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More Force Review!

  1. Oct 25, 2007 #1
    Two crates, of mass m1 = 40 kg and m2 = 125 kg, are in contact and at rest on a horizontal surface (Fig. 4-54). A 620 N force is exerted on the 40 kg crate. The coefficient of kinetic friction is 0.15.
    [​IMG]

    Find the acceleration:

    Calculate the force that each crate exerts on the other:


    I am counting my forces incorrectly.

    I had
    [tex]\sum F_x=ma[/tex]
    [tex]\Rightarrow -F_k+F_a=ma[/tex]
    [tex]\Rightarrow-mg*\mu_k+F_a=ma[/tex]
    [tex]\Rightarrow\frac{ -(125+40)(.15)+620}{125+40}=a=3.61[/tex]

    I think block 2 exerts a Force on block 1...but I am not sure how to determine its magnitude...would someone be so kind as to remind me?

    Also, for Normal Force, I used the COMBINED weight...is that correct?

    Casey
     
  2. jcsd
  3. Oct 25, 2007 #2

    Dick

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    All fine so far. The force block 2 exerts on block 1 is equal and opposite to the force block 1 exerts on block 2. To find the force on block 2, you know the acceleration of block 2 and the mass of block 2. This should let you find force on block 2.
     
  4. Oct 25, 2007 #3
    I think you just made a little mistake. When you're calculating the friction force, you have the formula correct, but I think you forgot the 'g' i.e. 9.8 in the calculation.
     
  5. Oct 25, 2007 #4

    Dick

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    Duh, right. I saw it in the formula. But I didn't notice it was absent in the numbers. Thanks.
     
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