# Can the coefficient of friction be as high as 15.5?

• EmanuelPaz
In summary, the problem involved finding the coefficient of friction for a 20kg box being pulled at an angle of 37 degrees with a force of 300N at a constant velocity. The equations used to solve it included finding the horizontal and vertical components of the pulling force, the normal force, and the sum of forces in the x direction. The result was a coefficient of friction of 15.5, which may not be a realistic value due to the approximate calculations used. It was also noted that the vertical component of the pull force may have been significant in this problem.
EmanuelPaz

## Homework Statement

I had a question on my test that went something like: "A 20kg box is being pulled at an angle of 37 degrees across the floor with a force of 300N at a constant velocity. What is the coefficient of friction?"

## Homework Equations

The equations I used to solve it were:
Fpx= Fp x cos(37
Fpy= Fp x sin(37
Where Fpx is the horizontal component and Fpy is the vertical component of the pulling force.
Then I solved for Fn(normal force) with the formula:
Fn= Fg(weight) - Fpy
Then for the sum of the forces in the x direction I did
Fpx - f(friction) = ma
But ma is 0 since constant velocity
so then I said Fpx = f

## The Attempt at a Solution

So from there I plugged in the value I got for Fpx which was 239.6 into f
And then I got:
f= mu(coefficient of friction) x Fn
So I got that Fn= 15.5 and I divided 239.5 by 15.5 and I got the coefficient of friction to be 15.5.

Right physics, wrong algebra/arithmetic.

EmanuelPaz said:

## Homework Statement

I had a question on my test that went something like: "A 20kg box is being pulled at an angle of 37 degrees across the floor with a force of 300N at a constant velocity. What is the coefficient of friction?"

## Homework Equations

The equations I used to solve it were:
Fpx= Fp x cos(37
Fpy= Fp x sin(37
Where Fpx is the horizontal component and Fpy is the vertical component of the pulling force.
Then I solved for Fn(normal force) with the formula:
Fn= Fg(weight) - Fpy
Then for the sum of the forces in the x direction I did
Fpx - f(friction) = ma
But ma is 0 since constant velocity
so then I said Fpx = f

## The Attempt at a Solution

So from there I plugged in the value I got for Fpx which was 239.6 into f
And then I got:
f= mu(coefficient of friction) x Fn
So I got that Fn= 15.5 and I divided 239.5 by 15.5 and I got the coefficient of friction to be 15.5.
Hello EmanuelPaz. Welcome to PF !

It's not a very realistic value for a coefficient of friction, but it appears that you worked the problem correctly.

I did not check your numerical values exactly, but I did an estimate using approximate values such as g ≈ 10 m/s2 , sin(37°) ≈ 0.6 , and cos(37°) ≈ 0.8 .

With those values I got μ = 12 . That may seem significantly different, but the approximate g value, together with the fact that the vertical component of the applied force nearly cancels the gravitational force means that my value for the normal force was somewhat too large. Therefore, I got a smaller value for μ .

Last edited:
EmanuelPaz
I seem to get the same answer as you

EmanuelPaz
μ = cos(θ) / (W/FP - sin(θ))

americanforest said:
Right physics, wrong algebra/arithmetic.
Can you please tell me what I did wrong in my algebra/arithmetic?

SammyS said:
Hello EmanuelPaz. Welcome to PF !

It's not a very realistic value for a coefficient of friction, but it appears that you worked the problem correctly.

I did not check your numerical values exactly, but I did an estimate using approximate values such as g ≈ 10 m/s2 , sin(37°) ≈ 0.6 , and cos(37°) ≈ 0.8 .

With those values I got μ = 12 . That may seem significantly different, but the approximate g value, together with the fact that the vertical component of the applied force nearly cancels the gravitational force means that my value for the normal force was somewhat too large. Therefore, I got a smaller value for μ .
Yes, I also do not think it is a very realistic value for the coefficient of friction. But if you say I worked it out correctly then even if I get it wrong I'll still get most of the points, so that's good. Thank you for the help.

EmanuelPaz said:
Yes, I also do not think it is a very realistic value for the coefficient of friction. But if you say I worked it out correctly then even if I get it wrong I'll still get most of the points, so that's good. Thank you for the help.
I don't think you made any mistake.

I think you can ignore the vertical component of the force entirely, and since the speed is constant the pull and friction forces are the same.

You still need to divide by the normal force to get the coefficient of friction so you can't ignore the vertical component.

My guess is that this problem started life as a simpler problem with a horizontal pull of 300N and someone changed it so the towing force was at an angle. The vertical component is almost big enough to lift the box off the ground.

dean barry said:
I think you can ignore the vertical component of the force entirely
By what logic? As CWatters points out, the vertical component of the pull is nearly enough to lift the box off the ground, so the normal force is quite small.
Of course, it could be that the pull angle is supposed to be below the horizontal.

haruspex said:
...

Of course, it could be that the pull angle is supposed to be below the horizontal.
The back of my envelope says that gives μ between 0.6 and 0.7 . -- much more reasonable.

## 1. Can the coefficient of friction really be as high as 15.5?

Yes, in some cases, the coefficient of friction can reach as high as 15.5. This typically occurs when there is a lot of friction between two surfaces, such as in the case of rubber on asphalt or rubber on rubber.

## 2. What factors affect the coefficient of friction?

The coefficient of friction can be affected by a variety of factors, including the smoothness of the surfaces, the amount of force applied, and the material properties of the surfaces.

## 3. Is a higher coefficient of friction always better?

Not necessarily. While a higher coefficient of friction may be beneficial in some situations, it can also cause excess wear and tear on surfaces and make it more difficult to move objects.

## 4. Can the coefficient of friction be greater than 1?

Yes, the coefficient of friction can be greater than 1. This indicates a very high level of friction between two surfaces.

## 5. How is the coefficient of friction measured?

The coefficient of friction is typically measured using a device called a tribometer, which applies a known force to two surfaces and measures the amount of friction between them.

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