Physics Questions?

Question

A 61 kg skier speeds down a trail, as shown below. The surface is smooth and inclined at an angle of theta = 18 degrees with the horizontal.

a. Draw a free-body diagram for the skier. (Explain how would you draw something like this?)

b. Calculate the normal force on the skier.

c. Assuming that the snow is frictionless, calculate the acceleration of the skier.

Please show/explain all work clearly please.

Answer 1

(a) To draw a free-body diagram, you need to identify all of the forces on the object. These come in two flavors: forces from things that are touching the object (contact forces) and forces from things that are NOT touching the object (field forces).

You've only met one field force; that's gravity. So gravity puts a force mg on the object, directed towards the center of the Earth.

The skier is in contact with the ground, so there's one contact force. If the contact is with a surface, this force is usually expressed in two components; perpendicular to the contact surface (the "normal" force - normal simply means "perpendicular") and the component parallel to the surface (created by friction). You're told that the snow is frictionless, so the contact force has only a normal (or perpendicular) component.

So your free-body diagram will have two arrows: one pointing down, with magnitude mg, and the other pointing 18 degrees from the vertical (that is, perpendicular to the surface) with magnitude N.

(b) Apply a Cartesian coordinate system to your free-body diagram. Align the x axis with the slope, so your coordinate system is tilted. We'll call the direction perpendicular to the surface "up" even though it's not truly up.

You have two contributions to the net "vertical" force: the normal force (up) and a component of gravity, mg cos 18, down. There's no net acceleration (and no net force) in this dimension. So Newton's Second Law in this dimension is

F = 0 = N - mg cos 18

Plug in m, g, and cos 18 and solve for N.

In the "horizontal" dimension, you have but one force: mg sin 18. So Newton's Second Law for this dimension is

F = ma = mg sin 18

ma = mg sin 18

a = g sin 18

Plug in g and sin 18 to get the downhill acceleration.