Two touching bars 1 and 2 are placed on an inclined plane forming an angle α with the horizontal (Fig,. 1.10). The masses of the bars are equal to m1 and m2, and the coefficients of friction between the inclined plane and these bars are equal to k1 and k2 respectively, with $${{k}_{1}}>{{k}_{2}}$$. Find:

(a) the force of interaction of the bars in the process of motion;
(b) the minimum value of the angle α at which the bars start sliding down.

Solution:

(a) Let us indicate the positive direction of x-axis along the incline (Fig.). Figures show the force diagram for the blocks.

Let, E be force of interaction between the bars and they are obviously sliding down with the same constant acceleration w.

Newton’s second law of motion in projection form along x-axis for the blocks gives:

$${{m}_{1}}g\sin \alpha -{{k}_{1}}{{m}_{1}}g\cos \alpha +R={{m}_{1}}w\,\,\,\,\,(1)$$

$${{m}_{2}}g\sin \alpha -R-{{k}_{2}}{{m}_{2}}g\cos \alpha ={{m}_{2}}w\,\,\,\,\,\,(2)$$

Solving Eqs. (1) and (2) simultaneously, we get  $$w=g\sin \alpha -g\cos \alpha .\frac{{{k}_{1}}{{m}_{1}}+{{k}_{2}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}$$ and  $$R=\frac{{{m}_{1}}{{m}_{2}}({{k}_{1}}-{{k}_{2}})g\cos \alpha }{{{m}_{1}}+{{m}_{2}}}$$          (3)

(b) When the blocks just slide down the plane, w = 0, so from Eqn. (3)

$$g\sin \alpha -g\cos \alpha .\frac{{{k}_{1}}{{m}_{1}}+{{k}_{2}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}=0$$ or,  $$({{m}_{1}}+{{m}_{2}})sin\alpha =({{k}_{1}}{{m}_{1}}+{{k}_{2}}{{m}_{2}})\cos \alpha$$

Hence  $$\tan \alpha =\frac{{{k}_{1}}{{m}_{1}}+{{k}_{2}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}$$.

## 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 !!