# How do I draw a free body diagram of a car on a slope?

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1. Jan 16, 2016

### Chozen Juan

1. The problem statement, all variables and given/known data
A car attempts to accelerate up a hill at an angle θ to the horizontal. The coefficient of static friction between the tires and the hill is µ > tan θ. What is the maximum acceleration the car can achieve (in the direction upwards along the hill)? Neglect the rotational inertia of the wheels.

2. Relevant equations
ΣF = ma
f_k = μN

3. The attempt at a solution
I first started by drawing a free-body diagram. I knew that there had to be a weight force on the car acting downward, so I drew it. The other obvious force is the normal force, which acts perpendicular to the surface. Now, I first thought that there should also be an external applied force on the car acting in the direction upwards along the hill. I also reasoned that in the direction of motion is upwards along the hill, then there should be a friction force in the direction directly opposite the driving force. I tried solving for a, the maximum acceleration of the car, but I couldn't because I didn't have enough information:

N-mgcos(θ) = 0 ⇒ N = mgcos(θ)

f_k = μN ⇒ f_k = μ(mgcos(θ))

(F_applied) - mgsin(θ) - μ(mgcos(θ)) = ma

a = ((F_applied) - mgsin(θ) - μ(mgcos(θ)))/(m)

The above equation for a is simply incorrect; even if I found out the value of (F_applied) or if I found out (F_applied) in terms of some other variables, the answer would be incorrect. Because of this, i suspected that perhaps the applied force of the engine is for some reason some internal force of the system that cancels out with another force to become zero. Then, I considered the possibility of friction acting in the direction of motion, opposite the direction of mgsin(θ). One reason why I thought kinetic friction would act in the direction of motion is that the rear tire has a tendency to rotate clockwise. I then tried solving the problem again with these new forces, and I got the correct answer:

N = mgcos(θ)
f_k = μ(mgcos(θ))

μ(mgcos(θ)) - mgsin(θ) = ma

a = (μ(mgcos(θ)) - mgsin(θ))/(m) ⇒ a = g(µ cos(θ) − sin(θ))

Although this is the correct answer, I still am not sure if my second approach to the problem is correct. If it is the correct attempt, I still don't conceptually understand why there is no applied force in the free body diagram. Also, I don't exactly fully understand why the friction is propelling the car forward; it would make more intuitive sense to me if friction were to be in the opposite direction of the car's motion.

How can I fully conceptually understand how the forces act on this system?

2. Jan 16, 2016

### TSny

Hello.

The second approach is correct. The acceleration of the car is due to external forces only. These are the forces "applied" to the car. What are the external forces acting on the car? You should be able to list these forces and state what objects external to the car create these forces.

If something tries to slide down a slope, what direction does friction act on the object?

3. Jan 16, 2016

### Chozen Juan

Ok, the external forces acting on the car would be Weight, Normal force, and friction force. The weight is the force of Earth on the car, the normal force is a force of the surface on the car, and the friction force is another force of the surface on the car. These are the only external forces, but I still don't understand why we don't consider the force of the engine on the car an external force.

Since the car has a tendency to slide down the ramp due to its weight, the friction force should be in the direction upwards along the hill.

But why is the force of the engine on the car not external?

Also, in my mind, it only makes sense to say that the friction acts upwards along the hill if the car is not moving upwards... but that is only if the car has zero applied force from the engine. I don't understand why the friction force would act in the same direction even if the car has a force from the engine propelling it upwards.

4. Jan 16, 2016

### SteamKing

Staff Emeritus
The engine of a car doesn't apply a direct force to the driving wheels; it supplies a torque, which is then transmitted by the axles to the driving wheels. The friction between the tires and the road surface permit this torque from the driving wheels to move the car without the wheels slipping on the road surface.

5. Jan 16, 2016

### TSny

The car is a complicated system of parts that move relative to one another. However, you are only interested in the motion of the center of mass of the car. Only external forces affect the motion of the center of mass of a system. By definition, external forces are forces due to objects outside the system.

However, the engine does affect the tendency of the wheels to slip on the ground. As the engine applies more torque to the wheels, there is a greater tendency for the wheels to slip which causes the static friction between the wheels and the ground to increase. So, the amount of external force of friction is affected by the internal forces.

It's similar to when you walk up a hill. If you stand on the slope of the hill, friction between your feet and the ground prevent you from sliding down the hill. As you start to walk up the hill, you push your foot so that you increase how much your foot pushes down the hill (parallel to the slope). The reaction force from the hill on your foot is up the hill.

The wheels of the car and your feet are acting similarly.

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Last edited: Jan 16, 2016