Forces and Friction Sample Problem Answers

1. A 10kg mass is resting on a flat surface, with a coefficient of static friction of 0.35. What force would it take to move the object?

FS = 34.418N

For this problem, since the normal force is equal to gravity, we need to find the force of gravity and multiply it by the coefficient of static friction.

FG = g*(5.98e24kg)*(10kg)/(6.37e6)^2
FG = FN = 98.337N
FS = 0.35 * 98.337N
FS = 34.418N

2. A 1500kg delivery truck is traveling at a constant speed up a 25° hill with a continual pull from its wheels of 900N.
a. What is the force of kinetic friction acting on the truck?
b. What is the coefficient of kinetic friction between the truck and the road?

a. 266.0726N
b. 0.1957

First, we need to break up gravity into its respective parts. Then, knowing that gravity and the force of friction need to balance the 900N pull, we simply subtract the x component of gravity from 900N to find the friction force.

FGX = 1500kg sin 25° = 633.9274N
FGY = 1500kg cos 25° = 1359.4617N
FK = 900N - 633.9274N = 266.0726N

Now, for part b, all we need to do is plug in the force of kinetic friction and the normal force, which is equal to the y component of gravity, and solve for the coefficient of kinetic friction.

266.0726N = * 1359.4617N
= 0.1957