Solved 1 Balance the meter stick on the knife edge. The | Chegg.com - Free Printable

Let's solve the problem step by step based on the instructions and Figure 5 provided.

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Step 1: Determine the Center of Gravity (CG) of the Meter Stick

- A meter stick is typically uniform, so its center of gravity is at its geometric center, which is at:
$$
\text{Center of gravity} = \frac{0 + 100}{2} = 50\ \text{cm}
$$
- So, the knife edge should be placed at 50 cm to balance the stick.

✔ Answer for Step 1: The center of gravity is at 50 cm.

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Step 2: Select Masses

- Two 200-gram masses
- One 100-gram mass

These will be used in subsequent steps.

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Step 3: Set Up the System as Described

We are told:

- Place a hanger at the 20-cm mark with mass $ m_1 = 200\ \text{g} $
- Place another hanger at the 65-cm mark with mass $ m_2 = 200\ \text{g} $
- We'll also place a third mass $ m_3 = 100\ \text{g} $ at some point (though not specified yet — likely to be determined).

But first, let’s find distances from the center of gravity (at 50 cm):

#### Calculate $ x_1 $: Distance from CG to $ m_1 $
- $ m_1 $ is at 20 cm
- CG is at 50 cm
- So,
$$
x_1 = 50 - 20 = 30\ \text{cm} \quad \text{(to the left)}
$$

#### Calculate $ x_2 $: Distance from CG to $ m_2 $
- $ m_2 $ is at 65 cm
- So,
$$
x_2 = 65 - 50 = 15\ \text{cm} \quad \text{(to the right)}
$$

So far:
- $ m_1 = 200\ \text{g} $ at $ x_1 = 30\ \text{cm} $ (left)
- $ m_2 = 200\ \text{g} $ at $ x_2 = 15\ \text{cm} $ (right)

Now, we need to consider torque equilibrium.

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Torque Balance Principle

For the system to be balanced (no rotation), the net torque about the pivot (CG) must be zero.

Torque $ \tau = r \times F = r \times mg $

We define:
- Clockwise torques as positive
- Counterclockwise torques as negative (or vice versa — just be consistent)

Let’s assume:
- Torque due to $ m_1 $ (left side): counterclockwise → negative
- Torque due to $ m_2 $ (right side): clockwise → positive
- Torque due to $ m_3 $: depends on position

But in this setup, only two masses are mentioned in Step 3. However, Figure 5 shows three masses ($ m_1, m_2, m_3 $). So likely, the experiment involves adding $ m_3 = 100\ \text{g} $ at some position to balance the system.

But let's see what happens without $ m_3 $ first.

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Check if System Balances with Only $ m_1 $ and $ m_2 $

Compute torques about the pivot (50 cm):

- Torque from $ m_1 $: $ \tau_1 = -m_1 g x_1 = -(0.2\ \text{kg}) \cdot g \cdot 0.3\ \text{m} = -0.06g\ \text{N·m} $
- Torque from $ m_2 $: $ \tau_2 = +m_2 g x_2 = +(0.2\ \text{kg}) \cdot g \cdot 0.15\ \text{m} = +0.03g\ \text{N·m} $

Net torque: $ \tau_{\text{net}} = -0.06g + 0.03g = -0.03g\ \text{N·m} $ → Not balanced!

So the system will rotate counterclockwise unless a third mass is added.

Thus, $ m_3 = 100\ \text{g} $ must be placed on the right side to counteract the excess counterclockwise torque.

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Find Position of $ m_3 $ to Balance the System

Let $ x_3 $ be the distance from CG to $ m_3 $, on the right side (since we need clockwise torque).

Set net torque = 0:

$$
\tau_{\text{total}} = -m_1 g x_1 + m_2 g x_2 + m_3 g x_3 = 0
$$

Plug in values (in grams and cm — since $ g $ cancels out, we can work in gram-cm units):

$$
-(200)(30) + (200)(15) + (100)(x_3) = 0
$$

$$
-6000 + 3000 + 100x_3 = 0
$$

$$
-3000 + 100x_3 = 0
$$

$$
100x_3 = 3000 \Rightarrow x_3 = 30\ \text{cm}
$$

So, $ m_3 = 100\ \text{g} $ must be placed 30 cm to the right of the center of gravity.

Since CG is at 50 cm:
$$
\text{Position of } m_3 = 50 + 30 = 80\ \text{cm}
$$

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✔ Final Answers for Data Table 1

| Quantity | Value |
|--------|-------|
| Center of gravity | 50 cm |
| $ m_1 $ | 200 g |
| $ x_1 $ (distance from CG) | 30 cm |
| $ m_2 $ | 200 g |
| $ x_2 $ (distance from CG) | 15 cm |
| $ m_3 $ | 100 g |
| $ x_3 $ (distance from CG) | 30 cm |
| Position of $ m_3 $ | 80 cm |

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🔍 Explanation of the Solution

- The center of gravity of a uniform meter stick is at 50 cm.
- Placing masses at different positions creates torques about the pivot.
- For equilibrium, sum of clockwise torques = sum of counterclockwise torques.
- $ m_1 $ at 20 cm (30 cm left) creates a large counterclockwise torque.
- $ m_2 $ at 65 cm (15 cm right) creates a smaller clockwise torque.
- To balance, a third mass ($ m_3 = 100\ \text{g} $) must be placed 30 cm to the right of CG (i.e., at 80 cm), providing enough clockwise torque to cancel the imbalance.

This demonstrates the principle of rotational equilibrium:
$$
\sum \tau = 0
$$

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📌 Summary

- Center of gravity: 50 cm
- $ x_1 = 30\ \text{cm} $, $ x_2 = 15\ \text{cm} $
- To balance, $ m_3 = 100\ \text{g} $ must be placed at 80 cm (so $ x_3 = 30\ \text{cm} $)

You can now enter these values into Data Table 1.