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Homework Help: Force with two blocks on a table.

  1. Sep 20, 2010 #1
    1. The problem statement, all variables and given/known data

    A force of 32.68 N pushes and pulls to blocks as shown in the figure below. The vertical contact surfaces between the two blocks are frictionless. The contact between the blocks and the horizontal surface has a coefficient of friction of 0.24. The acceleration of gravity is 9.8 m/s2 .

    (On a table) The weight of the first block is 4.3 kg. The force is pushing in the same direction as the table upon the first block. The second block weighs 8.6 kg. The force pulling this block, however, is at a 60 degree angle from the table.

    The coefficient of friction is 0.24.

    What is the magnitude a of the acceleration of the blocks?
    Answer in units of m/s2.

    2. Relevant equations

    3. The attempt at a solution
    I drew two free body diagrams. The first one has Force pushing from the left to the right, friction going in the opposite direction, natural force going up and w going down. The second has friction going from right to left, Force going from left to right at a 60 degree angle, w2 going down and natural force2 going up.

    I'm confused, however, because of the fact that the second block isn't being pulled straight forward. I understand that this affects the net Force of block 2 but does it also affect the force that block one puts on block two and that block two puts on block one?
  2. jcsd
  3. Sep 21, 2010 #2


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    Good so far (the 'natural' force is called the Normal force).
    You have to break up the second force into it's x and y components. The y component will reduce the normal force on that block (Use Newton 1 in that direction). Then use Newton 2 to calculate the acceleration in the horizontal direction.
  4. Sep 21, 2010 #3
    So all it does is change the normal force (which is what I meant, not natural. My brain had died)? How do you calculate that? Is it w-Fsintheta? Which means that the y component of the Force of the second block would not include the second force that is pulling the block? But that would make the acceleration in the y direction be zero. Is it supposed to be zero?

    I got 1.448 for acceleration in the x direction and if there is no acceleration in the y direction, that would be the answer. But it's not the answer. So either I need to find an acceleration in the y direction or I calculated the x direction acceleration wrong.

    Would the net Force of block two be N2 - W2 + Fsintheta=m2ay?

    I guess I'm confused on how the normal force is affected.
  5. Sep 21, 2010 #4
    But I guess if my normal force for the second block is incorrect, my friction for the second block is incorrect which makes my acceleration in the x direction incorrect.
  6. Sep 21, 2010 #5
    I got it! Thank you for your help. I went back and redid my friction using -W - Fsintheta instead of just -W and got the correct acceleration. I realized it said blocks plural so that meant it had to be x acceleration only.
  7. Sep 21, 2010 #6


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    it changes the normal force, and thus, since friction is a function of the Normal force, that is affected also. And acceleration in x direction changes.
    with no friction in between blocks, as stated, then yes, you are correct.
    It doesn't include the x component, but as you stated, it includes the y component (I don't quite understand what you mean).
    yes, as long as the block remains in contact with the table, there is no acceleration in the y direction.
    Look at the horizontal forces on the system of blocks in the x direction to calculate the acceleration in the x direction. Don't forget the friction forces.

    as i stated above, as long as the block is in contact withthe table, a_y =0. (If N turns out to be a negative number, then there will be acceleration in the y direction)
    Is this help making you more or less confused?

    Edit: You solved it as i was posting....so i guess you're all set, good work...
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