A rope of mass 5 kg is moving vertically in vertical position with an upwards force of 100 N acting at the upper end and a downwards force of 70 N acting at the lower end. The tension at midpoint of the rope is

Options

A. 100 N
B. 85 N
C. 75 N
D. 105 N

Answer

To find the tension at the midpoint of the rope, we can follow these steps:

### Step 1: Identify the forces acting on the rope
- The upward force acting on the upper end of the rope is \( F_{\text{up}} = 100 \, \text{N} \).
- The downward force acting on the lower end of the rope is \( F_{\text{down}} = 70 \, \text{N} \).
- The mass of the rope is \( m = 5 \, \text{kg} \), which gives a weight of \( W = mg = 5 \times 10 = 50 \, \text{N} \).

### Step 2: Calculate the net force acting on the rope
The net force \( F_{\text{net}} \) acting on the rope can be calculated as:
\[ F_{\text{net}} = F_{\text{up}} - (F_{\text{down}} + W) \]
Substituting the values:
\[ F_{\text{net}} = 100 \, \text{N} - (70 \, \text{N} + 50 \, \text{N}) = 100 \, \text{N} - 120 \, \text{N} = -20 \, \text{N} \]
The negative sign indicates that the net force is directed downward.

### Step 3: Calculate the acceleration of the rope
Using Newton's second law, \( F = ma \), we can find the acceleration \( a \):
\[ F_{\text{net}} = ma \implies a = \frac{F_{\text{net}}}{m} \]
Substituting the values:
\[ a = \frac{-20 \, \text{N}}{5 \, \text{kg}} = -4 \, \text{m/s}^2 \]
The negative sign indicates that the acceleration is downward.

### Step 4: Analyze the forces at the midpoint of the rope
At the midpoint, we consider the forces acting on the upper half of the rope. The forces acting on this section include:
- The tension \( T \) at the midpoint acting upward.
- The weight of the upper half of the rope, which is \( \frac{1}{2} mg = \frac{1}{2} \times 5 \times 10 = 25 \, \text{N} \).
- The downward force from the lower end, which is \( F_{\text{down}} = 70 \, \text{N} \).

### Step 5: Set up the equation for the upper half of the rope
Applying Newton's second law to the upper half of the rope:
\[ T - \left( \frac{1}{2} mg + F_{\text{down}} \right) = \frac{1}{2} m a \]
Substituting the known values:
\[ T - (25 \, \text{N} + 70 \, \text{N}) = \frac{1}{2} \times 5 \, \text{kg} \times (-4 \, \text{m/s}^2) \]
\[ T - 95 \, \text{N} = -10 \, \text{N} \]

### Step 6: Solve for tension \( T \)
Rearranging the equation gives:
\[ T = 95 \, \text{N} - 10 \, \text{N} = 85 \, \text{N} \]

### Final Answer
The tension at the midpoint of the rope is \( T = 85 \, \text{N} \).

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