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Homework Help: Frictional Forces, Static Friction

  1. Aug 12, 2010 #1
    1. The problem statement, all variables and given/known data
    Block A, with mass mA, is initially at rest on a horizontal floor. Block B, with mass mB, is
    initially at rest on the horizontal top surface of A. The coefficient of static friction between the two blocks is μs. Block A is pulled with a horizontal force. It begins to slide out from under B if the force is greater than:
    A. mA*g
    B. mB*g
    C. μs*mA*g
    D. μs*mB*g
    E. μs(mA +mB)g
    Ans: E

    2. Relevant equations
    Ff(static, max) = μs*NormalForce;


    3. The attempt at a solution
    So the answer is already given as E, but I really want to understand this.

    I'm confused on a couple of things. So if we pull on A too hard, then B will fall out behind it. Does this mean that the force of static friction is acting in the direction of the applied force F? And the normal force acting on mass B is just mB*g correct?

    That's all I got. I'm a little stuck from there. Any help would be great. Thanks a lot ahead of time.
     
  2. jcsd
  3. Aug 12, 2010 #2

    collinsmark

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    Homework Helper
    Gold Member

    Yes, the static friction is acting on block B, and it is in the same direction as the applied force F (where F is attached to block A). And it makes sense if you think about it. It has to be acting in the same direction, otherwise block B wouldn't be accelerating in the same direction as block A.
    Yes, that's right. :approve:
    Well, you seem to already have most everything you need. You know what the normal force acting on B is. So you can find the maximum static frictional force that can occur too (which is the normal force times μs).

    Using Newton's second law (FB = mBa), solve for the acceleration a, of block B.

    Invoke Newton's second law one more time to solve for the total applied force F, after realizing that blocks A and B are both accelerating at same rate a. :wink:

    (I'll leave all the appropriate substitutions up to you.)
     
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