Understanding Friction Force: A 15.0kg Box

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Homework Statement

A 15.0 kg box is released on a 30 degree incline and accelerates down the incline at .30 m/s^2. Find the firctiopn force impeding its motion. What is the coefficient of kinetic friction?

Homework Equations

simga F = ma
Force of friction = MU F_N

The Attempt at a Solution

SIGMA F_x = m a_x = F_g_x - F_fr_k = mg sin THETA - mg cos THETA MU

therefore

mg cos THETA MU = mg sin THETA - m a_x

I divided through by mass

g cos THETA MU = g sin THETA - a_x

I do not see what is wrong with this solution. This gave me 4.6 N. I'm suppose to get 69 N. I know that if I muliply 4.6 N by 15 (the mass that I canceled out) I get 69 N.

So apparently I don't know why I can not divide by the mass here and would like to know why...

Surpisingly enough when I rearanged for Mu I did this and got the right answer according to the back of the book

F_fr_k = 4.6 N = g cos THETA MU

therefore

MU = 4.6 N/(g cos THETA)

this gave me the right answer of .54

THANKS!

Those are subscripts by the way

F_fr_k
is the force of kinetic friction

and the all capital letters are one varialbe and are greek letters thansk!

 

[Doc Al]

SIGMA F_x = m a_x = F_g_x - F_fr_k = mg sin THETA - mg cos THETA MU

therefore

mg cos THETA MU = mg sin THETA - m a_x

I divided through by mass

g cos THETA MU = g sin THETA - a_x

I do not see what is wrong with this solution. This gave me 4.6 N. I'm suppose to get 69 N. I know that if I muliply 4.6 N by 15 (the mass that I canceled out) I get 69 N.

So apparently I don't know why I can not divide by the mass here and would like to know why...

There's nothing wrong with dividing by the mass when you're solving for μ. But the first part of the question asks for the friction force, which is μmgcosθ, not μgcosθ.