An engine is pushed on the floor. The engine weighs 32 lbs and the angle between
ID: 1329893 • Letter: A
Question
An engine is pushed on the floor. The engine weighs 32 lbs and the angle between it and the floor is 20°. The coeffiecient of static friction is 0.50 and the coefficient of kinetic friction is 0.45.
What is the minimum pushing force required to move the block?
The block is moving and is still being pushed with the force found in part A. What is the magnitude of the friction force?
The block is moving and is still being pushed with the force found in part A. What is the magnitude of the acceleration of the block?
An engine is pushed on the floor. The engine weighs 32 lbs and the angle between it and the floor is 20°. The coeffiecient of static friction is 0.50 and the coefficient of kinetic friction is 0.45.
Explanation / Answer
The angle 20 degree can either be the angle the force makes with the floor or the angle of incline on which mass is moved. I present both the solution here so read the question details and see which applies
Case 1 [Assumed that angle 20 degree is the angle of force with respect to floor.]
m = 32 lb = 14.515 Kg
normal force = mg - F sin 20
firctional force = 0.5 * normal = 0.5 (mg -F sin20)
To start movement: 0.5(mg - F sin 20) = F cos 20
=> F = 64.03 N
So minimum pushing force to move block = 64.03 N
frictional force once block starts moving = 0.45 (mg - F sin 20) =54.16 N
Force in direction of movement = F cos 20 = 60.17 N
so acceleration = (60.17 - 54.16) / 14.515 = 0.414 m/s2
Case 2 Assumed that the angle of incline is 20 degree.
Normal force = mg cos 20 = 133.67 N
firctional force = 133.67*0.5 = 66.83 N
Component of mass down the incline = mg sin 20 = 48.65 N
So total force required to move the block = 66.83+48.65 = 115.48 N
Once it starts moving, firctional force = 133.67*0.45 = 60.15 N
Net force in movement direction = 115.48 - 60.15 - 48.65 = 6.68 N
acceleration = 6.68 /14.515 = 0.46 m/s2