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Physics: Box sliding down an inclined plane

  1. Jun 4, 2004 #1
    i'm trying to learn this stuff on my own so please bear with me.
    here's the problem:

    you have two mass, a 10kg on a frictionless 35 degree inclined plane, which is attached to 20kg mass via a pulley cord to hang vertically. Whats the aceleration?

    Solution:

    for the box hanging freely they got this equation, 196N-T=20kg*acclerertion----i understand this

    To get the equation for the box on the plane they did this:
    98Nsin35=56N
    98cos35=80N

    out of these two results, 56n and 80N, they choose 56N to get the equation for the box which is T-56N=10kg*a
    my question is why they they choose to use 56N and not 80N which would have resulted in the equation T-80N=10*a

    continuing the solution the book has, they combined the two equations, 196N-T=20kg*a and T-56N=10kg*a to get this:
    140N30kg*a
    a=4.7 m/s/s I understand this part also.
     
  2. jcsd
  3. Jun 4, 2004 #2

    arildno

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    "To get the equation for the box on the plane they did this:
    98Nsin35=56N
    98cos35=80N

    out of these two results, 56n and 80N, they choose 56N to get the equation for the box which is T-56N=10kg*a
    my question is why they they choose to use 56N and not 80N which would have resulted in the equation T-80N=10*a"

    The tangent vector along the inclined plane has the form:
    [tex]\vec{T}=\cos35\vec{i}+\sin35\vec{j}[/tex]

    (Agreed?)

    Clearly, the weight of the block on the plane works in the vertical direction,
    that is:
    [tex]\vec{W}=-Mg\vec{j}[/tex]
    where M is the mass of the block.

    Now, clearly, only the component of the weight along the plane (i.e parallell to the plane) contributes to the block's acceleration.
    The component of the weight along the plane normal (90 degrees on the plane) is offset by a normal force from the ground.

    Try to establish the component of the weight along the inclined plane!
     
  4. Jun 4, 2004 #3

    BobG

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    I'm a little stumped as well. There's always a couple different points of view you can use to set these up, with one usually being easier to visualize than the others. A picture is worth a thousand words in a lot of cases, this being a good example. If the reader doesn't visualize the position of the pulley correctly, there is no way to solve the problem.

    If you cut the rope between the 10kg block and the pulley, there is only one force acting on the block - the force of gravity (9.8 m/sec^2) times the mass of the block to give you a 98 N force acting on the block with the direction of the force being straight down. The 10kg block can only move one direction - down the inclined plane.

    If you set the block's only possible direction of motion as the principal direction, the force is at a 55 degree angle to the only possible direction of movement. Multiplying the force times the cosine of 55 degrees (which happens to equal the sine of 35 degrees) provides the component of the force along the block's direction of movement. The force times the sine of 55 degrees gives you how much force is directed perpendicular to the plane's inclined surface and, being perpendicualr, it contributes nothing to the object's motion down the inclined plane. So, you could say the block is accelerating down the inclined plane at 3.5 meters/second^2.

    Since the author used the sine of 35 degrees as the force directed down the inclined surface of the plane, he obviously took a different point of view. Doesn't matter, sometimes the easiest point of view to take is a subjective matter.

    If you don't cut the rope, you have another force acting on the object. Without a picture, there's no way to know what direction that force is coming from.

    If the pulley is directly above the 10kg block, there is a 196N force straight up - directly opposite the force of gravity and there's no need to worry about the sine or cosine. So, I'm assuming this isn't the way the picture is drawn in the book (unless the author of the problem became a little confused).

    If the pulley is at the high end of the ramp, positioned so the rope runs along the inclined surface to the 10kg block, whatever force results from the 20kg block is directed parallel to the surface of the inclined plane from the bottom of the plane towards the top. Then the equations the problem used make sense. There is a 196N force pulling from the low end of the incline to the high end directly opposite the 56N force accelerating the 10kg block down the ramp.
     
  5. Jun 4, 2004 #4
    thanks alot, i understand it now
     
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