I need help on this physics question. I am not sure how to approach it.

Question

A student is trying to calculate the coefficent of friction between a box and a surface. She measures that the 80kg box will slide if the student pushes with a force greater than 408N.

Answer each of the following questions for the case in which the student is pushing as hard as possible without causing the box to slide.

How many forces are acting on the box?

What is the magnitude of the friction?

What is the magnitude of the normal force from the floor?

Which coefficient of friction can be determined in this case?

What is the value of this coefficient friction?

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(pretend that both the person and the box are on the floor)

Answer 1

Hi Emma,

Since we are considering a scenario where the student is pushing as hard as possible without causing the box to move, then we can assume that they are pushing with a force equal to but not greater than 408 N given what the problem tells us in the beginning.

There are 4 forces acting on the box:

1. The force of the student on the box (horizontally to the right, let's say)
2. The force of static friction acting in the opposite direction of the student's force on the box
3. why static and not kinetic? because the box is not moving in this scenario. If the student had pushed with more force and the box started moving, then we would be considering the kinetic friction, but remember this box is NOT moving!

c. The force of gravity on the box, pointing down

d. The force of the floor on the box (the "normal" force), pointing up

The magnitude of the static friction force must in this case be equal exactly to the force that the student is applying. This is because we know that the box is NOT moving, so the resultant net force in the horizontal direction must be zero. For it to be zero, the static friction force magnitude must equal to the magnitude of the force applied by the student.

The magnitude of the normal force must be equal exactly to the force of gravity for similar reasoning. The box is not moving in the vertical direction, so the net vertical force must be equal to zero. This can only happen if the normal force and the gravitational force are equal in magnitude.

l F_N l = l F_g l = m*g (mass of the box times acceleration due to gravity (g))

Since this is an example of static friction (see point i above), this scenario allows us to calculate the static coefficient of friction

we know that l F_frictionl = l F_applied l = 408 N

and we can remember that F_friction = μ_s * F_N

and recall l F_N l = m*g

so putting this together:

μ_s * F_N = 408 N

μ_s*m*g = 408 N

now you can solve this for μ_s, the coefficient of static friction