A 65-kg skier speeds down a trail, as shown below. The surface is smooth and inclined at an angle of 22° with the horizontal.Find the direction and magnitude of the net force acting on the skier.

The motion of skier is parallel to the slope therefore the the direction of the net force is also parallel to the slope.

There are two forces acting on the skier one is Weight (Mg) in downward direction and another one is Normal force (N) as shown in the diagram.

Now lets resolve the weight into two vector along horizontal and on vertical direction as shown in the diagram.

As we know that there is no motion of body in vertical direction so we can say that net force along vertical direction is zero.

$\therefore \sum F_{y}\: =\: N-Mgcos22=0$

$\therefore \ N=Mgcos22$

$\therefore \ N=69\times 9.8\times 0.93$

$\therefore \ N=628.9 \: N$

Now net force in horizontal direction is given by

$F_{x}=Mgsin22$

$F_{x}=69\times 9.8\times sin22$

$F_{x}=253\: N$

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