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I can work with questions relating to a stationary body on an inclined plane but when it starts moving I can't seem to work it out. Take for example, a body weight W being pushed up a plane inclined at @° to the horizontal with a horizontal force say dN. How much force does the plane exert on the body, that's the normal reaction, R? I know if the body was at rest on the plane R would be the product of W and the cosine of the inclination angle, is it the same here?

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  • $\begingroup$ @Farcher please dumb it down a little, I still don't get it $\endgroup$ – lekarane Sep 22 '16 at 8:28
  • $\begingroup$ I have drawn you a FBD for the body going up the slope. $\endgroup$ – Farcher Sep 22 '16 at 12:20
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Consider the forces perpendicular to the plane.
There is the normal reaction and components perpendicular to the plane of the weight and the applied force.
Applying Newton's second law perpendicular to the plane will give you a connection between these three forces.


Update

Here is the FBD for the body going up the slope.

enter image description here

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Draw a free body diagram (FBD) for the object. The 4 forces on the object are the pull of gravity $W=mg$, the applied force $N$, the friction force $F$ along the incline and the normal reaction $R$.

If the applied force $N$ has a component perpendicular to the plane, this will affect $R$.

If the object is moving with constant velocity along the incline, the net force on it is zero. If it is accelerating, the net force is $ma$.

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