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Question regarding friction

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  1. Aug 5, 2015 #1
    1. The problem statement, all variables and given/known data
    A block weighing 300 N slides downwards on the angled surface shown at an initial velocity of 2.5 m/s. At t = 0, a force P with the given orientation is applied until the block stops. Knowing that it took 5 seconds for the block to stop and that the coefficient of kinetic friction μk = 0.1, determine the magnitude of P
    y9t5MfJ.png

    2. Relevant equations
    F = Fn x coefficient of friction s/k

    3. The attempt at a solution

    First i broke the forces into components. Added the Fn and Friction force on the diagram and labelled 300N force acting downwards. Once i broke the forces into components i got two equations, one for forces acting on the y-axis and the second for forces acting on the x-axis.

    -0.4226P + Fn - 281.90 = 0 ---------- Equation (1)
    0.9063P - F(f) +102.606 = 0 ---------- Equation (2) Note that F(f) is force of friction.

    Then i knew that F(f) = Fn x 0.1 from the question so i replaced it F(f) with it.

    -0.4226P + Fn - 281.90 = 0
    0.9063P - 0.1Fn +102.606 = 0


    After this i solved these two equations simultaneously and found the value of P.

    My question/confusion is regarding all the other information given in the question, such as the initial velocity, time etc. I feel like i have done something wrong here. Any guidance would be highly appreciated.
     
    Last edited: Aug 5, 2015
  2. jcsd
  3. Aug 5, 2015 #2

    Doc Al

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    Staff: Mentor

    Why did you set this equal to zero?
    (Also, what's the direction of the component of P?)
     
  4. Aug 5, 2015 #3
    I set that equal to zero because i think that in order for that block to stop the net forces should be euqal to zero.

    P is pointing towards the block so the y component would point downward and the x component would point towards right.
     
  5. Aug 5, 2015 #4

    Doc Al

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    Not so. If the net force was zero then the block would not change speed.

    Hint: That's why they gave you all that other info--so you could calculate the acceleration.

    Right. So fix your equation. (You have the x components of P and the weight pointing in the same direction.)
     
  6. Aug 5, 2015 #5
    Ok i fixed my equation. Though i am still kind of confused about the first part that you posted. What am i suppose to do once i calculate the acceleration?

    Also there was a mistake i made in the diagram. It has been fixed. P is 25 degrees from the inclined surface, not from the horizontal.
     
  7. Aug 5, 2015 #6

    Doc Al

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    Show the corrected equation.

    If you fixed the equation properly, the acceleration will be part of it.
     
  8. Aug 5, 2015 #7
    Ok so what i understand is that since the block is not moving in the y direction, i can put the equation concerning with forces in the y direction to zero. However since the block is moving in the x direction, i cannot set it to zero and there the acceleration ( which i calculated it to be 0.5m/s) should be put somewhere in the equation.

    So this is what i have come up with so far. I apologize before hand if i keep on repeating the same mistakes again and again. These concepts are a little challenging to me. I hope you understand my situation :-)

    Y direction:

    Fn - 281.90 - 0.4226P = 0

    X direction:

    F=ma

    (0.9063P - F(f) - 102.606)/30 = 0.5 (?) where 0.5 is the acceleration and 30kg is the mass of the block
     
    Last edited: Aug 5, 2015
  9. Aug 5, 2015 #8

    Doc Al

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    Good.

    Good. What's the direction of the acceleration?

    I didn't check your calculations, but the basic idea looks good.

    What you are trying for here is Newton's 2nd law: ΣF = ma. (Figure out the block's mass.)

    Pay attention to the direction of the forces and the acceleration. That determines the signs they have in your equation. (State the direction of each.)
     
  10. Aug 5, 2015 #9
    F=m.a

    (0.9063P - F(f) - 102.606)/30 = -0.5 where 0.5 is the acceleration and 30kg is the mass of the block. Acceleration is -0.5 since it is going to the left.
     
  11. Aug 5, 2015 #10
    And there from there on i put everything together, move everything to one side and solve the y and x equations simultaneously i believe?
     
  12. Aug 6, 2015 #11
    Where does it say the block weighs 30 kg? - Nevermind stupid question
     
  13. Aug 6, 2015 #12
    It doesn't say anything about the mass directly but since w=mg and we know the weight, we can find the mass.
     
  14. Aug 6, 2015 #13
    What did you get for your final answer?
     
  15. Aug 6, 2015 #14

    Doc Al

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    Which way does each force act? (You need to check your signs.)

    The block is moving to the left, but it is slowing down. (If the acceleration acted to the left, it would speed up.)
     
  16. Aug 6, 2015 #15
    I see, so the acceleration is going towards right. Am i right?
     
  17. Aug 6, 2015 #16
    And also the friction force is acting towards right?
     
  18. Aug 6, 2015 #17

    Doc Al

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    Yes.
     
  19. Aug 6, 2015 #18

    Doc Al

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    Yes, because the block is sliding to the left and kinetic friction will oppose that motion.
     
  20. Aug 6, 2015 #19
    Yes, sorry for that mistake. It makes sense now. Just wanted to say I have huge respect for you guys, taking out time, trying to help others in their course work. Thank you very much.
     
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