# A Ladder Rests Against Frictionless Vertical Wall, With Its Upper End 6 M Above The Ground And The Lower End 4 M Away From The Wall. The Weight Of The Ladder Is 500 N And Its Centre Of Gravity At (1/3) Rd Distance From The Lower End. What Will Be The Wall's Reaction?

Sol:

N1 = Normal force from the wall to the ladder.

N2 = Normal force from the ground to the ladder.

Let friction force due to the ground f pass through the contact point at the ground.

sinA = $$\frac{3}{\sqrt{13}}$$

By balancing forces:

$$\Rightarrow N_{2} * 6 = 500 * \frac{4}{3}$$

Here 4/3 m is the normal distance of the ground contact point to the line of weight action force.

$$\Rightarrow$$ N2 = 111.11 N.

Therefore, the wall’s reaction will be 111.11 N.

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