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
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 $
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 $\displaystyle \mu$:
$\displaystyle \mu = \frac{3600-450 \cdot 9.81 \cdot \frac12\ N}{450 \cdot 9.81 \cdot \frac12 \sqrt{3}\ N} \approx 0.3643$
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