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Some trouble with my force calculations

  1. Jun 12, 2008 #1
    1. The problem statement, all variables and given/known data
    I have 2 problems I'm having trouble with right now.(actually more but I'll ask 2 and try to figure out the others for now.

    Two blocks are in contact on a frictionless table. A horizontal force F is applied to M2, as shown. If M1 = 1.06 kg, M2 = 3.38 kg, and F = 4.75 N, find the magnitude of the contact force between the two blocks.
    I calculated the acceleration of the entire system to be 1.069m/s^2 but don't know where to go from there.

    A window washer pulls herself upward using the bucket-pulley apparatus shown in figure below.

    How hard must she pull downward to raise herself slowly at constant speed? The mass of the person plus the bucket is 61.5kg.

    Correct, computer gets: 3.01e+02 N

    If she increases this force by 10.3 percent, what will her acceleration be?
    I answered the first part, and for the second part have determined that if she increases her force ten percent to ~331N, that the acceleration upward would be 662N-601N/mass, but this is incorrect.

    2. Relevant equations
    F=ma?
     
    Last edited: Jun 12, 2008
  2. jcsd
  3. Jun 12, 2008 #2
    I keep doing the second one like this, and keep getting .98m/s/s

    301x1.1=331.1x2=661-601=60.1/61.5kg=.98m/s/s which is not right
     
    Last edited: Jun 12, 2008
  4. Jun 12, 2008 #3

    Kurdt

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    Draw a free body diagram of each block separately and analyse the forces acting on it.
     
  5. Jun 12, 2008 #4
    OK I just did so and got the 4.75N initial force acting on M2, 3.6132N force acting due to acceleration, and on m1 I got 1.13314N in the same direction, so would I do 4.75+3.6132-1.13314? Or does the force from the push carry over to M1 as well?
     
  6. Jun 12, 2008 #5

    Kurdt

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    I'm not sure what you mean with that but 3.613N is the answer. No further action needed.
     
  7. Jun 12, 2008 #6
    I entered that and it said it was wrong? Weird
     
  8. Jun 12, 2008 #7

    Kurdt

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    Sorry, it was very close to what I got as an answer but not exactly the same. Try calculating again.
     
  9. Jun 12, 2008 #8
    I tried, all my tries are up so I missed it. Can anyone help on the other one?
     
  10. Jun 12, 2008 #9

    Kurdt

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    I'm surprised they were so strict, it was only 0.003 off.
     
    Last edited: Jun 12, 2008
  11. Jun 12, 2008 #10
    Ya thats weird. [​IMG]
     
  12. Jun 12, 2008 #11
    can you guys see that picture?
     
  13. Jun 12, 2008 #12

    alphysicist

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    Hi big_d,

    I believe the answer for the first problem would be 1.13 N. The acceleration of the entire system is 1.07 m/s^2, and the contact force is the only horizontal force acting on m1 (1.06 kg), so using F=ma gives 1.13 N.

    If you want to get the same thing from m2, you would say that the force 4.75 is pushing in the direction of acceleration, the unknown contact force is pushing backwards, and the acceleration is 1.07 m/s^2, so using

    (net force) = m a

    gives

    (4.75) - (contact force) = (3.38) (1.07) ---> contact force = 1.13 N
     
    Last edited: Jun 12, 2008
  14. Jun 12, 2008 #13

    alphysicist

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    I think this is the right procedure. However, when you don't round off the intermediate values, I get a value that is about 3 percet different from yours. Is that enough for the software to say the answer is wrong?
     
  15. Jun 13, 2008 #14

    Kurdt

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    I'm terribly sorry, I didn't realise the initial force was on the second block. Thats unusual notation though.
     
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