anonymous
  • anonymous
a 94.0kg crate is placed on an inclined ramp. when the angle ramp makes with the horizontal is increased to theta , the crate begins to slide down. . calculate the equations representing the components of the forces acting on the crate?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
First, draw a free body diagram. Can you do that?
anonymous
  • anonymous
|dw:1332917116844:dw|
anonymous
  • anonymous
Great! Now, we define a coordinate system. Let's define n as being normal to the surface and p as being perpendicular to the surface. This coordinate system is more useful than x and y. |dw:1332917337319:dw| Now, we need to sum the forces in the n and p directions.

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anonymous
  • anonymous
are the components in equlibrium>?
anonymous
  • anonymous
The components in the n-direction will sum to zero, but since the box MOVES along the p-direction, the forces will not sum to zero in this direction.
anonymous
  • anonymous
how do we know that the in the n direction the net force is equal to zero
anonymous
  • anonymous
Consider the following expression for the forces in the n-direction of a box sitting on a flat surface. \[\sum F_n = m \cdot a = N - mg\]Since, \(N = mg\) by definition in this case, \[m \cdot a = 0 ~~~ \therefore \sum F_n = 0\]
anonymous
  • anonymous
i know the sum of the force in the x direction is mgsintheta- static friction=?
anonymous
  • anonymous
Essentially, we know this because gravity is is the only external force acting in the n-direction. The normal force between the box and incline opposes this gravitational force. The object object doesn't move in the n-direction for the same reason we don't move into the sidewalk as we walk down the street! Are we considering friction here? If we are\[\sum F_p = m \cdot a \rightarrow m \cdot a = mg \sin(\theta) - F_f\]
anonymous
  • anonymous
u set it equal to ma bc the object is moving
anonymous
  • anonymous
Correct. Remember we discussed this last weekend.

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