Question

Three blocks of equal masses (each 3kg) are suspended by weightless strings as shown in the figure above. If the applied force is 100N, then $T_1$ is equal to: $(g=10m/s^2)$

• A. 130 N
• B. 190 N
• C. 100 N
• D. 160 N

Answer 1

The correct option is A 130 N

$\text{Step 1: Identify all the forces and predict the direction of motion}$

As all blocks are connected to ceiling with rope, So acceleration of system will be zero.

$\text{Step 2: Draw F.B.D. of the block required and apply Newton's Second law}$

Applying Newton's second law on on block $A$ (Taking downward direction as positive)

$\Sigma F=ma$

$\Rightarrow 100N+mg-T_1=ma$ $....\left(1\right)$

$\text{Step 3: Solving above equation to get the required value}$

Given: $m=3kg$ and $a=0m/s^2$

From equation $\left(1\right)$

$100N+3kg\times 10m/s^2-T_1=3kg\times 0m/s^2$

$\Rightarrow 100+30-T_1=0$

$\Rightarrow T_1=130N$

Hence, Tension in the lowermost string $T_1$ is $130N.$

Option A is correct.