Question

The three blocks as shown in the Figure , move with constant velocities. Find the velocity of each block at the instant when the relative velocity of A with respect to C is 400 mm/s upward and the relative velocity of B with respect to A is 300 mm/s downward.​

Answer 1

Given :

The three blocks are moving with constant velocities .

Relative velocity  of A with respect to C  = 400 mm per sec upward

Relative velocity of B with respect to A = 300 mm per sec  downward

To Find :

At  the give instant or situation , velocities of each block  = ?

Solution :

Since the length of string connecting mass A and pulley is constant , so the velocity of pulley is  equal and opposite to A .

So , with respect to that pulley we can write :

$x_1 + x_2 =constant$   -(1)

Differentiating eq -(1)  w.r.t. we get :

$\frac{dx_1}{dt}+ \frac{dx_2}{dt} =0$

Or , -( velocity of B relative to pulley ) + (velocity of C relative to pulley )  = 0

Or,$- ( V_P + V_B ) + ( V_C -V_P) = 0$

Now since $V_P =V_A$  , so :

$V_C - V_A = V_B + V_A$

Or, $2V_A = -V_B + V_C$     -(2)

Since , it is given that :

$V_A + V_C= 400$     -(3)

$V_B -V_A = -300$  -  (4)

Now adding eq 2 and 3 :

$2V_A +V_A + V_C = -V_B + V_C + 400$  

Or, $3V_A + V_B = 400$     -(5)

On subtracting eq  4 from 5 :

$3V_A + V_B - V_B + V_A = 400 - (-300)$

Or ,$V_A = \frac{700}{4} = 175 \frac{mm}{s}$

Putting the value of $V_A$ in eq (3) :

$175 + V_C= 400$

So, $V_C = 225 \frac{mm}{s}$

Putting $V_A$ in eq 4 :

$V_B - 175 = -300$

So, $V_B = -125 \frac{mm}{s}$

So, the velocities of the three blocks at given instants are :

Velocity of block A = 175 mm per sec

Velocity of block B = -125 mm per sec

Velocity of block C = 225 mm per sec