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happyjoe
Oct24-07, 06:08 PM
Hey everyone, i'm new here, and just wanted to say hi first. Second, I was wondering if anyone would so kind as to help me with this problem.

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

An 80kg skiier has a force of 200N exerted on him down the slope. Now i've already found the acceleration of 2.5m/s^2, however, now i need to work out whether the slop is more than 45 degrees. I understand I need to use the gsincos equation, but i'm getting nowhere in using it. If anyone could explain how I'm able to find whether the slope is greater than 45 degrees or not, then i'd really appreciate it. Thanks


2. Relevant equations

Not sure if im on the right track but.

Sincos = a/g = 2.5/9.8

3. The attempt at a solution

pooface
Oct24-07, 06:15 PM
looks like you are on the right track.

you can also solve this by using the forces. weight^2=net force^2+normal force^2 for inclines.

EDIT: sorry had made a mistake in the formula.

happyjoe
Oct24-07, 06:40 PM
looks like you are on the right track.

you can also solve this by using the forces. weight^2=net force^2+normal force^2 for inclines.

EDIT: sorry had made a mistake in the formula.


Thanks for your help! :D Im not sure if im doing it right still though as when I do the first equation of sincos = acceleration/gravity I get 0.255, and when I do the second equation that you gave me I get weight ^2 = 614,656kg, net force ^2 = 40,0000 + normal force^2 = 40000. The answer I keep getting is 7.68 which may be correct, but Im not sure if it is.

Thanks! :D

pooface
Oct24-07, 07:07 PM
do arcsin of a/g:

Check your equation in the notes again. I dont know what this 'sincos' is.

kg is not a weight quantity, it is mass. To convert kg to weight in newtons you must first multiply the mass in kg by acceleration due to gravity which is 9.81m/s^2.

The second formula is the pythagorean formula. The forces create a triangle.
see this diagram. http://www.cheops-pyramide.ch/khufu-pyramid/great-pyramid/inclined-plane.gif

Gm is the weight, N is normal force.