Question

Rod ab carries three loads of 30 n, 70 n and 100 n at distances of 20 mm, 90 mm and 150 mm respectively from



a.Neglecting the weight of the rod, the point at which the rod will balance is

Answer 1

Let us suppose the distance of the balancing point be ''y''cm from Z.

Balancing moments of loads,

(15-y) 10 = (x-9) 7 + (y-2) 3

150 - 10y = 10y - 6

20y = 219

y = 219/20

y = 10.45cm from Z

Hope it helps.

Answer 2

Answer:

109.5 from A

Explanation:

Moment at point X should be zero for stability of the rod.

$100 \times (150 - x) + 70 \times (90 - x) - 30 \times (x - 20) = 0 \\ 21900 - 200x = 0 \\ 21900 = 200x \\ x = (21900 \div 200) \\ x = 109.5mm \: from \: a$