Question

in the figure M is fixed wedge masses m1 and m2 are such that m2/m1°in the pulley are light and smooth the connecting strings are light the inclined plane is also smooth find acceleration of the mass m2​

Answer 1

$a=\frac{(10m_2-3m_1)g}{5(m_1+2m_2)}$

1) We have,

fixed wedge of Mass M with two blocks as shown in figure.

Now,we  make FBD of both the masses (blocks) as shown .

Marking Tensions as shown in figure.

Applying Newton's First law on both the blocks, we get

$m_2g-T=m_2a\\ \\=>2m_2g-2T=2m_2a----(1)$

2) Now , again on mass m1

$2T-m_1gsin(37^{\circ})=m_1a----(2)$

3) Adding both the equations 1 and 2, we get

$a=\frac{(2m_2g-m_1gsin(37^{\circ}))}{5(m_1+2m_2)}\\ \\=\frac{(10m_2-3m_1)g}{5(m_1+2m_2)}$

And if m_2 is greater than m_1,then 'a' is positive.

That is, m_2 goes downward.

Answer 2

SOLUTION.

Let the mass of block hanging be m

Let the mass of the block on wedge be M

Tension in the string of small m be T

Tension in the string of big block is 2T

Component of gravity along the wedge= gsin37°

Let the small block travels down wards with velocity v and bigger block move upward with velocity u. Than using power method.

2T(u) + T(-v) = 0.....(power by the internal forces is zero.)

2u = v

differentiate it with respect to time

2(A) = a

A = a/2

acceleration of larger block = acceleration of small block/2

ACC to NLM

For smaller block.....

mg - T = ma........(1)

For larger block....

2T - Mgsin37° = Ma/2......(2)

Multiply first equation by 2 and add with the 2 equation

2mg - Mgsin37° = (2m + M)a/2

2mg - Mg(3/5) = (2m + M)a/2

g(10m - 3M)/5 = (2m + M)a/2

2g(10m - 3M)/5(2m + M) = a

#answerwithquality

#BAL