Question

A uniform rod AB of length $1.0m$ and mass $5.0kg$ leans on a frictionless vertical wall and a rough horizontal floor with end B touching the wall and end A at a distance of $40cm$ from the wall as shown in figure then find the normal reaction of the wall and normal reaction by the floor

• A. $10.9N,50N$
• B. $50N,10.9N$
• C. $10.9N,10.9N$
• D. $50N,50N$

Answer 1

The correct option is A $10.9N,50N$

Let normal reaction be wall be N${}_1$ and normal reaction by Floor be $N_2$ and frictional force acting at floor be f .

And let the angle that rod makes with horizontal be $\theta$

Cot$\theta =\frac{0.4}{\sqrt{1^2-{0.4}^2}}\simeq 0.436$

Equation of force in vertical direction $\to mg=N_2\to N_2=50N$

Equation of force in horizontal direction N${}_1$= f

Eqution of torque about centre $N_1\times 0.5\times sin\theta +f\times 0.5\times sin\theta =N_2\times 0.5\times cos\theta \to N_1=\frac{N_{2}Cot\theta }{2}\simeq 25\times 0.436=10.9$

Thus A is the correct answer .