Question

Three blocks of masses $2\text{kg}$, $4\text{kg}$ and $6\text{kg}$ arranged as shown in figure connected by string on a frictionless incline of ${37}^{\circ }$. A force of $120\text{N}$ is applied upward along the incline to the uppermost block. The cords are light. The tensions $T_1$ and $T_2$ in the strings are $(g=10{\text{m/s}}^2,\sin {37}^{\circ }=\frac{3}{5})$

• A. $T_2=60\text{N}$
• B. $T_2=36\text{N}$
• C. $T_1=8\text{N}$
• D. $T_1=20\text{N}$

Answer 1

Answer: (A), (D)

$T_1=20\text{N}$

Given,
Force applied along the inclined plane, $F=120\text{N}$.

Considering the blocks $2\text{kg}$, $4\text{kg}$, $6\text{kg}$ and string as a part of single system.


Net upward force acting along the inclined plane is,

$F_{net}=F-12g\sin {37}^{\circ }$

$\Rightarrow F_{net}=120-120\times \frac{3}{5}=48\text{N}$

So, the net upward acceleration,

$\Rightarrow a=\frac{F_{net}}{M_{system}}=\frac{48}{12}$

$\therefore a=4{\text{m/s}}^2$

All the blocks will have some acceleration $a$ such that length of every string remains constant.

$\text{F.B.D of 4 kg}:$


Balancing force along the inclined plane:

$T_2-T_1-4g\sin {37}^{\circ }=4a...\left(1\right)$

Similarly for $6\text{kg}$ block we can write,

$F-T_2-6g\sin {37}^{\circ }=6a$

$\Rightarrow 120-T_2-6\times 10\times \frac{3}{5}=6\times 4$

$\Rightarrow 120-T_2-36=24$

$\therefore T_2=60\text{N}$

Substituting the value of $T_2$ in equation $\left(1\right)$, we get

$60-T_1-4\times 10\times \left(\frac{3}{5}\right)=4\times 4$

$\Rightarrow 60-T_1-24=16$

$\therefore T_1=20\text{N}$

Hence, options (c) and (d) are the correct alternatives.