Question

Three blocks of mass $m_1=2.0,m_2=4.0$ and $m_3=6.0$ $kg$ are connected by strings on a frictionless inclined plane of $60^{\circ },$ as shown in the figure. $A$ force $F=120$ $N$ is applied upward along the incline to the uppermost block, causing an upward movement of the blocks. The connecting cords are light. The values of tension $T_1$ and $T$, in the cords are

• A. $T_1=20N,T_2=60N$
• B. $T_1=60N,T_2=60N$
• C. $T_1=30N,T_2=50N$
• D. $T_1=20N,T_2=100N$

Answer 1

Answer: (A)

The free body diagram are given below:

For the motion of mass $m_1,m_2$ and $m_3,$ we have
$T_1-m_{1}g\cos 30^{\circ }=m_{1}a\dots$ (i)
$T_2-T_1-m_{2}g\cos 30^{\circ }=m_{2}a\dots$ (ii)
$F-T_2-m_{3}g\cos 30^{\circ }=m_{3}a\dots$ (ii)
Adding Eqs., (i), (ii) and (iii), we get
$F=\left(m_1+m_2+m_3\right)g\cos 30^{\circ }=\left(m_1+m_2+m_3\right)a$
$\Rightarrow \frac{F}{m_1+m_2+m_3}=g\cos 30^{\circ }=a$
$\Rightarrow \frac{120}{12}-10\times \frac{\sqrt{3}}{2}=a$
$\Rightarrow a=10\left(1-\frac{\sqrt{3}}{2}\right)$
Now, $T_1=m_1\left(a+g\cos 30^{\circ }\right)$
$=2\left(10-\frac{10\sqrt{3}}{2}+10\frac{\sqrt{3}}{2}\right)=20N$
$T_2=a\left(m_1+m_2\right)+\left(m_1+m_2\right)g\cos 30^{\circ }$
$=6\times 10=60N$