Question

Find the acceleration of rod A and wedge B in the arrangement shown in figure if the mass of rod equal
that of the wedge and the friction
and the friction between all contact surfaces is negligible and rod A is free to move
downwards only. Take angle of wedge as
37°

​

Answer 1

Answer:  

$\frac{9g}{25} ,\frac{12g}{25}$

Explanation:

1)Let us take the angle of wedge be $\theta=37\degree$

Figure is attached .

Let the acceleration of rod and wedge be 'a' and 'b' respectively (see figure)

2) By constraint motion , we can say

$acos(\theta)=bsin(\theta)\\ \\=>tan(\theta)=\frac{a}{b}$

Now, using Newton's Law on Rod A,we get

$mg-Ncos(\theta)=ma \\ \\Ncos(\theta)=mg-ma$

Also, on wedge B

$Nsin(\theta)=mb$

3) Dividing above equations,

$tan(\theta)=\frac{b}{g-a}\\ \\=\frac{1}{g/b-a/b}\\ \\=\frac{1}{g/b-tan(\theta)}\\ \\=>tan(\theta)=\frac{1}{g/b-tan(\theta)}\\ \\ =>\frac{g}{b} tan(\theta)-tan^2(\theta)=1\\ \\=>\frac{g}{b} tan(\theta)=tan^2(\theta)+1\\ \\=>\frac{g}{b} tan(\theta)=sec^2(\theta)\\ \\=>b=gtan(\theta)cos^2(\theta)=gsin(\theta)cos(\theta)$

4)Using $tan(\theta)=\frac{a}{b}$

$a=gsin^2(\theta)$

Now, Substituting $\theta=37^{\circ}$

$a=\frac{9g}{25},b=\frac{12g}{25}$

Hence, acceleration of Rod A is 9g/25 and that of wedge B is 12g/25.

Answer 2

Answer:  

Explanation:

1)Let us take the angle of wedge be  

Figure is attached .

Let the acceleration of rod and wedge be 'a' and 'b' respectively (see figure)

2) By constraint motion , we can say

Now, using Newton's Law on Rod A,we get

Also, on wedge B

3) Dividing above equations,

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