GS-1 Physical Sciences Question 34

[borgcollec2919]

I don't understand. The answer explanation says that the maximum frictional force decreases, but the question only asks for the actual frictional force on the car, which I think should vary according to the angle of inclination:

From the previous question, the car exerts a force of 300N, and the maximum frictional force for the car in level ground is 4000N.

Now when the car goes on an incline, the maximum frictional force will be less than 4000N. Depending on the angle of inclination it may or may not fall below 300N. If it doesn't fall below 300N, then the frictional force on the car will still be 300N.[/i]

[mcat_premed3832]

There are several issues:

To begin with, you must realize that the equation for friction (mew times the force normal) refers to maximal frictional force. Also, it is important to recognize that when we are referring to a moving object then the frictional force, whether or not it is stated, IS the maximal kinetic frictional force.

Thus the actual frictional force on the car IS the maximal frictional force and cannot be otherwise (at least not in MCAT/introductory level physics!).

The 300 N in the previous problem does not refer to a moving object (static friction). And, thus the 300 N was shown to be far below the maximal frictional force and so the number 300 is no longer relevant in this problem where the vehicle is actually moving.

You are correct to say that the frictional force will decrease according to the angle of inclination. If there is no inclination, then it would be 4000 N but as it goes from 1 degree to say, 45 degrees, the frictional force (and thus the normal force) progressively decreases.

[borgcollec2919]

I am not sure I quite get it. The previous question says that the car "exerts" a force of 300N, so I assumed that that force was exerted by the car, on the ground, in a backwards direction.

(i.e. the wheels of the car exerts a force of 300N backwards at its point of contact with the ground. When that force is below the maximal frictional force, the ground supplies an opposite 300N frictional force, which propels the car forward. If the force exerted by the car exceeds the maximal frictional force, then the car wheels will begin to slip and the frictional force becomes a kinetic frictional force.)

But if I understand you correctly the 300N force was not exerted by the car but actually on the car in a forward direction so that it is modelled as a single object moved by an external force forward, and the frictional force is actually opposing its movement (as if the car was kept "slipping" forward)?

[curious7133088]

I also seem to think there is a problem with this question. "mcat_premed3822" says the frictional force IS the maximal kinetic frictional force. What? The tires are rotating and the contact surface of the tire generally should NOT be sliding, thus be experiencing a STATIC frictional force. Yes, as everyone seems to agree, the maximal static frictional force will be less on an incline due to a decreasing Normal force, but this is only relevant if it is exceeded.

On a level surface, the forward force the engine has to provide is simply to overcome air friction slowing the car. Going up an incline, assuming we want to maintain a constant velocity (no acceleration, net force zero), the engine has to provide additional force to counteract the X component of the gravitational force acting to decelerate the car (you have to give the car more gas when going up hill or you will decelerate!). Since the force required to propel the car forward is increased, the STATIC frictional force between the tires and the road must increase.

[robertbai5188]

I think this question is very flawed in that the question is unclear to the readers. The official explanation seem to be flawed as well. Why are we talking about the maximal frictional force here? The only reason we would talk about static maximal frictional force would be if the car is starting to slide, which is completely irrelevant to the question.

Being under the impression that the car is still exerting 300 N (from the previous question), wouldn't it make sense that the static friction force is 300 N as well (due to Newton's third law) and consequently make choice A (No Change) correct?

[jellywing_2058]

Frictional Force on the car depends on two factors;
- coefficient of kinetic friction between the two surfaces which depends on the nature of the surfaces AND
- the normal reaction on the car exerted by the surface

On level ground, Force of friction = μ N
(μ = Coefficient Of Kinetic Friction) and because the Normal reaction balances the Weight, N = M g
Therefore Force of Friction = μ N = μ M g

On the slope uphill, the Normal, which is always normal or perpendicular to the surface in contact is now at an angle and because it balances the Weight component normal to the surface, will be equal to
N = M g cos θ
And,
Force of Friction = μ N = μ M g cos θ

And, because cos θ < 1,
Force of Friction = μ M g cos θ < μ M g
Correct Answer: C