# Homework Help: Force, Skier

1. Sep 2, 2007

### lfwake2wake

1. The problem statement, all variables and given/known data

A 54 kg skier skis directly down a frictionless slope angled at 12° to the horizontal. Choose the positive direction of the x axis to be downhill along the slope. A wind force with component Fx acts on the skier. What is Fx if the magnitude of the skier's velocity is (a) constant, (b) increasing at a rate of 1.1 m/s2, and (c) increasing at a rate of 2.2 m/s^2

2. Relevant equations

Fx=M*ax
Fy=M*ay

3. The attempt at a solution

First I need to solve for the weight acting in the x direction. Putting the positive x direction relevant to the motion of the skier, I get:

a. Fx=M*g*cos(12)=517.64 N
b. Since having no net force on an object gives no acceleration, by setting the force of the wind equal to the force of gravity in the x direction, this is achieved.

Fx(wind)=517.64 in the negative x direction
Ftotal=Fxg-Fxw Since Ftotal=0, Fxg=Fxw

c. Now is where I'm having trouble. If it is increasing at a velocity of 1.1 m/s^2, then a=1.1 m/s^2. So to find the acceraltion in x, the net force is divided by the mass, I think.

F=ma ===>>> 517.64 N / 54 kg = 9.5858 m/s^2. So now, I think i need to subtract 1.1 m/s^2 from that answer and multiply again by 54 to find the solution. So 9.5858 +1.1=10.69 m/s^2 * 54kg=577.04 N

d. Same as above, but add 2.2 to 9.5858....636.44 N

I'm not sure if any of this is right. Thanks in advance for the help.

2. Sep 2, 2007

### lfwake2wake

Im sure you know this, but there is no (d), the answer i gave for (b) was a continuation of (a).

3. Sep 2, 2007

### rootX

"Fx=M*g*cos(12)=517.64 N"

but shouldn't that be sin?

4. Sep 2, 2007

### lfwake2wake

Yes, that definitely should be sin, other than that, does it look correct, I feel like I'm missing something on such a basic problem.

5. Sep 2, 2007

### rootX

c. Now is where I'm having trouble. If it is increasing at a velocity of 1.1 m/s^2, then a=1.1 m/s^2. So to find the acceraltion in x, the net force is divided by the mass, I think.

but net force is equal to m*1.1?

6. Sep 2, 2007

### lfwake2wake

I figured it out...Insted of adding the accelerations, they needed to be subtracted as Fx is the force of the wind in relation to the motion of the skier. Thanks for the help.

7. Sep 2, 2007

### learningphysics

instead of using accelerations like that, I'd stick with forces...

Find the net force you need for the given acceleration... then use the fact that:

Fgravity - Fwind = m * acceleration

It is just like the first part except instead of 0 on the right side, you have a different value...

8. Sep 2, 2007

### rootX

awesome!

never forget FBDs