[SOLVED] Laws of motion?

• Apr 30th 2009, 09:07 PM
fardeen_gen
[SOLVED] Laws of motion?
The upper portion of an inclined plane of inclination $\displaystyle \alpha$ is smooth and the lower portion is rough. A particle slides down from rest from the top and just comes to rest at the foot. If the ratio of smooth length to rough length is $\displaystyle m:n$, find the coefficient of friction.
• May 7th 2009, 10:48 PM
Work-energy
Hello fardeen_gen
Quote:

Originally Posted by fardeen_gen
The upper portion of an inclined plane of inclination $\displaystyle \alpha$ is smooth and the lower portion is rough. A particle slides down from rest from the top and just comes to rest at the foot. If the ratio of smooth length to rough length is $\displaystyle m:n$, find the coefficient of friction.

We do this by finding the work done by all the forces acting on the particle, and equating the total to zero.

Assume the total length of the plane is $\displaystyle (m+n)d$, the coefficient of friction is $\displaystyle \mu$ and the mass of the particle is $\displaystyle M$.

The acceleration of the particle at right angles to the plane is zero. So, resolving in this direction, the normal contact force $\displaystyle N$ between the particle and the plane is given by

$\displaystyle N = Mg\cos \alpha$

and, since the particle moves at right angles to this force, it does a zero amount of work.

During the second stage the friction force acting on the particle is $\displaystyle \mu N=\mu Mg\cos\alpha$. In this phase, the particle moves through a distance $\displaystyle nd$ down the plane, and so the work done on the particle by this friction force is therefore $\displaystyle -\mu Mgnd\cos \alpha$.

The particle falls through a total height $\displaystyle (m+n)d\sin\alpha$. The work done on the particle by its weight is therefore $\displaystyle Mg(m+n)d\sin\alpha$.

Since the particle starts and finishes at rest, the gain in its KE is zero, and therefore the total amount of work done on the particle is zero. So we have:

$\displaystyle -\mu Mgnd\cos \alpha + Mg(m+n)d\sin\alpha = 0$

$\displaystyle \Rightarrow \mu =\frac{(m+n)\tan\alpha}{n}$
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