A uniform meter rule of mass 120 g is pivoted at the 60 cm mark. At what point on the meter rule should a mass of 50 g be suspended for it to balance horizontally?
am confused
the center-of-mass of the rule is the 50 cm mark
... 10 cm from the pivot
the suspended mass is on the other side of the pivot
the moments are equal ... 120 g * 10 cm = 50 g * d
To find the point on the meter rule where a mass of 50 g should be suspended for it to balance horizontally, we need to consider the principle of moments.
The principle of moments states that for an object in rotational equilibrium, the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point.
In this case, we have a uniform meter rule pivoted at the 60 cm mark, with a mass of 120 g. We want to find the position where a mass of 50 g should be suspended to balance it horizontally.
Let's assume that the 50 g mass is suspended at a distance x from the pivot point.
The clockwise moment is generated by the 120 g mass at a distance of 60 cm from the pivot:
Clockwise Moment = mass × distance = 120 g × 60 cm = 7200 g·cm
The anticlockwise moment is generated by the 50 g mass at a distance of x cm from the pivot:
Anticlockwise Moment = mass × distance = 50 g × x cm = 50x g·cm
According to the principle of moments, the clockwise moment must be equal to the anticlockwise moment:
7200 g·cm = 50x g·cm
To find x, we can rearrange the equation:
x = (7200 g·cm) / 50 g
x = 144 cm
Therefore, a mass of 50 g should be suspended at the 144 cm mark on the meter rule to balance it horizontally.