A 65 kg skier speeds down a trail, as shown in Figure 5-24. The surface is smooth and inclined at an angle of θ = 20° with the horizontal.

(a) Draw a free-body diagram for the skier. (Do this on paper. Your instructor may ask you to turn in this diagram.)

(b) Determine the magnitude of the normal force acting on the skier.

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  1. Draw the skier as a block on the slope with an origin at the center of the block.

    There is a force down at that origin of mg=65*9.8 = 637 Newtons.
    we want the components of that force in the normal to the slope direction and the down the hill parallel to the surface direction. The normal component is 20 degrees from the weight down vector.

    Component normal to the slope = 637 cos 20. That is down on the surface of the hill. The hill pushes up with an equal and opposite force so there is no acceleration normal to the slope.

    By the way the component parallel to the slope is 637 cos 70 = 637 sin 20 and that is not balanced by any other force so the skier will accelerate parallel to the slope down the hill.

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