Question

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

Answer 1

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Relation between acceleration of rod and acceleration of

wedge placed on ground, is

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1)Let us take the angle of wedge be 37

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

2) By constraint motion, we can say

Alcosa2sine

Tan = a1/a2

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

Mg-Ncos = ma1

Ncose = mg - ma1

Also on wedge

Nsin -ma2

Dividing above equations,

Tana2/(g-a1)

= 1/(g/a2 - a1/a2)

Tan = 1/(g/a2 - tano)

Solving this we get;

A2=gsin cose

4)Using Tan = a1/a2

A1 = gsin^2 0

A1 = 9g/25,

A2 = 12g/25

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Answer 2

Explanation:

)Let us take the angle of wedge be 37

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

2) By constraint motion, we can say

Alcosa2sine

Tan = a1/a2

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

Mg-Ncos = ma1

Ncose = mg - ma1

Also on wedge

Nsin -ma2

Dividing above equations,

Tana2/(g-a1)

= 1/(g/a2 - a1/a2)

Tan = 1/(g/a2 - tano)

Solving this we get;

A2=gsin cose

4)Using Tan = a1/a2

A1 = gsin^2 0

A1 = 9g/25,

A2 = 12g/25

\boxed{I \:Hope\: it's \:Helpful}