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Free-rolling ramp with crate

  1. Nov 23, 2009 #1
    1. The problem statement, all variables and given/known data
    I am doing a practice problem out of the text book in order to improve my skills. The problem states the following: A free-rolling ramp has a weight of 120 lb. If the 80-lb crate is released from point A (top of the ramp), determine the distance the ramp moves when the crate slides 15 ft down the ramp and reaches the bottom pt B.

    I created an image for convenience:

    http://img689.imageshack.us/img689/2213/rampo.jpg [Broken]

    2. Relevant equations

    Kinematics (maybe), Work-Energy and Impulse-Momentum equations apply. However we are mostly dealing with Work-Energy and Impulse-Momentum. If we do not have to use kinematics, then we should use work-energy. In other words, the simplest approach is the best.

    3. The attempt at a solution

    I had trouble even beginning to attempt to solve the problem. My problem lies within the concept. I understand impulse and momentum, however I don't know why I can't extend it to problems like this (quirky ones). My first step was to establish a coordinate system. I chose a normal-tangential coordinate system. My positive x direction is parallel to the surface of the ramp and points in the direction the crate is moving down the ramp. My positive y-direction was chosen as perpendicular and "upwards" to/from the surface of the ramp.

    The next step I took was to draw the free-body diagrams of the crate and the ramp, separately, and then again together. On the crate, I had the weight force and the normal-force of the ramp on the crate. On the ramp, I had the normal force of the crate on the ramp, the weight of the ramp, and the reaction forces of the wheels.

    Next, I asked myself whether momentum is being conserved, and in which direction is it being conserved. This is where I came to a roadblock. I know there is a change in momentum because of the simple fact that the crate is released from rest, and ends up with some velocity. The problem wants me to find the distance the ramp travels... but how can the ramp even travel a certain distance if the normal forces between the ramp and crate cancel, and there is no friction to move the ramp in the other direction? I don't want to apply equations to the problem without knowing what is going on first.

    In other words, how do I proceed to solve this problem? I asked my instructor but he was very vague. He showed me a momentum equation which did still not make sense. I assume the problem involves relative velocities.

    Anyways, please help! And if my approach is wrong, please tell me how you would approach it and what coordinate system you would use?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 23, 2009 #2
    In the y direction gravity is acting on the sliding block but in the x direction there is no external force acting on the objects so momentum of the system will be conserved in the x direction.
     
  4. Nov 23, 2009 #3
    This makes sense. I solved for the velocity of crate A when it reaches the bottom of the ramp, which I obtained using mgh = (1/2) * m * v^2 where h can be found using the hypotenuse and the fact that the triangle is a 3-4-5 triangle.

    Is this correct to do?

    Also, I have a question regarding the momentum equation. I proceeded to write down the momentum equation for the x direction (pointing along the ground) and this is what I have:

    0 = m_c * v_c + m_r + v_r where 'c' and 'r' are the crate and ramp respectively. Now, do I use the velocity of the crate relative to the ramp for v_c? Or should I be using the velocity of the crate with respect to the ground?

    And, is the velocity I solved for above the velocity relative to the ramp? Or is it relative to the ground?
     
  5. Nov 23, 2009 #4
    The velocity you found is with respect to the ground. The velocities you use for the conservations of momentum are the velocities with respect to the ground.
     
  6. Nov 23, 2009 #5
    Okay. So now I have all these velocities, and I have no clue what the next step is. Not even a hint to give myself. My approach would normally be to find a relation between the distance the crate moves from the original position to the distance the ramp moves from the original position. However I have no clue how in the world you would do that. I'm not sure what other questions I should be asking myself here to lead me in the right direction. I'm not trying to get the answer handed out to me, but I am really struggling with this in my mind for some reason.
     
    Last edited: Nov 23, 2009
  7. Nov 23, 2009 #6
    You could find the acceleration of the small block and the time it takes for it to reach the bottom. With that you could find the distance the big block travels.

    Or you can use the center of mass.
     
  8. Nov 23, 2009 #7

    I thought about using the center of mass, which I have no idea how to even do for a problem like this. No such examples in my textbook exist nor has my instructor hinted at how to do this.

    As for the acceleration... This would make sense if the force causing the acceleration of the small block is also exerted on the ramp... I have no clue why this is even true though. The block has acceleration in the direction parallel to the surface of the ramp, so how could we even use this acceleration? How do we even find this acceleration? What integrals am I using?
     
  9. Nov 23, 2009 #8
    Acceleration acts down the ramp and is gsin0. So if you find how long it takes to reach the bottom you can use that to find the acceleration of the ramp (V-Vo)/t.
     
  10. Nov 23, 2009 #9
    Okay so I went through and solved the problem. However I get the wrong answer. I get 8 feet as the displacement... so here were my steps. I solved for the velocity of the crate at the bottom and got v_crate = -SQRT(18 * g) where g is gravity (g = 32.2 here).

    I wrote out the momentum equation in the x direction, and got 0 = mass_crate * velocity_crate * cos(theta) + mass_ramp * velocity_ramp_xdirection... I plugged all the given information in, and solved for v_ramp_x = 12.84 feet/sec... again my positive direction is from left-to-right. Then, I found out how long it takes for the crate to reach the bottom. I know its velocity is simply -SQRT(18 * g) and I know its acceleration is -g * sin(theta). I used v = v_0 + a * t in the x-direction to solve for t... I obtained t = 1.246 seconds... then I applied the same equation to the ramp in the x-direction, except this time plugging in the t we got, the velocity of the ramp and then solved for acceleration. The acceleration I got as 10.30 ft/s^2 in the x-direction. Then, I applied the equation s = s_0 + v_0 * t + (1/2) a * t^2 ... so s = 0 + 0 + .5*10.30*1.246*1.246... which gives 7.995 feet, or approximately 8 feet displacement...

    However the back of the book's answer is different, the back of the book obtains an answer of s = 4.8 feet. How did this happen? What did I do wrong? was it wrong to assume that the velocities in the momentum equation were relative to the ground?
     
  11. Nov 23, 2009 #10

    rl.bhat

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    Homework Helper

    Identify the forces acting on the block.
    i) mg*sinθ along -ve x-axis
    ii) mg*cosθ along -ve y-axis
    iii) normal reaction N due to the ramp along +ve y-axis.
    So the acceleration along x-axis = ax = - g*sinθ
    Αnd force along y-axis = m*ay = mg*cosθ - N.------(1)
    Now forces acting on the ramp are
    i) Weight of the ramp vertically downwards.
    ii) Normal force N due to the block along -ve y-axis.
    iii) Normal reaction N' from the ground.
    Resolve the normal force N into two components.
    Horizontal component = N*sinθ
    Vertical component = Ν* cosθ.
    N' = mg + Ν* cosθ. So there is no motion along this direction.
    Horizontal component produces an accelerstion in the ramp.
    So N*sinθ = Μ*α.---(2) Hence the sliding block pushes the ramp towards right.
    When the block displaces a distance y along -ve y -axis, ramp moves a distance X horizontally such that y = X*sinθ. Or ay = a*sinθ.---(3)
    From eq. 1, 2 and 3, eliminate N and find a.
    You have already found t. Using the value of a you can find X.
     
  12. Nov 24, 2009 #11
    nocturnus,

    I have worked this problem before. It was a test question when I was in dynamics.

    I made a youtube video about how to solve it.

    I hope this helps.

    http://www.alexpleasehelp.com/online/problems/rollingramp [Broken]
     
    Last edited by a moderator: May 4, 2017
  13. Nov 24, 2009 #12
    Thank you. I don't understand the notation. Well, I do, but honestly it's complicating things for me. Further more, you used the wrong gravitational constant. It's 32.2 for U.S. units, and 9.81 for SI.

    And to the previous poster... Thank you for showing me the F=ma version of the problem. However I am still trying to solve it using momentum and impulse equations.

    Allow me to make a slight modification to my earlier problem. Suppose I chose my positive x direction to be from left to right and along the ground, and the y direction to be positive upwards, perpendicular to the x direction. How is momentum conserved? and why? I do not understand what errors I am making. If you can help me out conceptually please do!
     
    Last edited by a moderator: May 4, 2017
  14. Nov 24, 2009 #13
    Okay, actually, regarding the F=ma version. I still get the wrong answer. I think I know why this method doesn't work, but then that leaves me with still no right way to solve this problem other than using relative velocities and the center of mass, which I do not know how to do.

    Here is why F=ma does not work in this case. If the ramp were fixed, then finding the time 't' for the crate to reach the bottom of the ramp can easily be found using g*sin(theta) as acceleration along the ramp (net acceleration, as it so happens) and then we can integrate to find v, then integrate again to find s (displacement), then if our x direction is positive from left to right parallel to the ramp, then we set -15 = g * sin(theta) * t .. However, the ramp is NOT fixed in this case. When the crate moves one way, the ramp will move the other way. That means the time it takes for the crate to reach the bottom of the ramp is actually LESS than the t we solved for in the fixed-ramp case I described.

    So then, now what? What is the most intuitive way for me to go about this? I guess I'm trying to figure out how I would figure this out on my own on a test or something. What steps would you take to solve this problem? What questions are you asking yourself?
     
  15. Nov 24, 2009 #14

    rl.bhat

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    According to me the acceleration is found by
    a = mgcosθsinθ/( M + msin^2θ)
    When I substituted the given values I got a = 8.31 ft/s^2.
    You are correct regarding the time t.
    Net acceleration along x-axis = ax' = (mg*sinθ + a*cosθ)
    Now find the time t and then X. I got the answer 4.8 ft.
     
  16. Nov 24, 2009 #15

    rl.bhat

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    You have seen two methods to solve this problem. Here is a simple third method using center of mass.
    Consider the ramp as a thin rectangular lamina of mass M.
    Let co-ordinate of B, A and c are (0,0), ( 12, 9) and (12,0)
    x-component of the CM ( centroid) of the ramp is ( 0 + 12 + 12)/3 = 8 ft.
    x-component of the CM of the block is 12.
    Hence x-component of the CM of the ramp and block system is
    CMx = (8*M + 12*m)/(M + m) ------(1)
    When block slides from A to B, ramp and block move through a distance x.
    Now the CMx is
    CMx = m*x + M(8 + x)/( M + m) ------(2)
    Since CMx remains the same, equate eq. 1 and 2 and solve for x.
     
    Last edited: Nov 24, 2009
  17. Nov 24, 2009 #16
    I used the correct value for g, I just used more decimal places than just 9.81. It's really 9.80665, but typically rounded to 9.81.

    I didn't use the English g in ft/s^2 because I prefer to work in the metric system.

    Ultimately it doesn't matter which system you use as long as you stay consistent throughout the problem. The answers at the end of my video are correct.


    what about my notation is confusing you?


    - Alex

    http://www.alexpleasehelp.com
     
  18. Nov 24, 2009 #17

    Okay. Whoa there. How did you get 0 + 12 + 12 over three? What process gave you these numbers? I know that the centroid of a triangle is 1/3rd the base distance from one of the sides, so it's 2/3rds of the base distance from the bottom left corner of the ramp. But that's how I calculated it. What are you doing right there?
     
  19. Nov 24, 2009 #18
    You are using relative velocities. The only reason I know you're doing it the right way is because any other way I tried it did not work. Therefore it would be extremely helpful if someone could explain it this way. This is the way the professor wants it. And this is the way I should be looking at it on a test. It is certainly less painful than F=ma. I don't understand WHY you're using relative velocities. I can't picture it. Can you please help me out with this? I'm struggling with this problem more conceptually than I am numerically.

    What coordinates are you using? What frame of reference? We suddenly switch from a diagram to relative velocities? I know the center of mass does not move, but if our frame of reference were the center of mass, we would simply see the crate have velocity v_c * cos(theta) in the x direction and v_r for the ramp in the x-direction. You wouldn't need to subtract anything. What is the error in my reasoning?
     
    Last edited: Nov 24, 2009
  20. Nov 24, 2009 #19

    rl.bhat

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    The co-ordinates of a centroid is [( x1+x2+x3)/3, (y1+y2+y3)/3]. The above numbers are x-cordiality of B, A and C.
     
  21. Nov 24, 2009 #20
    I said "relative velocities" because that's what you have to use to figure out what the ABSOLUTE velocity of the crate is.

    v.crate.x = v.crate*cos(theta) - v.ramp

    you have to subtract out the velocity of the ramp relative to the crate.

    Ultimately i'm using the absolute velocities of both the ramp and crate.
     
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