Question

The tension in the string (in N) is:

• A. $4.6$
• B. $3.9$
• C. $6.2$
• D. $8.4$

Answer 1

The correct option is B $3.9$

After drawing the free body diagrams of the system , we can get four equations of motion:
$T=m_1{a}_1=1.3a_1..........\left(1\right)$
$m_3{a}_3=T-T\cos 37+N\sin 37=T-\frac{4}{5}T+\frac{3}{5}N$
$3.45a_3=\frac{T}{5}+\frac{3}{5}N.........\left(2\right)$
$m_{2}g\cos 37=m_2{a}_3\sin 37+N$
$12=0.9a_3+N......\left(3\right)$
$m_2(a_1+a_3)=m_{2}g\sin 37+m_2{a}_3\cos 37-T$
$1.5(a-1+a_3)=9+1.2a_3-T.........\left(4\right)$

$\left(2\right)\times \frac{5}{3}+\left(3\right)$ , then we get,
$\frac{T}{3}+12=6.65a_3.......\left(5\right)$

Using$\left(1\right),\left(4\right)$ becomes,
$9-2.15T=0.3a_3......\left(6\right)$

Now using $\left(6\right),\left(5\right)$ becomes,
$12+\frac{T}{3}=6.65\left(\frac{9-2.15T}{0.3}\right)=199.5-47.66T$
$48T=187.5$
$T=3.9N$