A ruler (meter stick) is balanced with the fulcrum at the 50 cm mark as shown in the diagram. If a 130 g mass?
A ruler (meter stick) is balanced with the fulcrum at the 50 cm mark as shown in the diagram. If a 130 g mass was placed at the 25 cm mark, and a 20 g mass at the 8 cm mark, where should a 500 g mass be placed to balance the system?
- 1 decade agoFavorite Answer
At the 58.18 cm (approx at the 52 cm) mark. The moment on the forces on both sides of the pivot should be equal. The forces are weight of the 130 gram and 20 gm mass on the left side (if the ruler started from 0 on the left to 100 at the right tip) and 500 gram on the right. The distances FROM THE PIVOT are 25 cm for the 130 gram, 42 cm (50-8) for the 20 gm and X for the 500 gm. So
(130G)25 + (20G)42 = (500G)X,
giving a value of X as 8.18, but this is from the pivot to the right so it should be placed at the 50+8.18 = 58.18 mark.
P.S.: the "G" stands for acceleration due to gravity
Use consistent units ALWAYS, here we use grams for mass and cm for length, a better idea is to convert these to SI units, more complex problems might require such standard units be used.
- 4 years ago
180*24=500*D D=180*24/500=4320/500=8.64 cm. from fulcrum position=58.64