Question

In the system shown in figure, block A is released from rest. Find the following:
(a) The acceleration of both the blocks A and B.
(b) Tension in the string.
(c) Contact force between A and B​

Answer 1

Answer :

(a) We know that, In this type of questions we need FBD of this diagrams

FBD for small block

So, Now, Look in (2) and (3) diagrams

According to that We can make FBD for (2),(3) diagrams

N - 0 = ma = $Σ f_x$ _____(1)

mg - T = ma = $Σ f_y$ ______(2)

Now Add Equation (1) and Equation (2)

→ (N - 0) + (mg - T) = (ma) + (ma)

→ mg + N - T = 2ma _____(3)

Now, FBD For big block

[Look (3) attechment]

T - N = ma ______(4)

Now, Add equation (3) and (4)

→ (mg + N - T) + (T - N ) = (2ma + ma)

→ mg = 3ma

→ $\sf \frac{mg}{3m} = a$

→ $\sf \frac{\cancel{m}g}{3\cancel{m}} = a$

→ $\sf a = \frac{g}{3}$

So, Now we can say that

For Block (B) = $\sf a = \frac{g}{3}$

And For (A)

• $\sf a_N = \sqrt{a² + a²}$

• $\sf a_N = \sqrt{2a²}$

• $\sf a_N = \sqrt{2}a$

Hence, we have Acceleration of

Block (B) = $\underline{ \boxed{\bf a = \frac{g}{3}}}$

Block (A) = $\underline{\boxed{ \bf a_N = \sqrt{2}a }}$

(b) For Tension,

We can use equation (2)

mg - T = ma ______(2)

We Can put $\sf a = \frac{g}{3}$

$\sf mg - T = \frac{mg}{3}$

$\sf mg - \frac{mg}{3} = T$

$\sf \frac{3mg - mg}{3} = T$

$\sf \frac{2mg}{3} = T$

$\sf T = \frac{2mg}{3}$

Hence, Tension in string is $\underline{\boxed{\bf T = \frac{2mg}{3} }}$

(c) Contact Force between A and B

$\sf F = ma$

We can put a = $\sf a = \frac{g}{3}$

$\sf F = \frac{mg}{3}$

Hence, Contact Force between A and B is $\underline{\boxed{\bf F = \frac{mg}{3}} }$

I hope it helps you...

Answer 2

Explanation:

answer refer to the attachment

Hope it helps you