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Pushing blocksI suck at forces

  1. Oct 15, 2003 #1
    I've added a picture as an attachment.

    Four blocks of mass m = 10 kg are arranged as shown in the picture, on top of a frictionless table. A hand touching block 1 applies a force of F1h = 90 N to the right. The coefficient of friction between the blocks is sufficient to keep the blocks from moving with respect to each other.

    What is the total force exerted on block 3 by block 2 ?

    ---------------------------------------------------------

    Ok. This is what my force diagram for object 3 looks like.

    Imagine a box. There is an arrow pointing from the top of the box upwards. This is the normal force put on block 3 by block 2.
    I have a normal force pointing down put on block 3 by block 4. There is also a W (weight of block 4) pointing down also.

    To the right, I have a friction force put on block 3 by block 2, and to the left, I have another friction force put on block 3 by block 4.

    Also, I did a force diagram with block 4. A normal pointing up from block 3, a friction force to the right from block 3, and a W pointing down.

    Through some analysis, I know that the normal force on block 3 by block 2 is equal to twice the normal force put on block 3 by block 4.

    Likewise, the friction force put on block 3 by block 2 is equal to twice the friction force put on block 3 by block 4.

    It's twice the normal and friction because block 4 has to put force through two blocks to get to block 2 (hope that makes sense).

    And I know that w=m*g=(10kg)(9.8m/s/s)=98 N.
    ---------------------------------------------
    Now I'm pretty much stuck. I know that I could combine block 3 and block 4 so it is one big block 3, but after that, I don't know how to calculate the net force onto block 2. I'm still trying to get the hang of forces.
     
  2. jcsd
  3. Oct 15, 2003 #2
    the picture

    Here's the picture.
     

    Attached Files:

  4. Oct 15, 2003 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    This problem is actually simpler than it looks.
    If the question is simply "What is the total force exerted on block 3 by block 2 ?" then most of the information is irrelevant. Certainly the "90 Newton" force by the hand is irrelevant since it is horizontal. Also the force by block 1 on block 2 is irrelevant. The fact that blocks 3 and 4 are separate blocks is irrelevant!!

    The only thing relevant is that there is a mass of 20 kg on top of block 2. 20 kg is 20(9.8)= 196 Newtons weight. Block 3 is exerting 196 Newtons force (the weight of blocks 3 and 4 together) on block 2.

    I actually did this whole problem thinking that the question was the more interesting question "What is the force block 1 exerts on block 2?" Since I refuse to throw away that work, here it is:

    Since we are told "The coefficient of friction between the blocks is sufficient to keep the blocks from moving with respect to each other" we can treat blocks 2, 3, and 4 as one block of mass 30 kg. Since all block move together we can start by treating this as a single block of mass 40 kg. Since this "block of 40 kg" is being pushed by a force of 90 Newtons, it accelerates (F= ma so 90= 40a), 90/40= 9/4 m/s2.

    Now look at block 1 alone. In order for it, alone, to move at 9/4 m/s2, it would have to have a net force of (again F= ma so
    F= 10(9/4)) 90/4= 22.5 Newtons. That means that blocks 2, 3, and 4 must be pushing "back" on block 1 with force 90- 22.5= 67.6 Newtons
     
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