Question

A triangular prism is placed in gravity free space.
(i) Make its $FBD$
(ii) Find resultant force on block.
(iii) Find magnitude of force to be applied on prism to make it in equilibrium.

Answer 1

$\left(i\right)FBD=$ The prism has $2$ angle equal to ${60}^0$

So$,$ third angle$=$ ${180}^0-{120}^0={60}^0$

So$,$ the triangle becomes equilateral $△.$

Resultant force $=$ $\sqrt{{\left(100\right)}^2+{\left(100\right)}^2+2\left(100\right)\cos {60}^0}$

$\sqrt{3\times {\left(100\right)}^2}=100\sqrt{3}N$

$\left(ii\right)$ Both forces are equal $,$ so Resultant force will bisect the angle$.$ Resultant force $=100\sqrt{3}N$ at ${30}^0$

$\left(iii\right)$ Magnitude of force to be applied to make it in equilibrium$.$

Same resultant force $=100\sqrt{3}N$ but in opposite direction