## How do I solve this incline problem?

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

### How do I solve this incline problem?

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
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jeff
Posts: 39
Joined: Mon Mar 05, 2018 11:54 pm

### Re: How do I solve this incline problem?

Hi AbakbakaJam,
AbakbakaJam wrote:
Tue Aug 14, 2018 12:11 pm
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
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.)

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.)

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

### Re: How do I solve this incline problem?

yes show the picture pls
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jeff
Posts: 39
Joined: Mon Mar 05, 2018 11:54 pm

### Re: How do I solve this incline problem?

AbakbakaJam wrote:
Tue Aug 14, 2018 8:12 pm
yes show the picture pls
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?

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

### Re: How do I solve this incline problem?

I got 5 and its correct

thx
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