Question

a uniform rod of 30cm is pivoted at its center.a 40N weight is hung 5cm from left.from where 50N weight be hung to maintain equilibrium?

Answer 1

The 50N weight be hung at 23 cm to maintain equilibrium

Step-by-step explanation:

Given;

length of the uniform rod = 30 cm

center of the uniform rod = 15 cm

weight of 40N is hung at 5 cm mark

weight of 50 N will be hung at ?

0------5cm-----------------15cm-------------P---------30cm

↓ 10cm Δ xcm ↓

40N 50N

Take moment about the pivot point and apply the principle of moment

50N (x cm) = 40N (10 cm)

x = (400) / 50

x = 8cm

P = x cm + 15 cm

P = 8 cm + 15 cm

P = 23 cm

Therefore, the 50N weight be hung at 23 cm to maintain equilibrium