Question

(a) The stick is supported by a sharp point at the middle. On the left side, a weight of 100 g is suspended at 40 cm from the middle point. On the right hand side, a weight of 200 g is suspended. When the stick is balanced, where should the 200 g weight be suspended (how many cm from the middle point)?

Answer 1

Answer:

20cm

Explanation:

Hello!

remember that the condition for a body to be at rest is that the sum of its moments and its forces be zero,

To solve this problem you must draw the free body diagram of the stick (attached image) and sum up moments at point 0 (where the sharp is located), which results in the following equation

(100g)(40cm)=x(200g)

$X=\frac{(100)(40)}{(200)} =20cm$