How Do You Calculate Frictional Force for Moving Boxes?

[vrobins1]

Homework Statement

Three identical boxes sit on concrete floor, each box weighs 30 lbs (133 N). Each has a coefficient of static friction of 0.74 and a coefficient of dynamic friction of 0.64. Horizontal force of 80 N is applied to Box 1, horizontal force of 100 N is applied to Box 2, and horizontal force of 120 N is applied to Box 3.

a.) Do any of the boxes move?
b.)Calculate the actual frictional force elicited between the ground and box in each case. If the boxes move, compute the dynamic friction that resists the sliding. If the boxes do not move, compute the static friction that resists the applied force.
Note: frictional force is not always the maximum.

Homework Equations

a.) F = ma
R_Normal- W = ma
R_N = W

Fmax = \mu_s (R_N)

b.) not sure

The Attempt at a Solution

a.) I used R_N = W to determine that R_N = 133 N

Then I used Fmax = \mu_s (R_N)
Fmax = 0.74 (133)
Fmax = 99 N, the force required to move the box

So Box 1 would not move, Boxes 2 and 3 would move.

b.) This is where I am most confused, I am not sure how to calculate the static and dynamic friction, but I tried anyways:
For Box 1: static friction = \mu_s (F_applied)
=0.74 (80 N)
=59.2 N

I did the same for Box 2 and Box 3, but I multiplied their applied forces times the coefficient of kinetic friction given, 0.64.

Box 2 = 64 N
Box 3 = 76.8 N

Can anyone offer any insight into what I did wrong, or if I did this correctly? Thank you.

 

[Doc Al]

a.) I used R_N = W to determine that R_N = 133 N

Then I used Fmax = \mu_s (R_N)
Fmax = 0.74 (133)
Fmax = 99 N, the force required to move the box

So Box 1 would not move, Boxes 2 and 3 would move.

Good.

b.) This is where I am most confused, I am not sure how to calculate the static and dynamic friction, but I tried anyways:
For Box 1: static friction = \mu_s (F_applied)

Hints:
Static friction ≤ μ_sR_N
Dynamic friction = μ_kR_N