Question

A meterstick balances at its center. If an \(86-g\) necklace is suspended from one end of the meterstick, the balance point moves \(8.2 \mathrm{~cm}\) toward that end. (a) Is the mass of the meterstick greater than, less than, or equal to the mass of the necklace? Explain. (b) Find the mass of the meterstick.

Short answer

(a) The mass of the meterstick is greater than the necklace. (b) The mass is approximately 524.39 g.

Step-by-step solution

Understanding the Balance Point Shift

The meterstick balances at its center initially, meaning the torque from both sides is equal. When the necklace is hung at one end, the balance point shifts 8.2 cm towards the necklace, indicating an imbalance caused by the necklace's weight.

Relating Torque and Balance

The meterstick is initially balanced at its center, implying the torques acting on either side are equal. The necklace creates additional torque on one side of the pivot, requiring an examination of the mass of the stick to maintain balance.

Analyzing the Equal Torque Points

Let's assume the center of mass of the meterstick balances the torque created by the necklace. Initially, the torque is zero when the meterstick's weight is evenly distributed. With the necklace, the torque from the meterstick mass is balanced 8.2 cm from its original position, towards the necklace.

Calculating Mass of the Meterstick using Torque Equation

To find the mass of the meterstick, set the torque by the meterstick equal to the torque by the necklace. The torque is the product of force and distance. If the stick's mass is \(m\), then the position change of 8.2 cm reflects balancing torques: \( m imes 8.2 = 86 imes 50 \).Solve for \( m \):\[ m = \frac{86 imes 50}{8.2} \approx 524.39 ext{ g} \]

Concluding the Comparison of Masses

The mass of the meterstick is approximately 524.39 g, which is greater than the 86 g necklace, explaining the shift in balance due to the heavier torque contribution by the stick.

Key concepts

Torque and Balance

To understand the concept of torque and balance, imagine a seesaw. For it to be balanced, the torque exerted by weights on either side of the pivot must be equal. Torque, in simple terms, is the rotational equivalent of force. It depends on two factors: the amount of force being applied and the distance from the pivot point at which it's applied. Therefore, the farther a force is from the pivot, the greater its torque. When an 86-g necklace is hung on one side of the meterstick, it disrupts the balance by creating additional torque. This results in a shift of the meterstick's balance point towards the necklace. This simple disruption helps us explore how torque plays a critical role in balance and equilibrium scenarios.

Center of Mass

The center of mass of an object is the point at which it can be perfectly balanced. For a symmetrical object like a meterstick, the center of mass is in the middle when all forces are evenly distributed. Initially, the meterstick mentioned in our problem is balanced at its center, illustrating that its center of mass is at this point. When a weight, like the necklace, is added, the center of mass shifts towards the added weight—hence the new balance point moves 8.2 cm towards the necklace. Understanding the center of mass is crucial in solving balance-related problems. It helps us determine where an object will naturally balance when influenced by external forces.

Torque Equation

The torque equation allows us to mathematically determine when forces and distances result in balance on a pivot or lever. The essential formula to keep in mind is:
\[ \text{Torque} = \text{Force} \times \text{Distance} \]
In our physics exercise, the torque from the meterstick must equal the torque from the necklace to maintain balance. This is expressed in the equation:
\[ m \times 8.2 = 86 \times 50 \]
Where \( m \) is the mass of the meterstick we are looking for. After solving this equation, we find that the mass of the meterstick is approximately 524.39 g. This demonstrates the power of the torque equation in identifying unknown forces or weights in balance problems.

Balancing Forces

Balancing forces means ensuring that the forces in a system are even so that no movement occurs. In our meterstick scenario, the balancing forces are the gravitational forces acting on both the necklace and the meterstick at their respective distances from the pivot or fulcrum.
The necklace exerts a downward force on one side, shifting the meterstick's balance point closer to it. To achieve a state of equilibrium, the torque generated by the meterstick's mass (as calculated by its weight and distance from the pivot) needs to counterbalance the torque exerted by the necklace.
Once balanced, we see that the position shift accurately reflects the differing masses and distances involved. Balancing forces ensures that objects remain stable and in equilibrium, providing a clear insight into how forces interact in various scenarios.