Question

Mass of wedge as shown in figure is M and that of the block is m. Neglecting friction at all the places and mass of the pulley, calculate acceleration of the wedge. Thread is inextensible.

• A. $\frac{mgsin\theta }{M+m(1-cos\theta )}$
• B. $\frac{mgsin\theta }{M+2m(1-cos\theta )}$
• C. $\frac{M+m(1-cos\theta )g}{msin\theta }$
• D. mg sin$\theta$

Answer 1

The correct option is B

$\frac{mgsin\theta }{M+2m(1-cos\theta )}$

Try to think in which direction the system moves

Okay

The wedge, seems it will go left and the block will slide on the incline

So far so good

Let's assume the wedge moves to the left by x meters.

So every point on x alone moves left by x meters.

So if the block was fixed on the wedge the centre of block would have moved to 0' and 0-0' will be x meters.

But the block isn't fixed

It moves down too

But by how much

Hmmmm ........

I know the string length is conserved or fixed

So if wedge moves by x

The string length between the wall and pulley is decreased by x

Then there has to be an equal increase is string length between pulley and block.

So the block will go down by x

$\Rightarrow$ 0” - 0' = x meters

So by the time wedge moves x meters in horizontal the block has moved from 0 to 0”

So 0 has moved (x - xcos$\theta$) leftwards and x sin$\theta$ downwards in same time. So if wedge has acceleration of a leftwards then block has an acceleration of a (1-cos$\theta$) leftwards and a sin $\theta$ downwards. oh now it's all easy after finding constraints.

Just draw free body drawing and write equations.

mg cos$\theta$ - N = masin$\theta$ ...(i)

mg sin$\theta$ - T = ma (1-cos$\theta$) ...(ii)

T (1 - cos $\theta$) + N sin $\theta$ = Ma ....(iii)

Substituting T & N from equation (i) & (ii) in (iii) (mg cos$\theta$ - ma sin$\theta$)sin$\theta$ - ma (1-cos$\theta$)} (1 - cos$\theta$) = Ma

mgsin$\theta$ - ma$si{n}^2\theta$ - ma$(1-cos\theta {)}^2$ = ma

mg sin$\theta$ + 2ma (cos$\theta$ - 1) = ma

$a=\frac{mgsin\theta }{\{M+2m(1-cos\theta \left)\right\}}$