The following parameters of the arrangement of Fig. 1.11 are available; the angle α which the inclined plane forms with the horizontal, and the coefficient of friction k between the body m1 and the inclined plane. The masses of the pulley and the threads, as well as the friction in the pulley, are negligible. Assuming both bodies to be motionless at the initial moment, find the mass ratio $$\frac{{{m}_{2}}}{{{m}_{1}}}$$ at which the body m2. (a) starts coming down
(b) starts going up;
(c) is at rest.

Solution:

Solution: At the initial moment, obviously the tension in the thread connecting m1 and m2 equals the weight of m2.

(a) For the block m2 to come down or the block m1 to go up, the conditions is

$${{m}_{2}}g-T\ge 0$$ and  $$T-{{m}_{1}}g\sin \alpha -fr\ge 0$$

Where T is tension and fr is friction which in the limiting case equals  $$k{{m}_{1}}g\cos \alpha$$ . Then

Or  $${{m}_{2}}g-{{m}_{1}}\sin \alpha >k{{m}_{1}}g\cos \alpha$$  or  $$\frac{{{m}_{2}}}{{{m}_{1}}}>k\cos \alpha +\sin \alpha$$ .

(b) Similarly in the case

$${{m}_{1}}g\sin \alpha -{{m}_{2}}g>f{{r}_{\lim }}$$ or,  $${{m}_{1}}g\sin \alpha -{{m}_{2}}g>k{{m}_{1}}g\cos \alpha$$  or,  $$\frac{{{m}_{2}}}{{{m}_{1}}}<\sin \alpha -k\cos \alpha$$ .

(c) For this case, neither kind of motion is possible, and fr need not be limiting.

Hence,  $$k\cos \alpha +\sin \alpha >\frac{{{m}_{2}}}{{{m}_{1}}}>\sin \alpha -k\cos \alpha$$.

## Mechanics!

#### A point traversed half the distance with a velocity v0. The remaining part of the distance was covered with velocity v1 for half the time, and with velocity v2 for the other half of the time error: Content is protected !!