A student tries to raise a chain consisting of three identical links. Each link has a mass of 300g. The three-piece chain is connected to a string and then suspended vertically, with the student holding the upper end of the string and pulling upward. Because of the student's pull, an upward force of 12 N is applied to the chain by the string.
(a) Draw a free-body diagram for each of the links in the chain and also for the entire-chain considered a single body.
(b) Use the results of part(a) Newton's laws to find (i) the acceleration of the chain and (ii) the force exerted by the top link on the middle link.

Explanation

To solve the problem step-by-step, we will first address part (a) by drawing the free-body diagrams for each link and the entire chain. Then, we will use Newton's laws to find the acceleration of the chain and the force exerted by the top link on the middle link as required in part (b).

### Part (a): Free-Body Diagrams

1. **Free-Body Diagram for the Entire Chain:**
- The entire chain consists of three links, each with a mass of 300 g (0.3 kg).
- The total mass of the chain is \(3 \times 0.3 \, \text{kg} = 0.9 \, \text{kg}\).
- The weight of the chain (downward force) is given by \(W = mg = 0.9 \, \text{kg} \times 10 \, \text{m/s}^2 = 9 \, \text{N}\).
- The upward force applied by the student is \(F = 12 \, \text{N}\).
- The free-body diagram will show:
- An upward force of 12 N.
- A downward force of 9 N (weight of the chain).
- The net force acting on the chain is \(F_{\text{net}} = F - W = 12 \, \text{N} - 9 \, \text{N} = 3 \, \text{N}\).

2. **Free-Body Diagram for the Top Link:**
- The top link has a weight of \(W_1 = mg = 0.3 \, \text{kg} \times 10 \, \text{m/s}^2 = 3 \, \text{N}\).
- The forces acting on it are:
- An upward force of 12 N (from the string).
- A downward force of 3 N (its own weight).
- An upward tension \(T_1\) from the middle link.
- The equation of motion for the top link can be written as:
\[ F_{\text{net}} = F - W_1 - T_1 = 0.3a \]

3. **Free-Body Diagram for the Middle Link:**
- The middle link also has a weight of 3 N.
- The forces acting on it are:
- An upward tension \(T_1\) from the top link.
- A downward tension \(T_2\) to the bottom link.
- The equation of motion for the middle link can be written as:
\[ T_1 - W_1 - T_2 = 0.3a \]

4. **Free-Body Diagram for the Bottom Link:**
- The bottom link has a weight of 3 N.
- The forces acting on it are:
- An upward tension \(T_2\) from the middle link.
- The equation of motion for the bottom link can be written as:
\[ T_2 - W_1 = 0.3a \]

### Part (b): Using Newton's Laws

(i) **Finding the Acceleration of the Chain:**
- From the free-body diagram of the entire chain, we can use Newton's second law:
\[ F_{\text{net}} = ma \]
\[ 3 \, \text{N} = 0.9 \, \text{kg} \times a \]
\[ a = \frac{3 \, \text{N}}{0.9 \, \text{kg}} = \frac{10}{3} \, \text{m/s}^2 \approx 3.33 \, \text{m/s}^2 \]

(ii) **Finding the Force Exerted by the Top Link on the Middle Link (Tension T1):**
- Using the free-body diagram of the top link:
\[ 12 \, \text{N} - T_1 - 3 \, \text{N} = 0.3 \times \frac{10}{3} \]
\[ 12 - T_1 - 3 = 1 \implies 9 - T_1 = 1 \implies T_1 = 8 \, \text{N} \]

### Summary of Answers:
- (i) The acceleration of the chain is \( \frac{10}{3} \, \text{m/s}^2 \).
- (ii) The force exerted by the top link on the middle link is \( 8 \, \text{N} \).