Two objects are stacked on each other with a nonzero coefficient of static friction between them, and the bottom block sits on a frictionless table. A force is applied rightward to the bottom block, and the right block moves along with it due to the static friction they exert on each other. My question is how do you know the magnitude of this static friction? Static friction is equal to the coefficient of friction multiplied by the normal force, but what normal force are we talking about? The normal force applied by the table on the bottom object? The normal force applied by the bottom force on the top block, even when it’s the bottom block that’s being pushed by the external force? Does the reaction force applied by the top block on the bottom block count as a “normal force”?
2 Answers
The only friction in the problem is between the top and bottom block. There is no friction between the table and the bottom block, so the normal force at that interface is irrelevant.
The friction between the top and bottom block depends on the magnitude of the normal force at that interface. The force of the top block on the bottom block and the force of the bottom block on the top are equal and opposite by Newton's third law. The magnitude of the force is the same regardless of which one you choose to call a reaction force.
Which block is being pushed is also irrelevant. The issue at hand is the contact between the two blocks, not the contact between a block and some other object that's pushing it.
Keep this thing in your mind while solving friction problems.
Friction always wants to keep those two blocks to move together, no matter how. If it fails, slippng will occur.
Look, Maximum Friction is always equal to $\mu$$N$. Here, $\mu$ must be given to you. N i.e the Normal reaction is always the force which the "floor" applies on the block. Here floor is the block on which it is kept. Look, static friction will always be less than or equal to $\mu$$N$. That is if you use F=ma, and use $\mu$$N$ in place of F, you will get a value of a. This will be a max. Because there cannot be a greater force than $\mu$$N$ and this force can provide an acceleration of a-max. First, check what is the acceleration of the lower block. That obviously depends on the force you apply on the lower block.
Here, use F= ($m_1$+$m_2$)A . Here F is the force that you apply on the lower block of mass $m_2$ . The condition for them to move together is that A< or equal to a max. Because as A will increase the friction force will also increase until it reaches $\mu$$N$ and if A increases further, the friction force will remain the same and it will cause a < A and hence slipping would occur.
For your problem, first of all, find $\mu$$N$. Then use $\mu$$N$ = $m_1$.a max. Find a max and then use the equation F= ($m_1$+$m_2$)A . See if A is smaller than, equal to or greater than $a_1$. If it is smaller than a, then it means, friction will make the above box move, and you simply use F ( friction ) = $m_1$$A$ . This will give you the friction on the block. However, if a max = A or A > a max, then Friction will be limiting, it will cause slipping and its value will simply be $\mu$ $N$!
If you have any confusion regarding my answer, feel free to comment!
