Answer:
The weight that will balance the meter stick at 70 cm mark is 35N
Explanation:
Given;
first weight at 15 cm mark, W₁ = 20 N
second weight at 70 cm mark, W₂ = ?
A sketch of the question in diagram form;
The center of gravity of the meter stick is 50 cm
15cm 50cm 70cm
0---------------------------------------Δ------------------------------------------100cm
↓ ↓
20N W₂
<------------------>|<--------------->
35cm 20cm
Take moment about the pivot;
Clockwise moment = anticlockwise moment
W₂(20cm) = 20N(35 cm)
W₂ = (20 x 35) / (20)
W₂ = 35 N
Therefore, the weight that will balance the meter stick at 70 cm mark is 35N
