Question

A 2,200-kg car is moving down a road with a slope (grade) of 14% at a constant speed of 11 m/s. What is the direction and magnitude of the frictional force?
(define positive in the forward direction, i.e., down the slope)?

A 2,282-kg car is moving up a road with a slope (grade) of 10% at a constant speed of 12 m/s. What is the direction and magnitude of the frictional force?
(define positive in the forward direction, i.e., up the slope)?

Answer 1

Free body diagram is made and the sum of force are made in the direction of interest, ie the x axis.

The assumption that the car is rising, as shown in the diagram of free body, therefore the positive sign in profit caluclo of the friction force would indicate is that the direction of the force is supposed correctly, if the sign is negative it indicates that the car is going down and the direction of the force of serious friction in the diagram opposite direction.

Where:

Using the free-body diagram weight decomposes in both directions depending on the angle

The equation for calculating the friction force is:

Case 1:

m= 2200 kg

g= 9.81 m/s2

=14 grade.

Case 2

m= 2282 kg

g= 9.81 m/s2

=10 grade.

Taking into account the signs, so the comment at the beginning of the resolution of ejercios should be applied. The direction of the frictional force is contrary to the alleged in the free-body diagram.