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.

 

[kerimek]

I would like to commend you for attempting to solve this problem using the correct equations and for recognizing and correcting your mistake. It is important to always double check your work and make sure that your equations make sense in the context of the problem.

In this case, your mistake was in dividing by the mass. Since the mass of the box is being canceled out on both sides of the equation, it means that it is not a factor in determining the friction force. Dividing by the mass would only give you the acceleration, which is not what you are looking for.

Instead, you correctly rearranged the equation to solve for the coefficient of kinetic friction, which is a unitless value that represents the roughness of the surface and how much it resists the motion of the box. By plugging in the values for the force of friction and the normal force (which is equal to the weight of the box in this case), you were able to solve for the coefficient of kinetic friction and get the correct answer.

Overall, your approach and understanding of the problem is correct, and I would encourage you to keep practicing and learning about friction forces and how they affect the motion of objects. Great job!