A 60kg skier shown below is skiing down a 35 degree incline with a coefficient
of friction is 0.08. Determine the acceleration of the skier.
I tried solving with mgx = mgsin(35) and mgy = mgcos(35) . I don't understand how to use that to solve for the acceleration.
Theres a picture if i need to upload it
How do I solve this incline problem?
Re: How do I solve this incline problem?
Hi AbakbakaJam,
The other two forces that are acting on the skier is friction and the normal force. The next step is to write out the force equations for the x and y direction. Each force equation is:
(sum of all forces components in the x direction) = mass x acceleration in x direction
and the same for the y direction.
So for example, in the y direction, there is the normal force in the positive y direction, and the mgcos(35) in the opposite direction, so the y force equation is:
(sum of y forces) = mass times y acceleration
N  m g cos(35) = 0
so that gives us the normal force N! (which we need for the friction)
Do you see how to set up the x equation? Now that we have the normal force, the x equation will give us the final answer. What do you get for that?
(By the way, if you would like a picture showing how the forces are pointing just let me know.)
So at this point you have the components of the weight force that are parallel and perpendicular to the incline. (So the x direction is down the incline, and the y is perpendicular.)AbakbakaJam wrote: ↑Tue Aug 14, 2018 12:11 pmA 60kg skier shown below is skiing down a 35 degree incline with a coefficient
of friction is 0.08. Determine the acceleration of the skier.
I tried solving with mgx = mgsin(35) and mgy = mgcos(35) . I don't understand how to use that to solve for the acceleration.
Theres a picture if i need to upload it
The other two forces that are acting on the skier is friction and the normal force. The next step is to write out the force equations for the x and y direction. Each force equation is:
(sum of all forces components in the x direction) = mass x acceleration in x direction
and the same for the y direction.
So for example, in the y direction, there is the normal force in the positive y direction, and the mgcos(35) in the opposite direction, so the y force equation is:
(sum of y forces) = mass times y acceleration
N  m g cos(35) = 0
so that gives us the normal force N! (which we need for the friction)
Do you see how to set up the x equation? Now that we have the normal force, the x equation will give us the final answer. What do you get for that?
(By the way, if you would like a picture showing how the forces are pointing just let me know.)

 Posts: 3
 Joined: Sat Aug 11, 2018 7:56 pm
Re: How do I solve this incline problem?
sure, after you broke up the weight force into two components, it looks like this:
(I drew the skier as a box, sorry!) You can see in the y direction, the two forces have to cancel because there is no acceleration in that direction. That's how I got the equation in the previous post:
N  m g cos(35) = 0
which lets us find N.
We can do the same thing for the two forces in the x direction, but the forces won't cancel to zero! because there IS acceleration in the x direction. So we get, from F = m a in the x direction,
m g sin(35)  u N = m a
because the kinetic friction is equal to uN. And the rest is algebra! After you plug in the value for N in this last equation, the only unknown left is acceleration a, so you can solve for it.
What do you get for that?

 Posts: 3
 Joined: Sat Aug 11, 2018 7:56 pm