# Understanding Friction Force: A 15.0kg Box

• GreenPrint
In summary: The solution you came up with by rearranging the equation is correct and gives you the friction force of 4.6 N, and the coefficient of kinetic friction of 0.54.
GreenPrint

## 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!

GreenPrint said:
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.

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!

## 1. What is friction force?

Friction force is a force that resists the movement of an object when it comes into contact with another object or surface. It is caused by the roughness of the surfaces and the interlocking of microscopic bumps and ridges.

## 2. How is friction force calculated?

Friction force can be calculated by multiplying the coefficient of friction (a measure of the roughness of a surface) by the normal force (the force perpendicular to the surface). The formula is Ff = μFn, where Ff is the friction force, μ is the coefficient of friction, and Fn is the normal force.

## 3. How does the weight of an object affect friction force?

The weight of an object does not directly affect friction force. However, it does affect the normal force, which in turn affects the friction force. The greater the weight of an object, the greater the normal force and therefore the greater the friction force.

## 4. How does the surface area of an object affect friction force?

The surface area of an object does not directly affect friction force. However, it does affect the pressure between the two surfaces, which can affect the coefficient of friction. In general, a larger surface area results in a lower pressure and therefore a lower coefficient of friction, which reduces the friction force.

## 5. How can friction force be reduced?

Friction force can be reduced by using lubricants, such as oil or grease, between the two surfaces. These substances create a slippery layer, reducing the roughness of the surfaces and therefore reducing friction force. Another way to reduce friction force is by using smoother surfaces or by reducing the weight or surface area of the object.

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