Find the coefficients of static and kinetic friction between crate and floor

Question

 A dlock worker loading crates on a ship finds that a 20 kg crate, initially at rest on a horizontal surface requires a 75-N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 60 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor

Answer 1

Hi Amelia!

 

One of the keys to this problem is the definition of "set in motion."  When trying to move an object against static friction, we usually define "set in motion" as the point where the force trying to move the object (the actuating force) equals, in magnitude, the maximum possible force that static friction can generate.

 

If the object is on a horizontal surface, then it can only move horizontally.  Since friction can only act opposite the direction of motion (or attempted motion), it must be in the horizontal direction.  If the actuating force is also horizontal, then we can set up Newton's 2nd law in the horizontal direction:

 

F_net-h = F_p - F_s-max = ma = 0

 

where

 

F_p = actuating force

F_s-max = maximum possible force of static friction

 

These two must add up to zero if F_p equals F_s-max  in magnitude and is opposite in direction.  We can put in the expression for the maximum static friction force, and rewrite the Newton's 2nd law equation above as:

 

F_p - μ_sN = 0  --> F_p = μ_sN --> μ_s = F_p/N

 

where

 

μ_s = coefficient of static friction

N = normal force on object

 

We are told the actuating force, F_p, is 75 N to set the crate in motion.  If we know the normal force, then we are set.

 

In the vertical direction, without any other information, we can assume the only forces acting are the normal force and gravity, meaning  

 

F_net-v = N - mg = 0   (taking down to be negative)

 

There add up to zero because the crate is stationary (in static equilibrium) vertically.  Thus, N = mg (the weight), which can be immediately calculated.  Then you have the normal force, N, and can solve for μ_s by the expression above.

 

For the coefficient of kinetic friction, it is the same procedure, except you swap in the force of kinetic friction for the maximum force of static friction in the horizontal.

 

F_net-h = F_p' - F_k = ma = 0

 

These two must sum to zero because they are the only horizontal forces in play, and the crate is moving at constant speed, so the acceleration must be zero,

 

Put in F_k = μ_kN, with N being the same value as before (because the crate is still at rest vertically, even though it is moving horizontally), and you can solve for μ_k in exactly the same way as for μ_s above.

 

I hope this helps some, but let me know if you have any further questions or problems.