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A monkey of mass m clings to a rope over a fixed pulley . The opposite ends of the rope is tide to a weight of mass M lying on a horizontal plate neglecting friction, find the acceleration of both bodies relative to the plate and tension of the rope in the following case: (i)the monkey does not move with respect to rope.(ii)the monkey moves upwards with respect to rope with acceleration beta(iii) the monkey moves downwards with respect to the rope with acceleration beta |
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Answer» Solution :(i) If the monkey is at rest on the rope , then acceleratiion of both mass M and monkey are same , equal to a, (say) If T is tension in the rope , then the equation of motion of monkey is `mg-T=ma_(1)`….(1) and equation of motion of mass M is `T=Ma_(1)`....(2) Adding (1) and (2) mg`=(M+m)a_(1) `:. a_(1)=m/(M+m)G`....(3) (ii) When monkey moves upward with acceleration `beta`, the reaction m`beta` acts downward , THEREFORE EQUATIONS of motion now take the form (`a_(2)` is acceleration of mass, M RELATIVE to plate) `mg+mbeta-T=ma_(2)` (for monkey )....(4) `T=Ma_(2)`(for mass M)....(5) Adding , we get , `mg+mbeta=(M+m)a_(2)` `:. a_(2)=(m(g+beta))/(M+m)` `:.` The acceleration of mass relative to plate `a_(2)=(m(g+beta))/(M+m)`....(6) and acceleration of monkey M relative to plate `a_(2)=a_(2)-beta=(m(g+beta))/(M+m)-beta =(m(g+beta)-beta(M+m))/(M+m)=(mg+mbeta-mbeta-mbeta)/(M+m)=(Mg-Mbeta)/(M+m)` (iii) When monkey moves downward with acceleration `beta`, teh reaction `(R=mbeta)` acts upward, therefore the equations of motion take the form (`a_(3)` is acceleration of block relative to plate ) `mg-mbeta-T=ma_(3)`....(8) `T=Ma_(3)`....(9) Adding we get `mg-mbeta=(M+m)a_(3)` `:. a_(3)=(m(g-beta))/(M+m)` and acceleration of monkey relative to plate `a_(3)^(1)=a_(3)+beta=(m(g-beta))/(M+m)+beta=(mg-mbeta+Mbeta+mbeta)/(M+m)` `:. a_(3)^(1)=(mg+Mbeta)/(M+m)` |
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