Friction-related question

  • Thread starter legking
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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 [tex]2.5_{m/s^2}[/tex], 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!
 
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  • #2
Doc Al
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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 [itex]\mu N[/itex]--[itex]\mu N[/itex] is the maximum value of static friction. But, since you are trying to find the minimum value of [itex]\mu[/itex], 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! :wink: ) Note that you--correctly--didn't list that applied force when you described your free body diagram for the smaller box.
 

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