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 \( \tan \alpha =\frac{{{k}_{1}}{{m}_{1}}+{{k}_{2}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \).
In the arrangement of Fig. 1.9 the masses m0, m1, and m2 of bodies are equal, the masses of the pulley and the threads are negligible, and there is no friction in the pulley. Find the acceleration w with which the body m0 comes down, and the tension of the thread binding together the bodies m1 and m2, if the coefficient of friction between these bodies and the horizontal surface is equal to Consider possible cases
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