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Object on an Incline

  1. Oct 6, 2004 #1
    Need help on this one:

    An object is fired up a frictionless ramp at 60 degrees. If the intial velocity is 35m/s, how ong does the object take to return to the starting point?

    I figured out the Initial y and x velocities.

    What i dont understand is how to use acceleration in this problem...doesnt it have to be equal to F-parallel?
     
  2. jcsd
  3. Oct 6, 2004 #2

    Pyrrhus

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    Well do the force analysis. What forces are acting on the x-axis? You know it's weight, but which amount of it?
     
  4. Oct 6, 2004 #3
    Umm, dont exactly get what you mean...it doesnt give its mass
     
  5. Oct 6, 2004 #4

    arildno

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    Format, you really ought to formulate Newton's 2.laws in the tangential&normal reference frame, rather than "x" and "y"
    The calculations are much easier that way (in particular, you only need to work with the tangential component of Newton's 2.law)
     
  6. Oct 6, 2004 #5
    i think ill stick to the way i was taught...seeing how i dont exacly understand what you mean lol
     
  7. Oct 6, 2004 #6

    arildno

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    What is the component of gravity along the inclined plane?
     
  8. Oct 6, 2004 #7
    Thats wut i dont know how to figure out... dont you need to use Fg = ma to answer that?
     
  9. Oct 6, 2004 #8

    Gokul43201

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    Do you know how to draw free-body diagrams ?
     
  10. Oct 6, 2004 #9

    arildno

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    You know that the weight is mg in the downwards direction, right?
     
  11. Oct 6, 2004 #10

    Diane_

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    OK, try it this way: you know that any number can be represented as a sum of two other numbers, yes? For instance, you can represent 10 as 6+4. You can also represent it as 7+3, 8+2, and an infinite number of other pairs.

    Vectors are the same way, except that you have to take direction into account. Generally, the most convenient way to decompose vectors is into perpendicular parts. This is why you were taught to take them apart into x and y, but there are times when there are better ways to do it.

    In this case, you are working on an inclined plane. You know that the object isn't going to leap up off of the plane and go soaring through the air - it's either just going to sit there or (in this case) slide along the surface of the plane. The most convenient way to handle this, then, is to decompose things into components parallel to and perpendicular to the surface of the plane. This is what arildno is talking about.

    Doing this gains you several things immediately. First off, because you know it isn't going to go flying, you know that the net force perpendicular to the plane is zero. The normal force cancels out any other perpendicular components. Secondly, since the initial velocity is parallel to the plane, you don't need to decompose it at all. All you have to do is find the component of gravity that acts parallel to the plane, and the problem devolves into a one-dimensional kinematics problem.

    Does that make sense?
     
  12. Oct 6, 2004 #11
    Yea i think im getting it...

    So to find the force of gravity parallel to the incline do i do something like this

    mg = (sin60)mg

    g = (sin60)g
    g = 11.31?
     
  13. Oct 6, 2004 #12

    Gokul43201

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    What you've written down is not a sensible equation. The RHS is correct, but what does the LHS mean ?

    Do you see that there is a problem with the equation you've written down ?
     
  14. Oct 6, 2004 #13
    its the Fg directly downwards?
     
  15. Oct 6, 2004 #14

    Pyrrhus

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    Remember

    This vectorial equation can give an amount of scalar equations depending on the dimensions.

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = m \vec{a} [/tex]

    For this problem 2 dimensions

    [tex] \sum F_{x} = ma_{x} [/tex]

    [tex] \sum F_{y} = ma_{y} [/tex]

    Yes weight is directed downwards.
     
  16. Oct 6, 2004 #15
    Sorry, i cant see where i went wrong

    Fg(down) = (Sin60)Fg
    ma = (sin60)ma

    I know you guys dont like giving direct answers but i would appreciate a little bit more lol
     
  17. Oct 6, 2004 #16

    arildno

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    Don't you understand why this is sheer NONSENSE??
    What do you think "=" means?
     
  18. Oct 6, 2004 #17

    Gokul43201

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    Okay here's the long answer. I'm not sure it's going to help in the long run because there are more basic holes in your understanding of math and physics.

    The forces acting on the object are : (draw a picture showing the forces, as you read this)

    (i) its weight (also known as gravitational force) acting downwards,
    (ii) the normal reaction from the incline, acting perpendicular to the incline.

    Since these forces are not in opposite directions, they will not cancl each other off.

    Now, we resolve the forces along and perpendicular to the direction of the incline. The normal reaction N, is already perpendicular to the incline, so there's nothing more to do. The weight (mg) has a component along the incline given by mg*sin60, and a component perpendicular to it, given by mg*cos60.

    Now all the forces have been resolved.

    There are two forces perpendicular to the plane and in opposite directions. Since the object does not move away from the incline (ie : it's motion has no component perpendicular to the plane), the net force in this direction must =0. This means that N = mg*cos60. For this question, this is not needed.

    Along the plane, there's only one force, given by mg*sin60 acting downslope. This force must give rise to an acceleration in that direction given by F = ma. So, we have mg*sin60 = ma. Or a = g*sin60 is the downslope acceleration (or upslope deceleration) of the block.

    Now you know what 'a' is. Use it in the equation of motion that relates velocity, acceleration and time.
     
  19. Oct 6, 2004 #18
    ok im wrong...i realise that. That doesnt change the fact that i dont know how to solve this question.
     
  20. Oct 6, 2004 #19
    Ah there we go...made a mistake with mg. You saw my steps...could of just said that Fg != ma. Thankyou gokul
     
  21. Oct 6, 2004 #20

    Pyrrhus

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    Format, arildno and Gokul pointed it out, and i did too.
     
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