# Work-Energy: Determining x- and y-components of wind force

1. Feb 28, 2014

### annas425

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

2. Relevant equations
Constant acceleration equations: Δx = v,xi *t + 0.5*(a,x)*t^2
Δy = v,yi *t + 0.5*(a,y)*t^2
K = 0.5*m*v^2
U = mgh
W,nonconservative forces = ∫ ∑ F dx
Conservation of Energy: K1 + U1 = K2 + U2
∑F = ma

3. The attempt at a solution (Note that in my notation, I denoted a subscript with a comma preceding it)
F,w = wind force
F,wx and F,wy are constant
∑F,x = -F,wx = m*a,x
∑F,x = -F,wx - mg = m*a,y

Δx = v,xi *t + 0.5*(a,x)*t^2 --> x,f = d1*cos(θ)
v,xi = v,a *cos(θ,1) = 27.8 m/s = 83.6*v,xi

Solve for a,x --> a,x = -0.199 m/s^2
-F,wx = m*a,x (Solve for F,wx)

Δy = v,yi *t + 0.5*(a,y)*t^2 (Solve for y,f)

Now I am stuck! Any help would be greatly appreciated…thank you in advance! :)

Last edited: Feb 28, 2014
2. Mar 1, 2014

### Staff: Mentor

You know everything apart from ay, where is the problem?

To write subscripts, you can use [noparse]ay[/noparse] -> ay. Alternatively, you can use LaTeX.

3. Mar 2, 2014

### annas425

Is this work correct?? Thank you in advance!

x=97.9cos(32)=83.02 m
y=97.9sin(32)=51.88 m
Also resolve Va into its x and y components:
Vx=28cos(7)=27.79 m/s
Vy=28sin(7)=3.412 m/s

Motion in x direction
Let Ax be acceleration in x direction
x=(1/2)(Ax)(t)^2 + (Vx)(t)
83.02=(1/2)Ax(3.02)^2 + (27.79)(3.02)
-.906=(1/2)Ax(3.02)^2
-.20=Ax
F=ma
Wind force in x direction = 80(-.20) = -16 Newtons

Motion in y direction
Let Ay be acceleration in y direction (positive Ay is acceleration down)
y=(1/2)(Ay)(t)^2 + (Vy)(t)
51.88=(1/2)Ay(3.02)^2 + 3.412(3.02)
41.58=(1/2)Ay(3.02)^2
9.12=Ay
Acceleration due to gravity is 9.8
9.8-9.12 = .68 (acceleration up)
Wind force in y direction = 80(.68) = 54 Newtons (blowing up)

(Energy the skier has when he lands)
Vfx will be his landing velocity in the x direction.
Vfy will be his landing velocity in the y direction.

Vfx = (Ax)(t) + Vx = (-.2)(3.02) + 27.79 = 27.2 m/s
Vfy = (Ay)(t) + Vy = (9.12)(3.02) + 3.412 = 31.0 m/s

K = kinetic energy when landing = (1/2)mv^2 = (1/2)(80)(27.2^2+31.0^2)
K = 68,000 J

N = Normal force when landing due to skiers mass = mgcos(32)
f = friction force = (2.5)(80)(9.8)cos(32) = 1660 Newtons

Force due to horizontal component of wind = 16cos(32) = 14 Newtons
Force due to vertical component of wind = 54sin(32) = 29 Newtons

Total force on skier = 1660+14+29 = 1700 Newtons

(Work-energy theorem)
Fz = K + (mgz)sin(32) (The last term is because the skier will continue to acquire energy from the force of gravity as he descends down the slope)
1700*(d2) = 68000 + (80)*(9.8)*(d2)*sin(32)
1700*(d2) = 68000 + 415*(d2)
1285*(d2) = 68000
d2 = 53 meters

4. Mar 2, 2014

### Staff: Mentor

I think you are supposed to ignore the wind between B and C (otherwise you would get a different normal force, too).

The numbers don't fit, and you cannot add horizontal and vertical forces like that.

There are a lot of units missing. I did not check the numbers (that's something a computer can do), but the approach is good.

5. Mar 2, 2014

### annas425

The numbers are good, I just checked them! Looking at the prompt again, it says to ignore the b component of the wind force (the component perpendicular to the ramp). So would the sum of the nonconservative forces be (Fw)a + f (the a component of the wind force and the friction force)? (Essentially, the sum of the forces in the a direction--the direction parallel to the ramp)?

6. Mar 2, 2014

### annas425

Also, is this a correct conversion from x-y to a-b coordinates, given the two relative coordinate systems?

7. Mar 2, 2014

### Staff: Mentor

Ah, right.
All your forces are conservative here.
Wind just adds another force to consider, noting magical.

The conversion looks fine.