Determine the point of application of third force for which the body (rod) is in equilibrium, when forces of 20N and 30N are acting on the rod as shown in figure

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70cm from point A

70cm from point C

60cm from point A

60cm from point B

answer is A.

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Detailed Solution

From condition of translational equilibrium
20+F=30 $\Rightarrow$ F=10N
Let 10N act at a distance of x from A
Now, condition of rotational equilibrium
About point C.
F₁x₁ = F₂x₂
30x20 = 10xx₂
x₂=60cm from C
(or) 60+10 = 70Cm from A

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