of them is incorrect. If the box is not sliding the coefficient of static friction tells the maximum possible frictional force. If it is sliding, the coefficient of kinetic friction tells you the actual frictionl force.

[fro]

You appear to have a typo on the 8N/80N weight of the box. One of them is incorrect. If the box is not sliding the coefficient of static friction tells the maximum possible frictional force. If it is sliding, the coefficient of kinetic friction tells you the actual frictionl force.

Thanks for the hint. So, the gravitational force should be 8N since that is how much the box weighs. Force applied (by me) is 40N which should be counteracted by the normal force of 40N. Then, the frictional force should be 40N (normal force) X 0.2 (coefficient of static friction) = 8N - that would keep it stationary. The direction of the frictional force should be opposite to gravitational force. Am I right? Thanks for your help.

[OlderDan]

Thanks for the hint. So, the gravitational force should be 8N since that is how much the box weighs. Force applied (by me) is 40N which should be counteracted by the normal force of 40N. Then, the frictional force should be 40N (normal force) X 0.2 (coefficient of static friction) = 8N - that would keep it stationary. The direction of the frictional force should be opposite to gravitational force. Am I right? Thanks for your help.
You've got it. Note that the 40N is just barely enough to keep the box from sliding. If you reduced that force, the friction force would be reduced and the box would begin to slide. Since the kinetic friction is less than the maximum static friction, the box would accelerate to the floor.

[fro]

If I stopped pushing the box into the wall, should not the applied force, frictional force and the normal force go away? Then only the gravitational force should be acting on the box. I am assuming that frictional force will go away since nothing is pushing the box to the wall anymore. Am I thinking this correctly?

[OlderDan]

If I stopped pushing the box into the wall, should not the applied force and the normal force go away? Then only the gravitational force and frictional force should be acting on the box. If so, without knowing the normal force how would I then calculate the magnitude of frictional force?
If you completely stopped pushing, the normal force would be zero and the frictional force would be zero. If you just slightly reduced the pushing force to say 39.99N, the normal force would drop to 39.99N and the maximum static friction would no longer be sufficient to oppose the weight of the box, so it would begin to slide. The friction force would suddenly drop to 6N because of the differenece in the friction coefficients and the box would accelerate to the floor with a net force of 2N acting downward.

[fro]

If you completely stopped pushing, the normal force would be zero and the frictional force would be zero. If you just slightly reduced the pushing force to say 39.99N, the normal force would drop to 39.99N and the maximum static friction would no longer be sufficient to oppose the weight of the box, so it would begin to slide. The friction force would suddenly drop to 6N because of the differenece in the friction coefficients and the box would accelerate to the floor with a net force of 2N acting downward.

So you are saying that if I completely stopped pushing the box, then only gravitational force of 8N would be acting on the box? So there would be no external forces acting on the box at all?

[OlderDan]

So you are saying that if I completely stopped pushing the box, then only gravitational force of 8N would be acting on the box? So there would be no external forces acting on the box at all?
Gravity is an external force. But yes, that is the only force that would be acting if you stopped pushing.