I need a bit of help on this problem...
A horizontal meter stick has a mass of 203 g. Three weights ride on the meter stick: 255g at 47.1cm, 181g at 84.5cm, and 207g at 90.7cm. At what location on the meter stick would the system be in balance if it were suspended there?
A horizontal meter stick has a mass of 203 g. Three weights ride on the meter stick: 255g at 47.1cm, 181g at 84.5cm, and 207g at 90.7cm. At what location on the meter stick would the system be in balance if it were suspended there?
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The total weight = 203 + 255 + 181 + 207 = 846g (strictly speaking this is a mass and should be converted to newtons to be a force, but this is unnecessary here)
When balanced, there will be an upwards force of 846g at the balance point at a distance x from the end of the metre stick.
When balanced, we can take moments (torques) about any convenient point. Use the end of the ruler (0.0cm) as the point:
846x = (203*50.0) + (255*47.1) + (181*84.5) + (207*90.7)
You can complete the calculation to find x.
When balanced, there will be an upwards force of 846g at the balance point at a distance x from the end of the metre stick.
When balanced, we can take moments (torques) about any convenient point. Use the end of the ruler (0.0cm) as the point:
846x = (203*50.0) + (255*47.1) + (181*84.5) + (207*90.7)
You can complete the calculation to find x.