Inclined plane, finding normal force


by GRice40
Tags: force, inclined, normal, plane
GRice40
GRice40 is offline
#1
Sep26-11, 01:31 PM
P: 20
1. The problem statement, all variables and given/known data
A 1160 kg car is held in place by a light cable on a very smooth (frictionless) ramp, as shown in the figure . The cable makes an angle of 31.0 degrees above the surface of the ramp, and the ramp itself rises at 25.0 degrees above the horizontal.

Find the normal force along with the tension of the cable.



2. Relevant equations
Fnormal = M*A


3. The attempt at a solution
I've found the tension by taking the mass X 9.8sin(25), then taking that divided by cos(31) which gave me a tension of ~5610

However, I don't know how to go about finding the normal force
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kreil
kreil is offline
#2
Sep26-11, 03:09 PM
kreil's Avatar
P: 518
I think you will find this very informative,

http://en.wikipedia.org/wiki/Incline...inclined_plane
GRice40
GRice40 is offline
#3
Sep28-11, 07:55 PM
P: 20
Ok, I understand that in essence, the Normal Force is going to be

Fnorm= m * a(cos theta), in this case, the theta is 25 degrees, and the mass is 1160 kg.

So:

Fnorm = 1160 * 9.8(cos(25)) = 10302.91 N

That's all fine and well, except that the answer is suppose to be 7410 N. I believe it has something to do with the relationship of the weight of the car compared to the tension, and if it's not that, then I'm completely baffled =(

GRice40
GRice40 is offline
#4
Sep28-11, 08:04 PM
P: 20

Inclined plane, finding normal force


Alright, just had an epiphany. The tension is, indeed, a force in the Y direction.

So here's what I came out with:

F(up) + F(tension in the y direction) - F(down) = 0

So, F(up/normal force) = F(down) - F(tension in the y direction)

Fnorm = 1160(cos(25)) - (5610(sin(31)) = 7413.55

That number is right around the correct answer that it gave me of 7410 N.

Appreciate all the help guys!


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