# Thread: Friction problem help

1. ## Friction problem help

A box of bananas weighing 30N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.40, and the coefficient of kinetic friction is 0.20N.
a.) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on the box?
fs = (0.40)(30N)
fs = 12N?
b.) What is the magnitude of the friction force if a monkey applies a horizontal force of 7.0N to the box and the box is initially at rest?

7.0N?

c.) what minimum horizontal force must the monkey apply to start the box in motion?
12N?

d.) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity.
fk = (0.20)(30)
= 6N?

e.)if the monkey applies a horizontal force of 15N, what is the magnitude of the friction force....?
-

having trouble with these questions .. need help

2. Originally Posted by ^_^Engineer_Adam^_^
A box of bananas weighing 30N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.40, and the coefficient of kinetic friction is 0.20N.
a.) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on the box?
fs = (0.40)(30N)
fs = 12N?
First, the coefficient of friction is unitless, so $\mu_k = 0.20$, not 0.20 N.

You are over thinking this one. The static friction force is only as large as it needs to be to keep the object in place. So if there is no horizontal force applied, $f_s = 0~N$.

Originally Posted by ^_^Engineer_Adam^_^
b.) What is the magnitude of the friction force if a monkey applies a horizontal force of 7.0N to the box and the box is initially at rest?

7.0N?
The maximum possible static friction force between the box and the surface will be:
$f_{s, max} = \mu_sN = 0.4 \cdot (30~N) = 12~N$

Since the applied force of 7 N is less than this then, yes, the static friction force will be 7 N.

Originally Posted by ^_^Engineer_Adam^_^
c.) what minimum horizontal force must the monkey apply to start the box in motion?
12N?
Since the maximum possible static friction force is 12 N, then any applied (horizontal) force greater than this will make the box move. So yes.

Originally Posted by ^_^Engineer_Adam^_^
d.) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity.
fk = (0.20)(30)
= 6N?
Yes. Briefly, this comes from the fact that, once the box is moving, the sum of the forces in the horizontal direction must be 0 N to make the velocity constant.

Originally Posted by ^_^Engineer_Adam^_^
e.)if the monkey applies a horizontal force of 15N, what is the magnitude of the friction force....?
This applied force is greater than the static friction force of 12 N, so the box will be moving. Since it is moving the friction is kinetic, which has a constant value. So
$f_k = \mu_k N = 0.20 \cdot (30~N) = 6~N$

-Dan

3. ## thanks!

thank you topsquark