Calculating frictional force?

[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)

:yuck:

Hints:
Static friction ≤ μ_sR_N
Dynamic friction = μ_kR_N

 

[kerimek]

Your approach to part a) is correct. For part b), you are on the right track but there are a few things to consider. First, when calculating the static friction, you need to use the maximum possible value, which is given by Fmax = \mus (RN). In this case, for Box 1, the maximum static friction is 99 N, not 59.2 N. For Boxes 2 and 3, the maximum static friction would also be 99 N, since the normal force is the same for all three boxes.

Next, for the dynamic friction, you need to use the actual applied force (not the maximum) and the coefficient of dynamic friction. So for Box 2, the dynamic friction would be 0.64 (100 N) = 64 N. For Box 3, it would be 0.64 (120 N) = 76.8 N.

Overall, your calculations were mostly correct, but just make sure to use the maximum static friction and the actual applied force for the dynamic friction. Also, remember that the maximum static friction is not always the same as the actual static friction, so it is important to use the correct values in your calculations. Keep up the good work!