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Frictional force question

  1. Oct 15, 2007 #1
    1. The problem statement, all variables and given/known data

    An easy question, but I'm not getting it:

    A 3.5kg block is pushed along a horizontal floor by a force F of mag 15N at an angle of 40 deg with teh horizontal. The coefficiennt of kinetic firction b/t the block and the floor is 0.25. Find the frictional force on the block from the floor and the block's acceleration.

    2. Relevant equations


    3. The attempt at a solution

    So I used the Newton's second law as


    Since I wanted to find normal force first, and normal force is in y direction, a would be zero, so I subsituted 0 for a and made the equation equal to Fn.


    From then, I just put the mass, gravity (9.81) and angle into teh equation, getting 19.8. After that, I put that into the equation for fs (in the second part of the question) and got 4.9, which turns out to be wrong. What'd I do?

    Oh oh, the acceleration didn't turn out any better. The book says to use the equation Fcos(O)-muk*Fn=ma for acceleration and solve for a. What I don't get about that is why you have to take away muk*Fn to use the second law on the x-axis...surely you should just use Fcos(O)=ma, right? Thanks.

    EDIT: I put the equation f or static frictional force instead of kinetic
    Last edited: Oct 15, 2007
  2. jcsd
  3. Oct 15, 2007 #2


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    Staff: Mentor

    The weight of the block is normal to the floor and provides some friction [itex]\mu[/itex]mg.

    Now the 15 N force at 40° provides a horizontal component to push the box and a vertical component, which either adds (pushes down) or subtracts (pushes up) the block depending on whether the 15 N force is above or below the horizontal.

    The difference between the horizontal pushing force and friction determines the force available to accelerate the block (or keep moving at constant velocity if the net force is zero, one it gets moving).
  4. Oct 16, 2007 #3
    Yes, yes, I forgot to add that the force is at 40 degrees in the fourth quadrant, or -40 deg, or 320 deg, whichever you like. The force is being applied downward, below the horizontal, if you will.

    But to find the normal force shouldn't be any different, should it? Normal force is upwards, and it should be equal to the Fg after mg was taken away, right? Yet I'm still getting an incorrect answer to the whole question.
  5. Oct 16, 2007 #4


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    Staff Emeritus
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    Gold Member

    The angle of your applied force will most definatly affect the normal force, draw yourself a FBD to convince yourself.
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