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I am in a study group for my AP Physics class. We are studying for exams which are in a couple weeks. A part of that is we go over a ton of problems, situations and concepts and do them together as a group. However, when we try to do this difficult problem, we get 2-3 different answers and disagree on who is correct. We were hoping you could show us the correct answer & your steps. Complete answers including numbers would be ideal, but at the least your step-by-step would be okay.

The 2-part question is:
A bag of groceries (with a mass m) sits on the passenger seat of a car going around a left-hand curve, (with a radius r). The coefficient of static friction between the bag and the seat is represented by u.
1. Create an equation using the given variables (plus other standard variables) for maximum speed (shown by Vmax) the car can go around the curve without the bag sliding around the seat.
2. If the car goes 1.5 Vmax around the curve, the bag slides and hits the passenger door. What is the magnitude of the normal force exerted by the door on the bag in terms of the given variables and any other standard variable?

Thank you very much, we greatly appreciate you helping us study!

For this problem, we need to use the result that the acceleration of an object (particle) in uniform circular motion is

a_c = v^2/r

This acceleration is directed inward toward the center of the circle describing the turn (to the left for your example). Now, the bag of groceries has an inertia which tends to make it keep going straight. The bag thus must have a force to "push" it around the turn and remain stationary on the seat, in the reference frame of the moving/turning car. The magnitude of the lateral force required is obtained from the centripetal acceleration by multiplying by the mass of the bag  (F = ma), so

F_lat = m･v^2/r

In order for the seat to drag the bag with it, a frictional force is needed to "attach" them. The amount of friction needed to keep the bag from slipping horizontally away is provided by the vertical force of gravity, which in this situation is called a normal force. This normal force is

F_norm = mg

where g is the acceleration of gravity. The horizontal frictional force needed to counter the centripetal force defined above is

F_fric = μmg

where μ = coefficient of friction (unit less).

Equating these 2 forces for a velocity Vmax gives

m･Vmax^2/r = μmg

or, with a little algebra

Vmax = sqrt(μrg)

which is the answer for part 1.

For part 2, I interpret it to mean that the bag has slid over to the door and is "resting" against it. The door now has to exert a force on the bag to accelerate it around the curve. It is also called a normal force, but it is different from the one described above; it is normal in the sense that it is a force perpendicular to the surfaces of the bag and door that are in contact and is also directed towards the center of the turn.

The bag has slid to the door because now the centripetal force exceeds the frictional force calculated above due to the higher speed. We can use the same formula for centripetal force to calculate its magnitude,

F_door = m･(1.5V･max)^2/r.

Hope this helps.

Questioner's Rating
 Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment A fast, amazing answer; highly concise and descriptive. This is a truly perfect answer, I could not have asked for more; in fact, this reply went above my expectations. Thank you so much, you have helped us HUGELY!

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#### randy patton

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