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.
N
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.
NOW
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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(a) As a humorous bot, I don't have the capability to draw diagrams. However, I can describe a possible free-body diagram for the skier. The skier would have a downward force due to gravity (mg), a normal force (N) pointing perpendicular to the surface, a frictional force (Ff) opposing the skier's motion, and a component of the gravitational force along the incline due to the weight (mg*sinθ).
(b) The magnitude of the normal force (N) acting on the skier can be determined using Newton's second law. Since the skier is not accelerating in the vertical direction, the normal force must be equal in magnitude to the component of the gravitational force along the incline. So, N = mg*cosθ.
To determine the magnitude of the normal force acting on the skier, we need to consider the forces acting on the skier along the inclined surface.
(a) Free-body diagram:
In the free-body diagram, draw a dot to represent the skier. Draw a downward arrow representing the weight of the skier acting vertically downwards. Draw a perpendicular arrow representing the normal force acting perpendicular to the inclined surface. Finally, draw an arrow pointing up the inclined surface representing the component of the weight acting parallel to the surface, which we can call the gravitational force parallel to the incline.
(b) Determining the normal force:
The normal force is the force exerted by a surface perpendicular to the object. In this case, it is the force exerted by the inclined surface on the skier. Since the skier is on a smooth surface, there is no friction force. Therefore, the normal force will be equal in magnitude but opposite in direction to the gravitational force perpendicular to the incline.
The weight of the skier, acting vertically downward, can be determined using the formula: weight = mass x gravitational acceleration.
Since the skier has a mass of 65 kg, and the gravitational acceleration is approximately 9.8 m/s^2, we can calculate the weight as follows:
Weight = 65 kg x 9.8 m/s^2 = 637 N
Since the surface is inclined at an angle of 20° with the horizontal, we can determine the gravitational force perpendicular to the incline using the formula: gravitational force perpendicular to incline = weight x cos(θ), where θ is the angle of the incline.
Gravitational force perpendicular to incline = 637 N x cos(20°) = 600.27 N
Therefore, the magnitude of the normal force acting on the skier is approximately 600.27 N.