Park the Car in Harvard Yard

Well, not in the Yard actually, but on a hill nearby.

We have a car parked on a slope. Assuming that the car is symmetric and all four wheels have the same size and assuming that the friction on rolling is negligible, what is the condition that either the back brakes or the front brakes should be activated to stop the car rolling down the hill.

(We are assuming that only one set of brakes may be activated, the front brakes are those further up the hill)
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Author Commented:
We take note of the symmetry in the puzzle and reduce the problem to one in two dimensions.
The car rests on an incline with the (now) two wheels in contact with the rough surface at A and B. We draw a perpendicular through the center of gravity, G at some point above the road, and let it cut the road (slope) at C, which may be somewhere between A and B or elsewhere.

At A and B there will be a reaction, perpendicular to the slope, due to the weight of the car in that direction, which will cause a force through friction perpendicular to it, ie: along the road, in a direction which will stop the car from moving. In fact this frictional force must match the weight of the car resolved in that direction (action and reaction being equal and opposite).

Let these two forces be R and S at A and B.

Now the force will be that of a braked friction or a rolling friction dependant on whether the wheel is braked or not. We have been told to ignore rolling friction, so the weight of the car in the direction of the slope must match the braked friction at A or B, ie: µR or µS.

Since the car is in equilibrium we can take moments about the point C (which eliminates the weight of the car) and we must have - for balances forces - R.AC = S.CB otherwise the car would turn off the slope. Since the friction µR or µS matches the resolved weight of W*sin(theta) where W is the weight and theta the angle of inclination, the greatest possible value for theta for a given µ corresponds to the greater value of R or S, which means that the wheels at A or B should be locked according as AC or CB is the smaller. This is the required condition.

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RetiredCommented:
Uphill brakes (front) - if not, the uphill wheels could more easily move and swing the car around. using the braked downhill wheels as a pivot.
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Commented:
The downhill wheels will have much more weight transfer hence a better coefficient of friction with the road.
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Commented:
Oops, last post, coefficient of friction doesn't change, adhesion changes.
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Author Commented:
There is a condition as to which brakes should be effected, and it is that condition and proof thereof which gains the cheeses for this question!
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Commented:
Assuming that there is to the left of the parking spot a sidewalk or a small hill of grass, then parking on the left side of the road should be done with the front wheels pointing to the right, so that the back of the front left tire is pressed against the sidewalk/hill.

In order for the car to roll downhill, the back of the front left tire will have to either rise increasing its PE significantly. There would have to be a sufficient tangential force of the car down the road to provide the work necessary to increase the car's PE.

That alone with no brakes may be sufficient to keep the car in place. But by breaking the rear wheels, the static friction is used to negate some of the tangential force pointing down the road, making it less likely that the car can increase its PE.
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Author Commented:
>>There would have to be a sufficient tangential force of the car down the road to provide the work necessary to increase the car's PE.

It is high time you took a course in elementary Statics! Neither PE nor Work need come into consideration.
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Commented:
If the slope of the road is beta, then the gravity force pulling the car down the road is F = mg sin(beta). By angling the wheels at, say 45 degrees, as the car moves down slightly, the front wheels press against the sidewalk and the friction results in a force along the 45 degree wheel. The upward component of this force (that happens to equal the normal force of the sidewalk against the car- but that is not relevant) will cancel out the downward gravity force.
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Commented:
If there's a little static friction, Us, then that friction provides an upward force of mg cos(beta) Us.
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Author Commented:
What has the "sidewalk" got to do with it? The car is simply parked on a slope. The direction of the vehicle is immaterial, but since one needs to identify the brakes, let us say that the front is up the hill, the rear is down the hill.

Now identify the forces involved and write down the equations governing them.
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Commented:
This link has all the equations to determine whether a block on an inclined plane has enough frictional force to overcome the gravitational pull down the slope.   http://farside.ph.utexas.edu/teaching/301/lectures/node49.htmlBut, how that relates to a condition that determines whether the front or back brakes should be applied is not clear to me. I like the reason given in the first post; so I choose to use the break the front tires.
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Author Commented:
Now that is a really useful link, I couldn't have put it better. The last diagram is particularly useful in solving problems.

Now paulsauve's posting is not very clear in what he means exactly. My question however has three tag words, the first two have been "used" in solving the puzzle. It's now the turn of the third one!

Since I'm going on holiday on Saturday, I shall be closing this question tomorrow afternoon (European time)
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Commented:
The figure below is from the link  http://farside.ph.utexas.edu/teaching/301/lectures/node49.html

If µ is the coefficient of static friction, then when the slope exceeds some angle, theta, then the car will slide down the hill even if braked. In the figure, the gravitational pull down the hill is
f_g = -m g sin(theta)
and the force due to static friction is f = µ f_normal = µ m g cos(theta).

The car will slide down the hill if
f + f_g < 0 ==> f < | f_g | ==> µ m g cos(theta) < m g sin(theta)
==> µ cos(theta) <  sin(theta)
==> µ  <  tan(theta)
So, if theta > arctan(µ), then static friction is not sufficient to keep car from sliding.
Car-parked-up-a-hill.GIF
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Commented:
>> friction on rolling is negligible
In previous comment, µ, the coefficient of static friction, was the friction associated with translational motion. Given that rolling friction is negligible, then stability is an issue.

Suppose the back wheels have the brake on. If the car is parked so that the tires and car are parallel to f_g, then there are no rotational force components which can cause the car to rotate around the back tires. But this is an unstable configuration as the slightest force on the car may cause it to move from this parallel alignment (say a small gust of wind) causing the car to have a small angle ß ~ 0 with the force, f_g. The figure below shows a vertical downward force that represents f_g. The diagonal line represents one of the four tires which are assumed to be parallel to the longitudinal axis of the car. This diagonal line makes a ß degree angle with f_g.

/|
/ |
/ß|
/   |
/    |
|
v  f_g

f_g cos ß is the f_g force component distributed amongst the four tires and in the direction of the longitudinal axis of the car.

f_g sin ß ~ (ß * f_g) is the f_g force component distributed amongst the four tires in a direction perpendicular to the four tires. (For small ß, sin ß ~ ß.)

If the back wheels are braked (I really did not mean to break any tires in the previous posts), then consider these wheels to be a pivot point.

Let I = moment of inertia with respect to the back right tire (i.e., the tire closest to the bottom of the hill). Of the four perpendicular forces on the tires, the torque on this bottom left tire can be ignored since the distance from this pivot point to the perpendicular force is 0.

For the other three tires, the torque is as follows:
Torque = § (r × F) where § means sum over the three tires; and r is the distance from the back right tire to the i_th tire; and F is 1/4th of the total perpendicular forces on the tires, (ß * f_g).

Although negligible for a small ß, we still have
Torque = I * alpha, where alpha is the angular acceleration
alpha = § (r × F) / I

This small alpha indicates that the car starts a slow angular acceleration, thereby increasing ß, which causes an even greater net torque. ß eventually becomes large enough so that the perpendicular force must be represented as (f_g sin ß).

The car's angular acceleration results in the car rotating around the pivot point. As its angular velocity increases, the assumption that the pivot point can be maintained due to frictional forces becomes invalid, since the pivot point must maintain the centripetal acceleration of the car as it rotates. The faster the angular velocity, the greater is the necessary centripetal acceleration, a_c. When f_c = m a_c > static friction force, then the car will slide down the hill. Intuitively this slippage will occur well before the car makes a full 180 degree rotation.
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Commented:
Consider the same ß configuration as above, but now the car is braked in the front meaning that the pivot point is now the front left tire (i.e., the tire furthest from the bottom of the hill).

If ß is very large, then the torques are likely significant resulting in a large angular acceleration. As long as the pivot point holds, the car is behaving as a pendulum with ever increasing angular velocity with a maximum when the angle is zero degrees. It is possible that this results in a force on the pivot point which exceeds the static friction force, in which case the car will slide.

More likely, the driver parks the car fairly vertical (i.e., small ß). Again the car acts like a pendulum over a small angle. If there are no rotational friction forces, the car oscillates forever switching pivot points as beta passes zero and now the front right tire becomes the new pivot point.
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Author Commented:
From ID: 33539979

So, if theta > arctan(µ), then static friction is not sufficient to keep car from sliding.

Very good. In that the theta is called the "angle of friction".

The moment of inertia only needs to be taken into consideration when one takes moments about an axis where the inertia counts. Since that is dependant on the car's configuration and since you don't know that, it cannot, and does not, enter into the configuration.

I am going on holiday for three weeks as of tomorrow, so I must close my open questions. I shall publish the answer to this question next and accept it as a solution. If you think that is unfair object, I abide by any mnoderators decision.
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