Can the coefficient of friction be as high as 15.5?

[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.

 

[americanforest]

Right physics, wrong algebra/arithmetic.

 

[SammyS]

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 .

 

[andrevdh]

I seem to get the same answer as you

 

[andrevdh]

μ = cos(θ) / (W/F_P - sin(θ))

 

[EmanuelPaz]

Right physics, wrong algebra/arithmetic.

Can you please tell me what I did wrong in my algebra/arithmetic?

 

[EmanuelPaz]

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.

 

[SammyS]

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.

 

[dean barry]

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.

 

[CWatters]

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

 

[CWatters]

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.

 

[haruspex]

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.

 

[SammyS]

...

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.