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Homework Help: Motion on Inclined ramp with Friction Problem

  1. May 24, 2012 #1
    1. The problem statement, all variables and given/known data

    "On the 50 degree ramp below an empty barrel is sliding up it to get filled at a factory. The empty barrel has a mass of 24kg and has an initial velocity of 1.2m/s. The kinetic coefficient of friction between the barrel and the ramp is μ =0.48 Finally a horizontal force is acting on the barrel of 320N.
    1.)What is the barrel's acceleration?
    2.)What is the direction of the barrel's acceleration?
    3.) How far up the ramp does the barrel go before its velocity is zero?"

    First picture is of what is given

    2. Relevant equations
    F=ma (of course)
    accelerations will involve x or y components with regard to theta (sin(σ) and cos(σ))
    ?Kinetic formulas? Δx=

    3. The attempt at a solution
    Second picture attachment is a picture of my diagram so far. Is it missing a force going down the ramp? Parallel to the friction force?

    A.)Not sure where to start here.

    B.) The acceleration is down the ramp because it is slowing to a stop. Correct?

    C.) Solvable after I find acceleration in part A? Using Δx= Vi(t)+ 0.5(a)t^2 where Vi is initial velocity. Or on second thought perhaps this one if you can not find time: V^2=Vi^2 + 2a(xΔx)

    Please and thank you!
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. May 24, 2012 #2
    In your second diagram the direction of the frictional force should be in opposite way, pointing downward since it is sliding up.

    Then you can resolve(replace) the 4 forces in only direction of parallel and tangent to plane.
    From this diagram the net tangent forces to the plane is zero.
    The net parallel forces will accelerate the barrrel up or down.
  4. May 24, 2012 #3


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    You've identified all the forces acting; Now you have to break them up into their vector components parallel and perpendicular to the plane, then use the applicabe laws of Newton in the y and x direction separately to sove for N and a.
    As per question 3, this is apparently correct, although you'd have to crunch out the numbers to solve for the correct direction of a.
    this last one is good, but you have an extra x term in there. Don't get too fancy doing calculus when you don't need to.
    And don't forget to ask yourself what is going on in the y direction perpendicular to the plane.
  5. May 24, 2012 #4
    Ah, right friction force is opposite direction of motion.

    The extra x in that equation was a mistake, thanks.

    So First I solved for the force going up the ramp which would be cos(σ) component of the horizontal force of 320N. Then solving for acceleration since the mass is given:

    cos(50°)(320N) = 205.69N
    205.69N = ma
    a= 205.69/24kg = 8.53m/s^2

    I feel like this is the force if it was stationary or something. How do I account for the frictional force?

    The below?
    μ= kinetic friction constant

    So if the total force up the ramp would be the force up the ramp minus the frictional force pulling it down the ramp that would be 205.69-(205.69*μ) = 106.96N

    Then I would have to recalculate the acceleration since 106.96N is the 'new' non-zero net force up the ramp. 106.96N/24kg= 4.45m/s^2

    Are either of these correct? If 4.45m/s^2 is the correct acceleration of the barrel then

    V^2=Vi^2 + 2a(Δx)
    (Δx) = V^2-Vi^2/(2a)
    (Δx) = 0-(1.2^2)/(2*4.45) →(this equals negative but assuming negative accel. to come to a stop?)
    = .1617 meters with sigfigs: 0.16m

    What do you think?

    Attached Files:

    Last edited: May 24, 2012
  6. May 24, 2012 #5
    Now you remove the mg and replace this force with parallel to plane and another perpendicular or resolving mg to resultant of 2 forces. Now you have 2 forces downward parallel to plane, component of mg and friction.

    Likewise you have to replace the 320N force to 2 forces. Parallel and perpendicular to plane.

    You will see that friction depends on N and N is result of 2 component forces.
    Last edited: May 24, 2012
  7. May 24, 2012 #6
    So you are saying resolve the force going up the ramp into two forces; one parallel to the horizontal and one perpendicular to the the first force?

    Attached Files:

  8. May 24, 2012 #7
    Use your previous diagram with the mg pointing downward and 320N horizontally pointing to the right. The normal force and friction are in correct position.

    The only thing to do is erase the mg and replace with 2 forces, perpendicular and horizontal to plane.
    Equally do it to 320N force.

    What you should get 6 forces acting on the barrel.
    2 downward, 1 upward parallel to plane
    2 downward, 1 upward perpendicular to plane.

    add: fbd with 6 forces.
    Last edited: May 24, 2012
  9. May 25, 2012 #8
    Diagram now has 6 forces.

    Fh is the force parallel to the ramp/plane which is pulling the barrel up
    Fp is the force perpendicular to the ramp/plane. This is equal and opposite to the Normal force correct?

    Fh= cos(50°)320N
    Fp= cos(50°)mg

    How to solve for acceleration?

    Attached Files:

  10. May 25, 2012 #9
    Once you resolve a force, you remove it from the diagram. It does not exist by itself anymore.

    You've done correctly resolving to parallel of the plane.(320N = component #1 + component #2)
    You should do resolve perpendicular to the plane.
    Means you're replace 320N to 2 components.

    Do the same with mg.

    Edit: Sorry ,You're resolve Fp. But it has nothing to do with mg.
    Good you have done Fh,Fp(wrong calculation) and now add MGh and MGp
    Last edited: May 25, 2012
  11. May 25, 2012 #10
    Okay I think these are the six forces after resolving the 320N force into x and y components and the force mg into x and y components.

    Fh= cos(σ)(320N)
    Fp= sin(σ)(320N)
    MGh= sin(σ)mg
    MGp= cos(σ)mg


    So the sum of the forces up and down the plane equal Fh-Fμ-MGh?

    Attached Files:

  12. May 25, 2012 #11
    That's correct fbd.
    Friction should be μN.
  13. May 25, 2012 #12
    So the sum of the forces up and down the plane equal Fh-Fμ-MGh which is:

    Fμ = Fh*μ where μ is the kinetic friction constant to .48?

    If the above is correct then:
    cos(50°)(320N) - ((.48)cos(50°)(320N))- (sin(50°)(24kg)(9.8)
    = 205 - 98.7 - 180 which would be negative value for the force. Does this make since?
    -73.7N=24kg(a) solve for acceleration here?

    a=-3.07 m/s^2

    Negative acceleration because the barrel is slowing down?
  14. May 25, 2012 #13
    There's error in your calculation.
    Frictional force=μN NOT Fμ or Fhμ

    Find the value of N.
  15. May 25, 2012 #14
    I don't have the frictional force to solve for N do I? Unless its MGp the force down on the barrel?
  16. May 25, 2012 #15
    You see from the diagram there are 2 forces pulling/pressing the object down.
    By Newton 3'rd law there must be equal opposite force, where we call normal force.

    Or you see that the barrel no going down or up perpendicullaly , net force equal to zero.
  17. May 25, 2012 #16
    Yes the Normal for going up perpendicular to the ramp has a magnitude equal to the sum of the downward forces Fp and MGp right?

    So applying that to the forces parallel to the ramp:

    Fh = Nμ + MGh

    cos(50°)(320N) = N(.48) + mg(sin(50°)
    N= 53.163
  18. May 25, 2012 #17
    You have right statement. From this statement you should know the value of N.
    Last edited: May 25, 2012
  19. May 25, 2012 #18
    Okay so the acceleration of the barrel is a = Fh/m = cos(50°)(320N)/24 = 8.57m/s^2

    Putting that into the equation

    V^2=Vi^2 + 2a(Δx)
    (Δx) = V^2-Vi^2/(2a)
    (Δx) = 0-(1.2^2)/(2*8.57)
    = 0.084 meters
  20. May 25, 2012 #19
    Maybe you can show the net horizontal force along the plane in term of MGh, MGp, Fh,Fp, and μ variables. Refer to your fbd.
    Last edited: May 25, 2012
  21. May 25, 2012 #20
    Nμ + MGh + Fh .... or so I thought
  22. May 25, 2012 #21

    I hope you understand why i replace N with MGp+Fp.

    You have done good at resolving the forces.
    I think you have to know more about N or normal force and frictional force.
    And the rest refer to PhantomJay.
  23. May 25, 2012 #22
    I understand now, I kept thinking N was newtons times μ instead of the Normal force. The normal force is equal and opposite the downward forces Fp and MGp hence those to added together equal it.

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