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Normal Force

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DDS

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A block of mass m is placed on a wedge. The wedge is pushed along a horizontal surface with accelertation a, so that the block stays in place on the wedge, even though there is no friction between the block and the wedge. The normal force acting on the block during the acceleration is equal to:

mg/cos(theta)

i kind of guessed this answer but now when i go sti down and try to understand why is mg/cos(theta) not jsut simply mgcos(theta) i dont know why.

can anyone explain this to me
 

Doc Al

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Consider that for the block not to slide down the wedge it must be in vertical equilibrium.
 

DDS

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can anyone give me adetailed explanation its jsut my exam is in 2 days so not to bash you DOC but im looking for a detailed explanation
 

OlderDan

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DDS said:
can anyone give me adetailed explanation its jsut my exam is in 2 days so not to bash you DOC but im looking for a detailed explanation
Draw yourself a diagram of the block and the forces acting on it. There are only two. One is its weight and the other is the normal force acting perpendicular to the surface of the wedge. Without friction, the wedge cannot exert a force in any other direcion. The sum of those two forces is the net force acting on the block, causing it to accerlerate horizontally. What does that (and Doc's suggestion) tell you about the vertical component of the net force?
 

DDS

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it has a component i both the x and y plane
 

OlderDan

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DDS said:
it has a component i both the x and y plane
If the net force had a vertical component, the mass would accelerate in the vertical (y) direction. It is only accelerating in the horizontal (x) direction. That tells you the vertical component of the normal force cancels the vertical gravitiational force. The reusltant of the normal force and gravity must be the horizontal component of the normal force, and that force provides the acceleration of the block.
 

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