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In the arrangement shown, the apppreciable friction exisits only b/w the bodym and the wedge.The frictional coeff being equals to @. The masses of pulley and the thread are negligible. Show that the acc of body m w.r.t the horizontal surface on which the wedge slides is given by mg / [2m+@m+M]
In the arrangement shown, the apppreciable friction exisits only b/w the bodym and the wedge.The frictional coeff being equals to @. The masses of pulley and the thread are negligible. Show that the acc of body m w.r.t the horizontal surface on which the wedge slides is given by
mg / [2m+@m+M]
hi, fbd of m fbd of M let the acceleration of m in horizontal direction be a, then from fbd. N = ma..............................(1) the acce. of the M will also be a, so, T - N = Ma T- ma = Ma .....................................(2) from the constraint of the length of the cord, the vertical accel. of m will be the same as horizontal one, balancing the vertical forces on m , mg -T - @ma = ma mg -T -ma (@+1) =0 ......................................(3) putting T = Ma +ma from eq. (2) to (3) we get, mg - (M+m)a - ma (@+1) = 0 (2m+ M + @m) a = mg a = mg /(2m+ M + @m)
hi,
fbd of m
fbd of M let the acceleration of m in horizontal direction be a, then from fbd. N = ma..............................(1) the acce. of the M will also be a, so, T - N = Ma T- ma = Ma .....................................(2) from the constraint of the length of the cord, the vertical accel. of m will be the same as horizontal one, balancing the vertical forces on m , mg -T - @ma = ma mg -T -ma (@+1) =0 ......................................(3) putting T = Ma +ma from eq. (2) to (3) we get, mg - (M+m)a - ma (@+1) = 0 (2m+ M + @m) a = mg a = mg /(2m+ M + @m)
fbd of M
let the acceleration of m in horizontal direction be a, then from fbd.
N = ma..............................(1)
the acce. of the M will also be a,
so, T - N = Ma
T- ma = Ma .....................................(2)
from the constraint of the length of the cord,
the vertical accel. of m will be the same as horizontal one,
balancing the vertical forces on m ,
mg -T - @ma = ma
mg -T -ma (@+1) =0 ......................................(3)
putting T = Ma +ma from eq. (2) to (3) we get,
mg - (M+m)a - ma (@+1) = 0
(2m+ M + @m) a = mg
a = mg /(2m+ M + @m)
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