Name: Pulley-Block System
Uploaded: 2019-07-19
Duration: 10 min 30 s
Description: We must notice the pulley block system to lift, move anything in real life. Do you know how we can estimate the forces and mechanism? Let us analyse this in...

Question

In the arrangement shown in figure the masses of the wedge $M$ and the body $m$ are known. The appreciable friction exists only between the wedge and the body $m$ , the friction coefficient being equal to $k$ . The masses of the pulley and the thread are negligible. Find the acceleration of the body $m$ relative to the horizontal surface on which the wedge slides.

Accepted Answer

Free body diagrams of wedge and body are as shown below,

Constraint relations,

As the body is in contact with the wedge, hence these two will move with the same acceleration in the horizontal direction. Let this acceleration is ${\vec{w}}_{M}$ (Acceleration of wedge with respect to ground)

Let acceleration of body with respect to wedge is given by ${\vec{w}}_{mM}$ , then by using the method of virtual work,

Work done by tension on wedge $+$ Work done by tension on body $= 0$

$\Rightarrow T w_{M} - T w_{mM} = 0 \Rightarrow w_{mM} = w_{M} \dots (1)$

From Newton's second law for the wedge in the horizontal direction,
$T - N = M w_{M} \dots (2)$ ,

Force acting on the body in the horizontal direction,
$N = m w_{M} \dots (3)$
For body in the vertical direction, (as the body is sliding with respect to wedge then type of friction will be kinetic in nature)
$m g - T - k N = m w_{mM}$

Substituting the values from equations $(1) and (3)$ ,

$\Rightarrow m g - T - k m w_{M} = m w_{M} \Rightarrow m g - T = m w_{M} + k m w_{M} \dots (4)$

Simultaneous solution of the equations $(2), (3)$ and $(4)$ ,
$w_{M} = \frac{m g}{(k m + 2 m + M)} = \frac{g}{(k + 2 + \frac{M}{m})}$
Hence, using the equation $(1)$ ,
$w_{mM} = \frac{g}{(2 + k + \frac{M}{m})}$ .

Now as the body has equal acceleration in horizontal and vertical direction then for acceleration with respect to ground,

${\vec{w}}_{m} = {\vec{w}}_{M} + {\vec{w}}_{mM}$

Since ${\vec{w}}_{M} = {\vec{w}}_{mM}$ and perpendicular to each other as shown in diagram,

Then, $w_{m} = \sqrt{2} w_{M} = \frac{g \sqrt{2}}{(2 + k + \frac{M}{m})}$ .

Important Questions on Laws of Motion

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