A small body was launched up an inclined plane set at an angle \( \alpha =15{}^\circ \) against the horizontal. Find the coefficient of friction, if the time of the ascent of the body is \( \eta =2.0 \) times less than the time of its descent.
Solution:
Case 1. When the body is lauched up:
Let k be the cofficeint of friction, u the velocity of projection and \( \ell \) the distance traversed along the incline. Retarding force on the block \( =mg\sin \alpha +kmg\cos \alpha \) and hence the retardation \( =g\sin \alpha +kg\cos \alpha \) .
Using the equation of particle kinematics along the incline, \( 0={{u}^{2}}-2(g\sin \alpha +kg\cos \alpha )\ell \)
Or, \( \ell =\frac{{{u}^{2}}}{2(g\sin \alpha +kg\cos \alpha )}\,\,\,\,\,\,\,\,\,(1) \) and \( 0-u-(g\sin \alpha +kg\cos \alpha )t \) or, \( u=(g\sin \alpha +kg\cos \alpha )t\,\,\,\,(2) \)
Using (2) in (1) \( \ell =\frac{1}{2}(g\sin \alpha +kg\cos \alpha ){{t}^{2}}\,\,\,\,\,\,\,(3) \)
Case (2). When the block comes downward, the net force on the body \( =mg\sin \alpha -kmg\cos \alpha \) and hence its acceleration \( =g\sin \alpha -kg\cos \alpha \) .
Let, t be the time required then, \( \ell =\frac{1}{2}(g\sin \alpha -kg\cos \alpha ){{{t}’}^{2}}\,\,\,\,\,\,\,(4) \)
From Eqs. (3) and (4) \( \frac{{{t}^{2}}}{{{{{t}’}}^{2}}}=\frac{\sin \alpha -k\cos \alpha }{\sin \alpha +k\cos \alpha } \)
But \( \frac{t}{{{t}’}}=\frac{1}{\eta } \) (according to the question),
Hence on solving we get \( k=\frac{{{\eta }^{2}}-1}{{{\eta }^{2}}+1}\tan \alpha =0.16 \).
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
Hỏi Đáp Vật Lý! được xây dựng trên WordPress