A heavy iron bar, of weight $$W$$ is having its one end on the ground and the other on the shoulder of a person. The bar makes an angle $$\theta$$ with the horizontal. The weight experienced by the person is :

At equilibrium, the net moment on the bar would be zero.

Weight acts on the center of the rod (at length L/2 from either end).

Let the reaction force acting on the man's shoulder be R. This is the weight experienced by the man.

Balancing the Moment at the end of the rod on the ground:-

Moment due to the weight of the bar = W $$\times\ $$ $$\frac{L}{2}$$ $$\times\ $$ $$\cos\theta\ $$ (clockwise)

Moment due to reaction force on the shoulder = R $$\times\ $$ $$L$$ $$\times\ $$ $$\cos\theta\ $$ (anticlockwise)

At equilibrium, both of these moments would be equal

⇒ W $$\times\ $$ $$\frac{L}{2}$$ $$\times\ $$ $$\cos\theta\ $$ $$=$$ R $$\times\ $$ $$L$$ $$\times\ $$ $$\cos\theta\ $$

$$\therefore\ \ R\ =\ \frac{W}{2}$$ is the weight experienced by the man