You have a 120-g meterstick. Where should you place the meterstick atop your finger so it balances? Where is the balance point if you place an additional 100-g mass at the 20-cm mark and a 200-g mass at the 80-cm mark?
(1) In the middle of the meterstick.
(2) Require that the net torque about the balance point be zero.
100(x - 20) = 200(80 - x) + 120(60 -x)
Solve for x.
does x = 60?
To find the balance point of a meterstick, you need to consider the principle of torque. Torque, also known as moment, is the product of the force applied and the perpendicular distance from the point of rotation to the line of action of the force. In this case, the point of rotation will be your finger.
For the meterstick to balance, the torques on both sides of the pivot point (your finger) must be equal.
In the first scenario, where there are no additional masses, the meterstick is assumed to have uniform mass distribution. This means that the center of mass, and hence the balance point, is located at the geometric center of the meterstick. Since the meterstick is 120 cm long, the balance point will be at the 60-cm mark.
In the second scenario, the additional masses change the mass distribution of the meterstick. To find the new balance point, you need to calculate the total torque on each side of the pivot.
First, let's calculate the torque produced by the 100-g mass at the 20-cm mark. The torque is given by the product of the mass, the gravitational acceleration, and the perpendicular distance from the pivot:
Torque = (mass) * (acceleration due to gravity) * (distance)
T1 = (100 g) * (9.8 m/s^2) * (20 cm)
To avoid unit inconsistencies, it's better to convert the distance to meters:
T1 = (100 g) * (9.8 m/s^2) * (0.2 m)
Similarly, let's calculate the torque produced by the 200-g mass at the 80-cm mark:
T2 = (200 g) * (9.8 m/s^2) * (0.8 m)
To achieve balance, the total torque on one side of the pivot must equal the total torque on the other side. Mathematically, this can be expressed as:
T1 = T2
(100 g) * (9.8 m/s^2) * (0.2 m) = (200 g) * (9.8 m/s^2) * (0.8 m)
Now, let's solve for the unknown balance point:
(distance) = [(200 g) * (9.8 m/s^2) * (0.8 m)] / [(100 g) * (9.8 m/s^2)]
(distance) = 1.6 m
Therefore, the new balance point with the additional masses is at the 1.6-meter mark.
To summarize:
- In the absence of additional masses, the balance point is at the 60-cm mark on the meterstick.
- With a 100-g mass placed at the 20-cm mark and a 200-g mass placed at the 80-cm mark, the balance point shifts to the 1.6-meter mark.