Force needed to move a box.

[wcbryant87]

Homework Statement

The weight of the crate is 875 N. You push it with a force of magnitude 300 N but it doesn't move

What is the magnitude F_p of the minimum force you need to exert on the crate to make it start sliding along the floor? Let the coefficent of the static friction $$\mu$$_s betwen the crate and the floor be 0.56 and that of kinetic friction, $$\mu$$_k, be 0.47

Homework Equations

The Attempt at a Solution

I tried solving it by using the formula f_k = $$\mu$$_kn but it told me that I'm going about the problem wrong. It says that I need to study the condition for which the crate ceases to be at rest, that is, the condition for which static friction ceases to be valid. I used kinetic friction though, so I don't know what to do now.

 

[Delphi51]

Use the static coefficient of friction in your formula.
Once it has started moving, the lower dynamic coefficient comes into play and a smaller force will suffice to keep it moving.

 

[nlightner]

I would suggest approaching this problem by first understanding the concept of static and kinetic friction. Static friction is the force that prevents an object from moving when a force is applied to it, while kinetic friction is the force that opposes the motion of an object that is already in motion. In this case, the force you are exerting on the crate (300 N) is not enough to overcome the static friction between the crate and the floor, which is why it is not moving.

To determine the minimum force needed to make the crate start sliding, we need to consider the maximum value of static friction, which is given by the equation fs = \musn. In this case, the coefficient of static friction is 0.56, and the normal force (n) is equal to the weight of the crate (875 N). Therefore, the maximum static friction that can be exerted on the crate is:

fs = (0.56)(875 N) = 490 N

To make the crate start sliding, we need to apply a force greater than 490 N. This minimum force can be calculated by subtracting the maximum static friction (490 N) from the force you are already exerting (300 N):

Fp = 300 N - 490 N = 790 N

Therefore, to make the crate start sliding along the floor, you would need to exert a minimum force of 790 N. This is because the force you are exerting (300 N) is not enough to overcome the maximum static friction (490 N) between the crate and the floor. Once the crate starts moving, the force needed to keep it in motion will be less, as it will then be opposed by the kinetic friction, which is given by the equation fk = \mukn. The coefficient of kinetic friction (0.47) is less than the coefficient of static friction (0.56), which means that once the crate is in motion, it will require less force to keep it moving.