Question

A 75 kg man holds a 25 kg uniform rod at an angle 60° as shown. He leans backward at an angle 60° so that his centre of gravity (centre of mass) 'P' is 0.5 m to the right of his feet. The length of rod is : Rod hand ΣΤΟ Vcds6o 2 1560 n Centre of XP gravity of man 5 feet 360º 60% rough 05m" ground (B) 6 m (D) 8 m (A) 4m (C) 7 m 1

Answer 1

Given:

Mass of the man, M₁= 75kg

Mass of the rod, M₂ =25 kg

Length of the rod, X₁= ?

Horizontal distance between man's leg and Centre of gravity, X₂ = 0.5kg

By the law of conservation of mass, we have,

M₁X₁=M₂X₂

Substituting the corresponding values to the above equation,

(25kg) (X₁) = (75kg) (0.5m)

$X_{1}=\frac{(75 kg)(0.5kg) }{25 kg}$

                                                                $\Rightarrow$X₁ = 1.5m

The center of gravity for the pole is, therefore (1.5m+ 0.5m) = 2m to the left of the farthest right edge of the pole.

Using trigonometric ratios,

$cos 60^{^{0}}=\frac{adjacent side}{hypotenuse}$

$\Rightarrow \frac{1}{}2=\frac{4}{l}$

Therefore, the length of the rod, l= (4 x 2)= 8m