# Overcoming Friction Sliding Box.

• PEZenfuego
In summary, the problem involves finding the coefficient of kinetic friction between a 75 kg box and a ramp with an angle of 25 degrees. The sum of forces in the Y direction does not equal 0, causing confusion. However, by transforming the coordinates and using the correct angles, the correct answer of 0.061 can be found. In addition, a diagram can help simplify the problem.
PEZenfuego

## Homework Statement

A 75 kg box slides down a 25 degree ramp with an acceleration of 3.60 m/s/s

Find the coefficient of kinetic friction (I'll use "uk") between the box and the ramp.
Find the acceleration for a 175 kg box.

## Homework Equations

F=ma

Fg= mXg (g=9.81 m/s/s in my class)

uk= Fk/Fn

Solution: 0.061 and 3.61 m/s/s

This is kind of a curveball question compared to what we've had. The big difference is that the sum of the forces in the y direction does not equal 0...I am also pretty sure Fapp is 0 since there is no applied force.

## The Attempt at a Solution

I wish I could include a drawing because that could be where the problem is...here was my go:

Fnet=Fk+Fp

Fnet=m X a

m X a=Fk-Fp

(Fk=(m X a)+(Fp)

Fp=cos(theta)m X g

Fn=sin(theta)m X g

uk=Fk/Fn

uk=((m X a)+(cos(theta)m X g))/(sin(theta) X m X g)

uk=((75X3.6)-(cos(25)X75X9.81))/sin(25)X75X9.81

uk=3.01 which is obviously way off...

I haven't tried to find the answer to the second part yet.

Any help would be very much appreciated.

when you do these problems, put your x-axis flush with the ramp, that way the sum of the forces in the Y direction equals zero. It simplifies the problem.[PLAIN]http://img710.imageshack.us/img710/6459/blockp.png

If you do that than the forces in the Y = 0

Forces in Y = N - mg = 0

therefor N just equals mg, the x direction will than compensate for the coordinate transformation

Last edited by a moderator:
I know what I am doing wrong, but I do not know why I am doing it wrong.

I had two problems. Firstly, I should have subtracted Fk from Fp. I figured this out and it was a stupid mistake in the first place. My second problem was that I mixed up sin and cos in the problem. When I fixed those, I got the right answer. The problem is that I do not understand why that happened.

take a look
[PLAIN]http://img718.imageshack.us/img718/8605/blockvg.png

Fy = Fn - Fgcos(t) = 0
Fx = Fgsin(t) - Fk = ma

Fk = (uk)Fn = (uk)Fgcos(t)

Fx = Fgsin(t) - (uk)Fgcos(t) = ma

Last edited by a moderator:
Thank you so much. I need to pay more attention to my geometry, I switched those angles. Thanks for the diagram, that's exactly what I needed.

## 1. How does friction affect the sliding of a box?

Friction occurs when two surfaces rub against each other, creating resistance and making it harder for the box to slide. The amount of friction depends on the type of surface and the weight of the box.

## 2. What are some ways to reduce friction when sliding a box?

One way to reduce friction is by using a lubricant such as oil or grease on the sliding surface. Another way is by using wheels or rollers to decrease the surface area in contact with the ground.

## 3. Can the weight of the box affect the level of friction?

Yes, the weight of the box can affect the level of friction. The heavier the box, the more force is needed to overcome the resistance caused by friction. This is why it may be easier to slide a lighter box compared to a heavier one.

## 4. Does the type of surface the box is sliding on make a difference in friction?

Yes, the type of surface can make a difference in friction. Smooth surfaces tend to have less friction compared to rough surfaces. This is because rough surfaces have more bumps and imperfections that create more resistance when rubbed against.

## 5. How can I calculate the force needed to overcome friction when sliding a box?

The force needed to overcome friction can be calculated using the formula F = μN, where F is the force needed, μ is the coefficient of friction, and N is the normal force (weight of the box). The coefficient of friction can be found by dividing the force of friction by the normal force.

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