can anybody show me how to work out :

a mass of 450kg on a smooth concerete floor , a force of 3.6kn is required to move it up a slope of 30 degrees determine the coefficient of friction between the floor and mass .

many thanks

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- Apr 6th 2008, 03:16 AMjon_mrocoefficient of friction
can anybody show me how to work out :

a mass of 450kg on a smooth concerete floor , a force of 3.6kn is required to move it up a slope of 30 degrees determine the coefficient of friction between the floor and mass .

many thanks - Apr 6th 2008, 04:03 AMbobak
so a force of $\displaystyle 3600 N$ is required to move it up this hill at a constant velocity.

so all the forces must be balanced.

Make sure you understand how the friction force is directed. It is down the plane because friction opposed the direction of motion (which is up the plane in this case).

So the forces going down the plane are the masses weight and the friction force.

so we have $\displaystyle mg \! \ sin \theta + \mu mg \cos \theta = 3600 $

use exact values for sine and cosine and solve for $\displaystyle \mu $ - Apr 6th 2008, 04:55 AMjon_mrocoefficient of friction
thank you for your reply is the answer 17.01

- Apr 6th 2008, 05:28 AMearboth
How did you get this answer?

Take bobaks's equation, plug in all known values and solve for $\displaystyle \mu$:

$\displaystyle 450\ kg \cdot 9.81\ \frac{m}{s^2} \cdot \underbrace{\sin(30^\circ)}_{= \frac12} + \mu \cdot 450\ kg \cdot 9.81\ \frac{m}{s^2} \cdot \underbrace{\cos(30^\circ)}_{= \frac12 \sqrt{3}} = 3600\ N$

Solve for $\displaystyle \mu$:

$\displaystyle \mu = \frac{3600-450 \cdot 9.81 \cdot \frac12\ N}{450 \cdot 9.81 \cdot \frac12 \sqrt{3}\ N} \approx 0.3643$ - Apr 6th 2008, 06:34 AMbobak
also you should know that $\displaystyle \mu $ can only take values between 0-1. that should be written in your books somewhere.

- Apr 6th 2008, 08:20 AMMathstud28Thats not true is it
I thought that $\displaystyle \mu$ could be greater than one...such as soft rubber on soft rubber?

- Apr 6th 2008, 08:32 AMbobak
- Apr 6th 2008, 08:35 AMMathstud28Haha
Don't let what I say hold too much sway...I am only seventeen...and in honors physics...Math my is my forte...Physics...Is my nap time =D

- Apr 6th 2008, 09:20 PMtopsquark
You are correct. $\displaystyle \mu$ can be as large as we want. That is not to say that it is likely to be greater than 1. In most (ordinary) circumstances it won't be.

So does there seem to be an upper limit on $\displaystyle \mu$? Soft rubber on soft rubber does not give a valid $\displaystyle \mu$ (as I understand the concept here) because as the rubber "slides" it also deforms. Breakage of the materials, deformation, wearing, etc. while sliding means that we can't measure a $\displaystyle \mu$ for that pair of materials. And, of course, if it won't slide at all that doesn't mean that $\displaystyle \mu \to \infty$, it means that we can't measure $\displaystyle \mu$ for that circumstance.

-Dan