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Thursday, 23 February 2017

Law of Motion VI

Movable Pulley
The pulley which moves in itself is known as movable pulley

Key Point
In case of light movable pulley, acceleration of a body (pulley) goes on decreasing on increasing the number of strings attached to it. That is the body attached with two ropes moves with half the acceleration of the body attached with single rope.
Length of the string is constant z
x + 2y + z = L (Constant)
Differentiating both sides with respect to t (Time)
dx/dt + 2dy/dt + dz/dt = dL/dt
or, v1 + 2v+ 0 = 0 (z and L are constant) x
or, v1 + 2v= 0
Again differentiating both sides with respect to t
dv1/dt + 2dv2/dt = 0
or, a1 + 2a2= 0
or, a1 = - 2a2
That is acceleration of  m1 (body attached to a single string) is opposite and twice the acceleration of m2 (body attached to a double string)

Different Cases of Light Movable Pulley

i) Case - 1
Mass m1 is attached at one end of the string and the other end is fixed to a rigid support. Mass m2 is attached to the light movable pulley.
For vertical acceleration of m1
 m1g - T =  m12a (m1 is connected to a single string)
For vertical acceleration of m2
T– m2g = m2a
(m1 accelerates downwards and m2 accelerates upwards since m> 2m2)
For the clamp holding the first pulley
T1= 2T
For the clamp holding the movable pulley
2T - T2 = mpulleya
or, 2T - T= 0 (light pulley)
or, 2T = T2

ii) Case - 2

Mass m1 is attached at one end of the string and placed on a smooth horizontal surface and the other end is fixed to a rigid support after passing through a light movable suspended pulley. Mass m2 is attached to the light movable pulley.
For vertical equilibrium of m1
N = m1g
For horizontal acceleration of m1
T = m12a
For vertical motion of m2
m2g – 2T = m2a

iii) Case - 3
Mass m1 is attached to the movable pulley and placed on a smooth horizontal surface. One end of the string is attached to the clamp holding the pulley fixed to the horizontal surface and from its other end mass m2 suspended.
For vertical equilibrium of m1
N = m1g
For horizontal motion of m1
2T = m1a
For vertical motion of m2
m2g - T = m22a

iv) Case - 4
Mass m1 is attached to a movable pulley and placed on a smooth inclined surface. Mass m2 is suspended freely from a fixed light pulley.
For equilibrium of mperpendicular to incline plane T
N = m1gCosθ
For acceleration of mup the incline plane N T m2
2T - m1gSinθ = m1a
For vertically downward acceleration of m2
m2g - T = m22a

Newton’ 3 rd law or Law of Action and Reaction

Every action is opposed by an equal and opposite reaction.
or
For every action there is an equal and opposite reaction.
F12 is the force on the first body (m1) due to second body (m2)
F21 is the force on the second body (m2) due to first body (m1)


Numerical Application
Force on the first body due to second body (F12) is equal and opposite to the force on the second body due to first body (F21).
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