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Pully and Inclined Ramp

  1. Nov 5, 2016 #1
    1. The problem statement, all variables and given/known data

    100 kg box is held in place on a ramp that rises at 30° above the horizontal. There is a massless rope joint to the box that makes a 22° angle above the surface of the ramp. Coefficients of friction between the box and the surface of the ramp are μk = 0.40 and μs = .60. The pulley has a mass of 10 kg and a radius of 10 cm. However the pulley does not have a friction.

    A). What is the Max weight the hanging mass can have so that the box remains at rest.

    B). The system is at rest with the weight of the hanging mass found in Part A, but a speck of dust comes to rest on the weight on the hanging mass, making the system unstable. What is the acceleration of the box at the very moment it starts moving?


    See diagram
    https://i.imgur.com/xCWIHWb.png xCWIHWb.png

    2. Relevant equations


    3. The attempt at a solution

    So I think I got part A, and sorta got part B setup.. but I am not really sure where to go from there.


    Part A)
    Here is how I setup my equations:
    Force Gravity on Block = (100)(9.8)sin(30) = 490
    NF = 100cos(30) - Tension*sin(22) = 50√3 - Tension*sin(22)
    Mass on Ramp = Ma
    Mb =
    Hanging mass = Tension-(Mb)(9.8) = 0

    Net force on Box = 0, forces equation =
    Tension*cos(22) - Magsin(30) - ( Magcos(30) - Tension*sin(22) )μ = 0

    I then proceeded to plug in Ma, g = 9.8, and μ = .60
    Then using the equation above and the Hanging mass equation I solved for Mb

    I got
    Mb = ~ 88 kg

    Part B)
    I am fairly confident about my answer of Part A.. but now unsure exactly how to solve part B.
    I know the radius of the pulley is 10 cm and the mass is 10 kg.


    I had to find the equations for the two tensions. However, one thing I had to form a new equation for (at least I think ) is the Normal Force. Before I knew the vertical component of the normal force would just be 0, but now since it is moving it could also change.

    My new equation is:
    NF = 100cos(30) - Tension*sin(22) = Ma A
    NF = 100cos(30) - Tension*sin(22) - Ma A = 0

    I know I need to substitute this into my original equation, so I can use it to calculate the force of Friction
    Tension2*cos(22) - Magsin(30) - ( 100gcos(30) - Tension2*sin(22) - Ma A )μ = 0
    put this equation in terms of Tension2, but I must leave acceleration in terms of A
    Tension2 = 129A + 770
    My equation for tension1 was easier, just:
    Tension1 = Mb A - Mb A

    Now I plug them into a Torque equation taking into account the mass of the pulley and radius
    Tnet = [ Tension2 - Tension1]*radius = .5 m r^2 ∝

    Now here is where my struggles appear, I have two unknowns with A and Mb. I've gotta imagine that the intention of Mass B is that the speck of dust would represent just about nothing, and the point of it was to explain that the system would be in motion and I should just account for it as the same mass from before, but I don't know. ANy advice on this would be great!


    Let me know if there is anything I should elaborate on something.. ,thanks in advance!

     
    Last edited: Nov 5, 2016
  2. jcsd
  3. Nov 5, 2016 #2

    TSny

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    WELCOME TO PF!
    Part (A) looks correct.
    This equation says that NF = MaA which implies that the net force accelerating Ma is the normal force. Is that correct?
    Also, the middle expression in your equation is missing a factor of g somewhere.

    OK, except the expression for the normal force is incorrect.
    [EDIT: The right hand side should not be zero.]

    I think you have a couple of typographical errors here. The right-hand side reduces to zero.
    Also, did you mean "Tension1" instead of "Tension2"?

    Yes
     
    Last edited: Nov 5, 2016
  4. Nov 5, 2016 #3
    Woops, Thanks for the fast reply.
    Yeah, that's what I was implying. Looking at the diagram, I was trying to account for the Y side.. and figured that the Y would be changing as well as it is not at rest..


    I edited the original post, I stupidly forgot the G.

    Tension2*cos(22) - Magsin(30) - ( 100gcos(30) - Tension2*sin(22) - Ma A )μ = 0

    Thanks for the reply!
    Yeah.. ment Tension 1.. stupid mistake.. wrote this all to fast
     
  5. Nov 5, 2016 #4

    TSny

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    Not sure what you mean by the "Y side" or that the "Y side would be changing". I don't see the Y direction indicated in the diagram.
     
  6. Nov 5, 2016 #5
    Sorry, Not sure what I ment by that. I put 100gcos(30) - Tension2*sin(22) - Ma A
    in because I figured that when I calculated the normal force I would need to account of Acceleration as well since it is now moving.. now that I say that it doesn't make a ton of sense to me.. should I have just left it as 100gcos(30) - Tension2*sin(22)
     
  7. Nov 5, 2016 #6

    TSny

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    When the block accelerates along the incline, what is the component of acceleration in the normal direction?
     
  8. Nov 5, 2016 #7
    Hmm, not quite sure what that means, would it be:

    (100gcos(30) - Tension2*sin(22))/Ma= A
     
  9. Nov 5, 2016 #8

    TSny

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    Suppose we let the Y-axis be oriented perpendicular to the incline (that is, the Y direction is in the Normal direction).

    What do you get if you set up Newton's 2nd law in terms of components in the Y direction: ΣFY = maY
     
  10. Nov 5, 2016 #9
    Oh, okay.. i see what your saying.. wouldnt it be something like:
    100gcos(30) - (Tension2*sin(22) + 100gcos (30)

    The tension and normal force acting in one direction, with the force of direction in the other? other wise I am a bit lost here haha

    Thanks
     
  11. Nov 5, 2016 #10

    TSny

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    I don't see what this represents. You have identified the forces acting on Ma. For each force, find an expression for the Y-component of the force. Also, think about the value of the Y-component of the acceleration of MA. Then you can set up the equation ΣFY = MaaY to find the normal force.
     
  12. Nov 5, 2016 #11
    This was my Y component: 100gcos(30) - (Tension2*sin(22
    What should the Y component be then? a little lost.. the axis are tilted.. should it be 100gsin(30) - Tension2*cos(22)
     
  13. Nov 5, 2016 #12
    Otherwise.. gonna need some help here.. really not sure what the Y component could be
     
  14. Nov 5, 2016 #13

    TSny

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    Let's back up just a little to make sure we are on the same page. Can you list all of the forces that act on Ma?
     
  15. Nov 5, 2016 #14
    Could I instead use my original equation I used to solve for mass b instead? very lost..
     
  16. Nov 5, 2016 #15

    TSny

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    You know the general approach to problems that require the application of Newton's 2nd law:

    1. Draw a free body diagram for each mass. Each diagram shows the forces on the mass as well as a choice of an x-y coordinate system.

    2. Use each diagram to help set up the component form of Newton's second law: ΣFx = max and ΣFy = may for each mass.
    (For a non-zero-mass pulley, set up Σ##\tau## = I ##\alpha##)

    3. Solve the equations from step 2 for the unknowns.

    So, we are trying to do step 1 for the block on the incline. In order to draw a free body diagram for this mass, you must first identify all of the forces acting on this mass. Your work for part (a) looks good. In part (b) you have essentially the same forces (but the values of some of the forces will be different) .

    Can you show (or describe) the free body diagram for the block on the incline?
     
    Last edited: Nov 5, 2016
  17. Nov 5, 2016 #16
    R8OEB.png

    Okay, so here is what I think the force diagram is for the block, ignore the other stuff though.

    So my equation for part 1 is Tension*cos(22) - Ma g sin(30) - ( Ma g cos(30) - Tension*sin(22) )μ = 0
    sometimes I think about it such as the equation is parallel and perpendicular.

    Would this be the Y equation:
    Normal Force + Tension sin(22) - mgcos(30) = Ma


    So trying to think about how my equation would be different from the work in Part 1. Besides the fact that there is acceleration because it is not at rest.
    Thanks for all your help on this, just a little lost on what from part 1 I need to do differently for part 2, other than account for acceleration
     
  18. Nov 5, 2016 #17

    TSny

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    Your diagram looks good for MA. As you say, you should think about the forces parallel and perpendicular to the incline. So, it is helpful to rotate the coordinate system so that the x-axis is parallel to the incline and the y-axis is perpendicular to the incline. Then when you set up ΣFx = max and ΣFy = may, you are working with the components parallel and perpendicular to the incline.
    upload_2016-11-5_21-38-3.png

    Try setting up ΣFy = may. This will allow you to get an expression for the normal force. Think about the value of ay.
     
  19. Nov 5, 2016 #18
    This would be my equation for may,
    may * acceleration = Normal Force + Tension sin(22) - mgcos(30)

    I guess I could only solve for the normal force within the whole system of equations, or I could put this in terms of the normal force and solve for this as Normal force in the other equation? If that is is the case I could do
    Normal Force = may * acceleration - Tension sin(22) + mgcos(30)

    And then plug into the parrallel equation:
    Tension2*cos(22) - Magsin(30) - ( may * acceleration - Tension sin(22) + mgcos(30 )μ = Ma

    right idea? or am I lost?
     
  20. Nov 5, 2016 #19

    TSny

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    Good. Definitely the right idea. In what direction does the block accelerate? So, what is the value of ay?
     
  21. Nov 5, 2016 #20
    The block is going to accelerate up the ramp...
    I guess to calculate Ay at this point I would use something like this:
    Tnet = [ Tension2 - Tension1]*radius = .5 m r^2 ∝ ?
     
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