Question

Relation between acceleration of rod
acceleration of wedge placed on ground, is
​

Answer 1

Answer:

A1 = 9g/25,

A2 = 12g/25

Explanation:

1)Let us take the angle of wedge be 37

Let the acceleration of rod and wedge be 'a1' and 'a2' respectively (see figure)

2) By constraint motion , we can say

A1cosΘ = a2sinΘ

Tan Θ = a1/a2

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

Mg – NcosΘ = ma1

NcosΘ = mg – ma1

Also on wedge

NsinΘ=ma2

Dividing above equations,

Tan Θ = a2/(g – a1)

= 1/(g/a2 – a1/a2)

Tan Θ = 1 / (g/a2 – tanΘ)

Solving this we get;

A2= gsinΘcosΘ

4)Using  Tan Θ = a1/a2

A1 = gsin^2 Θ

Now, Substituting Θ = 37

A1 = 9g/25,

A2 = 12g/25