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Force of a block on an incline above a horizontal

  1. Mar 4, 2017 #1
    1. The problem statement, all variables and given/known data
    The block shown in figure #1 above has a mass of 4.30 kg. The applied force F has a magnitude of 31.2 N and is directed at 38.0˚ above the horizontal frictionless surface the block in on. What is the magnitude of the normal force that acts on the block?

    2. Relevant equations
    Fnormal = m*g*cos(theta)

    3. The attempt at a solution
    I get 40.2 N. The answer of our solutions page is 22.9N

    What am I doing wrong?
     
  2. jcsd
  3. Mar 4, 2017 #2

    kuruman

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    We could help you if you posted the figure so we wouldn't have to guess.
     
  4. Mar 4, 2017 #3
    how do i upload the homework sheet?
     
  5. Mar 4, 2017 #4

    kuruman

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    First get a good quality and easily legible picture. Then click the "UPLOAD" button on the lower right.
     
  6. Mar 4, 2017 #5
    ok this is the sheet. Its problem 1
     

    Attached Files:

  7. Mar 4, 2017 #6

    kuruman

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    Can you show us how you got this answer? We cannot tell what you are doing wrong just by looking at one number.
     
  8. Mar 4, 2017 #7
    I actually figured it out. I did: (4.30)(9.80)-F31.2sin(38) = 22.9N
    How would I acquire the net force?
     
  9. Mar 4, 2017 #8
    31.2N is the applied force given in the problem.
     
  10. Mar 4, 2017 #9

    kuruman

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    Very good.
    Add all the forces in the horizontal direction to get the horizontal component of the net force.
    Add all the forces in the vertical direction to get the vertical component of the net force.
     
  11. Mar 4, 2017 #10
    Ok so i figured out all of number 1 and parts a,b,c, and d on number 2. Part e on number 2 is asking for the firctional force that acts on the block, i did 21.5N(cos(29.4)) = 18.7N. the answer is -18.7N. I know how to work it, but i do not understand why the frictional force is negative?
     
  12. Mar 4, 2017 #11

    kuruman

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    The force of kinetic friction is always opposite to the velocity. Here the velocity is to the right, in the conventional positive direction. Therefore, the force of kinetic friction is in the negative direction to the left.

    When you are calculating the net force you do a vector addition. For example, if vector A is to the right and vector B is to the left and they add up to zero, you would write ##\vec{A} + \vec{B} =0##. Now it is also true that ##\vec{A} = A \hat{i}## and ##\vec{B} = -B \hat{i}## where A and B are the magnitudes of the vectors. Thus, you have ##A \hat{i} +(-B \hat{i})=0## which simplifies to the following relation between magnitudes ##A - B = 0##. See how it works?
     
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