Determining Coefficient of Kinetic Friction on an Inclined Slide

[just.karl]

A child goes down a playground slide with an acceleration of 1.16m/s^2. Find the coefficient of kinetic friction between the child and the slide if the slide is inclined at an angle of 31.0 Degrees below the horizontal.

I'm lost as to how to even set up the equations and what I need to find to get the answer.

 

[just.karl]

sorry I realized I posted in the wrong section, I can't find how to delete this but help will still be greatly appreciated.

 

[rohanprabhu]

first find out the force acting on the boy as he slides down. Next find out the force that would have been provided if only the gravitational force were used.

The difference in the forces will give the force exerted by friction. Use the formula:

<br /> f_s = \mu N<br />

to find \mu

 

[||spoon||]

well first off draw a diagram and make all the forces parallel and perpendicular to the direction of motion.

Remember that f=ma, if you still have trouble setting up the equation post what you have so far and we can go from there ;)

 

[just.karl]

how do I find the force acting on the boy if I don't know his mass?

 

[||spoon||]

if you just leave mass as m you will see later that it cancels out. Acceleration due to gravity on a plane is independent of mass.

 

[just.karl]

So the force if only gravitational force would be W=mg and the force acting on the boy would be F_k=-ma ?

 

[||spoon||]

not quite. The childs weight will be mg. This is acting vertically downwards and is not parallel or perpendicular to the childs motion. By resolving this vector into it's components para and perp to the childs motion you will get mgcos31 acting perp which will be equal to N. And you will get mgsin31 acting parallel to the motion.

Because the child is moving down the slide we know that mgsin31 is greater than the frictional force (mu*N)

So we can now set up the equation f=ma as follows.

Mgsin31-(mu*N) = 1.16m

Now solve for mu.

 

[just.karl]

when I did the algebra I came out with .345=u but I still have m/s^2 though so I must have done something wrong

 

[just.karl]

the second answer I got is 1.76 which doesn't seem right to me either..

I finally just understood why we take the sin and cos which are the x and y which then the slant can be made up for. That didn't make much sense what I said but works in my head. lol

 

[||spoon||]

hmm I get .463

Do you know what the correct answer is?

 

[just.karl]

Um.. huh.. no I don't know what the correct answer is. I think I'll go with my first answer since it makes more sense than the 1.76. I really appreciate help, thanks.

Are you by chance going to be on much longer? In case I need more help on my other problems?