Question

A uniform metre rule is pivoted at its centre and weight of 5n and 12n are hung at the 3cm and 5cm marks respectively. how far from the pivot must a 25n weight be hung to balance the metre rule horizontally

Answer 1

Explanation:

consider the centre of ruler be origin

for equilibrium sum of all torques about origin must be zero.

torque of weight 5n is 5n(47). { as distance is 47cm}

torque of weight 12n is 12n(45). { as distance is 45cm}

torque of weight 25n is 25n(x). {let x be the distance for equilibrium}

now for equilibrium,

5n(47) + 12n(45) = 25n(x)

235+540=25x

775=25x

x= 31cm

so 25n weight must be hung 31cm away from the pivot.

Mark it as brainliest if you find it easy solution and correct answer.

thank you