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Homework Help: Block on a frictionless, accelerating, inclined plane

  1. Sep 26, 2010 #1
    1. The problem statement, all variables and given/known data
    A 2.0 kg block rests on a frictionless wedge that has a 60 deg incline and an acceleration [tex]\vec{a}[/tex] to the right such that the mass remains stationary relative to the wedge. a) Draw the free-body diagram of the block and use it to determine the magnitude of the acceleration.

    2. Relevant equations
    3. Picture on right. My attempted diagram on left
    http://img707.imageshack.us/img707/2420/physicsm.jpg [Broken]

    Uploaded with ImageShack.us

    3. The attempt at a solution
    So my main problem is determining what forces are acting on the block, and drawing the diagram. When I tried it, I thought that the normal force would be increased due to the acceleration of the wedge, but I am certain I have done this wrong. So, I drew a diagram of forces acting on the block which I said were the normal force, perpendicular to the wedge, and the force of gravity acting downwards. My coordinate system was setup so gravity pointed in the -y direction, and x was pointing straight to the right. I said that [tex]\sum(F_x)=F_n*\cos(30)=(2.0 kg)*\vec{a}[/tex] and [tex]\sum(F_y)=F_n*\sin(30)-F_g=0[/tex]. I used the summation of forces in y to find the normal force, which I plugged into the first to find acceleration. I got a normal force of 39.24N and an acceleration of 16.9914 m/s/s.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 26, 2010 #2
    Uh, we're going to need your question and attempt at the very least.

  4. Sep 26, 2010 #3
    Sorry, I hit enter too soon
  5. Sep 26, 2010 #4
    So you need to calculated the force parallel to the ramp.

    F = (m*g) / sin(theta) = (2*9.8)/sin60

    Now you have the force acting down the ramp, you can use F=ma to get the acceleration.

  6. Sep 26, 2010 #5
    So you're saying that since gravity is the only force pulling the block down, the acceleration component must be opposite that? I understand that. using that equation, you get an acceleration down the ramp of 8.495709211 m/s/s. Using trigonometry, a = ax/cos(60) where ax is 80495719211 m/s/s. that gives me 16.9914 m/s/s which is the same answer I already got. So is that the right answer?
  7. Sep 26, 2010 #6
    F = (m*g) / sin(theta) = (2*9.8)/sin60 = 16.97N of force acting down the ramp - this is the force acting down parallel to the ramp.

    F = ma = 16.97 = 2*a


    16.97 / 2 = 8.49m/s^2

    Acceleration = 8.49m/s^2

    I don't know why you do that final trig at the end. using sin(theta) above compensates for the slope of the ramp.

    Your free body diagram needs some modification to show the force acting parallel to the ramp downwards, which is what is / you are using to calculate acceleration.
  8. Sep 26, 2010 #7
    A useful tool that will save you when you get into more complex physics is called "translating your coordinate system". You can put your coordinate system anyway you want, it does not change the physics of the problem. With that said, if you place your x axis on the slope of the triangle block and your y axis perpendicular to your x, having your block at the origin, than you can simplify the problem a good amount.

    [PLAIN]http://img801.imageshack.us/img801/3856/blocke.png [Broken]

    Summing the forces is quite easier isn't it

    Forces in Y = N - mgcos(t) = 0

    Force in x = mgsin(t) = ma;

    now solve
    Last edited by a moderator: May 4, 2017
  9. Sep 26, 2010 #8
    Thanks fizzynoob, like I said, he should also show the force acting down the ramp as that is what you are using to calculate the acceleration. Just me being picky with the drawing styles.
  10. Sep 26, 2010 #9
    >>force acting down the ramp
    Did i not incorporate that with the force of gravity? gravity is the only force that has an x component.

    >>Just me being picky with the drawing styles.
    Yeah i realize that everyone on here has there own style, which corresponding to there professors style. It would be nice if all the professors would just create a standard so theres no confusion
  11. Sep 26, 2010 #10
    There is a standard, the force acting down the ramp does involve gravity, but it is the amount of force out of the weight which acts parallel down the ramp.

    This link shows the forces of such a block on a ramp:

    This is the way you should represent the vector forces. He doesn't need the normal force so needn't show it, but he's using the shear force and so it does need to be on the diagram.
    Last edited by a moderator: Apr 25, 2017
  12. Sep 26, 2010 #11
    >>There is a standard, the force acting down the ramp does involve gravity, but it is the amount of force out of the weight which acts parallel down the ramp.

    yes which i put

    >>Force in x = mgsin(t) = ma;

    I didn't put the different components because it was just a FBD to show the magnitudes of the forces.

    >>He doesn't need the normal force so needn't show it
    It does not contribute because there no friction, but why start developing bad habits now? Its good practice to draw out the forces, even if they are irrelevant. Plus they help you understand the physics of the problem
    Last edited: Sep 26, 2010
  13. Sep 26, 2010 #12
    I see your x now in the bottom. Apologies.
  14. Sep 26, 2010 #13
    I understand that the 8.4957 m/s/s is the acceleration of the block down the ramp. I don't believe that is the acceleration that I want, though. The question is asking for the acceleration of the wedge parallel to the ground. That's why I did the trigonometry at the end
    Last edited: Sep 26, 2010
  15. Sep 26, 2010 #14
    Well in that case, look at your FBD (free body diagram). The only force that contributes to the movement of the wedge in the x direction is from the normal of the block onto the wedge.

    sum the forces in the x for the wedge

    Forces in X = -Nsin(t) = -ma; // a being the acceleration of the wedge

    Now the normal is relevant to the problem.

    To find the normal you can sum the forces in the Y direction

    Forces in the Y = Nsin(t) - mg = 0;

    solve for N, than plug that normal into the force equation for the x direction
  16. Sep 26, 2010 #15
    What about component of the acceleration parallel to the ramp? That will be the greatest here and should be less than the figure parallel to the ramp, not more. The acceleration figure we have already does not have the normal included but it has a horizontal component. Using my phone at the moment, will be back later hopefully.
  17. Sep 26, 2010 #16
    I was referencing his fbd in his first post, translating the cooridnate system wont help when dealing with the wedge. Therefor the only component that results in the wedge moving is from the normal force due to the block

    thats why I put, -Nsin(t) =-ma
  18. Sep 27, 2010 #17
    Thanks for the replies. @Fizzynoob- that's the approach I took when I did it the first time, except I summed up forces for the block because the mass of the wedge is unknown.
  19. Sep 27, 2010 #18
    Neither the block or the wedge is moving.

    The question states an acceleration is in place to hold the block so it neither ascends or descends.

    The acceleration we have calculated is how fast it would descend if allowed to freely drop down the ramp. To keep the block steady you would have to apply that acceleration in the opposite direction. In your drawing, the wedge is the wrong way around. It should be with the right angle corner on the right hand side, so the acceleration is pushing on the block up the ramp.

    In this case the acceleration down the ramp is 8.4957m/s^2, so you would have to apply -8.4957m/s^2 up the ramp to compensate and keep it steady.

    Now you need to calculate the right hand horizontal component. Remember, it has to compensate for the vertical acceleration also.
    Last edited: Sep 27, 2010
  20. Sep 27, 2010 #19
    You can find the mass of tge wedge by summing the forces in the y direction I believe
  21. Sep 27, 2010 #20
    The block or the wedge doesn't move. The acceleration applied has to counteract the acceleration down the ramp.
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