What Is the Smallest Coefficient of Friction to Prevent Slippage Between Boxes?

[legking]

Just as a bit of background: I'm taking a Grade 12 Uni prep correspondence course, and not only do the notebooks not explain the subjects NEARLY well enough, my teacher lives an hour away and I'd be lucky to see her once a week.

That being said, here's the question:

A small box is resting on a larger box, which in turn sits on a table. When a horizontal force is applied to the larger box, both boxes accelerate together. The small box does not slip on the larger box.

a) Draw a free body diagram of the small box as it accelerates.

b) What force causes the small box to accelerate?

c) If the acceleration of the pair of boxes has a magnitude of $$2.5_{m/s^2}$$, determine the smallest coefficient of friction between the boxes that will prevent slippage.

My answers:

a) My FBD includes 3 forces: gravity (acting downwards on the box), normal force (acting opposite to gravity, perpendicular to the horizontal plane), and friction (acting in the same direction as the applied force on the larger box).

b) Friction, specifically static friction.

c)I'm not quite sure how to proceed from here. To prevent slippage, the force of static friction must be at least equal to the applied force. Therefore,

F(F)=F(A)
mu(S)F(N)=ma
mu(S)mg=ma
mu(S)=ma/mg
mu(S)=a/g
mu(S)=(2.5m/ss)/(9.8m/ss)
mu(S)=0.26

Does this look right? Sorry about the scripting - I'm learning Latex, but I can't tell how I'm using it from the "preview post" feature. I'm just wondering whether I can even include an applied force in the equation since it's being applied to the larger box, not the small box. Any help with this question, or any other questions I will undoubtedly have over the coming months, would be greatly appreciated!

 

[Doc Al]

My answers:

a) My FBD includes 3 forces: gravity (acting downwards on the box), normal force (acting opposite to gravity, perpendicular to the horizontal plane), and friction (acting in the same direction as the applied force on the larger box).

b) Friction, specifically static friction.

Perfect.

c)I'm not quite sure how to proceed from here. To prevent slippage, the force of static friction must at least equal to the applied force. Therefore,

F(F)=F(A)
mu(S)F(N)=ma
mu(S)mg=ma
mu(S)=ma/mg
mu(S)=a/g
mu(S)=(2.5m/ss)/(9.8m/ss)
mu(S)=0.26

Careful with your terminology. You said "applied force", which might be taken to mean the force that was applied to the bottom box. But what you actually did was correct: You recognized that static friction was the net force and set that equal to "ma" per Newton's 2nd law (where "m" is the mass of the smaller box).

Another thing to be careful about. In general, static friction does not equal $\mu N$--$\mu N$ is the maximum value of static friction. But, since you are trying to find the minimum value of $\mu$, you want to set static friction to its maximum value.

With that understanding, your answer is correct.

I'm just wondering whether I can even include an applied force in the equation since it's being applied to the larger box, not the small box.

Ah... now you're thinking. The applied force does not act on the small box, so you can't include it. (But you didn't really include it! ) Note that you--correctly--didn't list that applied force when you described your free body diagram for the smaller box.

 

[m2003]

Your answers are mostly correct. Here are some additional explanations:

a) Your FBD is correct, but it is important to label the forces and show their directions. The gravity force should be labeled as "mg", the normal force as "N", and the friction force as "Ff". The direction of the friction force should be opposite to the direction of the applied force on the larger box.

b) Yes, the force causing the small box to accelerate is friction. More specifically, it is the static friction force between the two boxes.

c) Your approach is correct, but there is a small mistake in your calculation. The coefficient of static friction should be equal to the maximum possible value of static friction, which is equal to the applied force divided by the normal force. So the correct equation is:

mu(S) = F(A)/N

Substituting the values, we get:

mu(S) = (2.5m/s^2 * m)/(9.8m/s^2 * kg)

mu(S) = 0.255

So the smallest coefficient of friction that will prevent slippage is 0.255. Your answer of 0.26 is very close, so good job! Just remember to be careful with units and make sure they cancel out correctly in your calculations.

As for your question about including the applied force in the equation, you can think of it this way: the applied force causes the larger box to accelerate, and because the smaller box is resting on the larger box, it also experiences the same acceleration. So the applied force indirectly causes the smaller box to accelerate through the larger box. Therefore, it is correct to include the applied force in the equation.

I hope this helps and good luck with your course! Remember to always label your forces and use correct units in your calculations. If you have any other questions, feel free to ask.