Question

If the surface shown in the figure is frictionless, and $T_2=6\sqrt{3}N$ , find $F$.

• A. $120N$
• B. $100N$
• C. $120\sqrt{3}N$
• D. $100\sqrt{3}N$

Answer 1

The correct option is A $120N$

Let the common acceleration of the system be $a$.
FBD of the $3kg$ body


$T_2=3a$
$a=\frac{T_2}{3}=\frac{6\sqrt{3}}{3}=2\sqrt{3}m/s^2$

From the FBD of combined system of mass ($15+12+3=30kg$ ), $F\cos {30}^0$ will become the external force.


$F\cos {30}^{\circ }=ma=30\left(2\sqrt{3}\right)$
$\Rightarrow F=120N$