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    This is a past paper question from M1 OCR Jan 2005. Q.1
    A box of weight 100 N rests in equilibrium on a plane inclined at an angle α to the horizontal. It is given that sin α = 0.28 and cos α = 0.96. A force of magnitude P N acts on the box parallel to the plane in the upwards direction (see diagram). The coefficient of friction between the box and the plane is 0.25.

    (i) Find the magnitude of the normal force acting on the box.

    How do i find the magnitude of the normal force? Is there a specific formula?
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    (Original post by Chelsea12345)
    This is a past paper question from M1 OCR Jan 2005. Q.1
    A box of weight 100 N rests in equilibrium on a plane inclined at an angle α to the horizontal. It is given that sin α = 0.28 and cos α = 0.96. A force of magnitude P N acts on the box parallel to the plane in the upwards direction (see diagram). The coefficient of friction between the box and the plane is 0.25.

    (i) Find the magnitude of the normal force acting on the box.

    How do i find the magnitude of the normal force? Is there a specific formula?
    Just resolve it, so the weight is 100N, therefore the force acting on the box due to the weight is going to be 100cos(angle) which is 100x0.96 which is 96N.

    The normal reaction force is just R, which acts perpendicular to the box, since it’s in equilibrium, and not moving, that force must equal the magnitude of the resolved weight perpendicular to the box, acting downwards.
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