A uniform rod is `4m` long and weights `10kg`. If it is supported on a kinetic edge at one meter from the end, what weight placed at that end keeps the rod horizontal.

Options

A. `8 kg`
B. `10 kg`
C. `12 kg`
D. `15 kg`

Explanation

To solve the problem of finding the weight that keeps the rod horizontal, we need to analyze the torque acting on the rod. Here’s the step-by-step solution:

### Step 1: Understand the System
We have a uniform rod that is 4 meters long and weighs 10 kg. It is supported at a point that is 1 meter from one end. This means that the distance from the support point to the center of mass of the rod (which is at the midpoint, 2 meters from one end) is 1 meter.

### Step 2: Identify Forces and Distances
- Weight of the rod (W_rod) = 10 kg × 9.81 m/s² = 98.1 N (acting downward at the center of the rod, which is 2 meters from the left end).
- Let the weight placed at the end (W) be the unknown force we need to find. This weight will act downward at the right end of the rod (4 meters from the left end).

### Step 3: Calculate Torques
To keep the rod horizontal, the sum of the torques about the pivot point (the support point) must be zero.

- Torque due to the weight of the rod (clockwise):
- Distance from the pivot (1 meter from the left end to the center of the rod) = 1 meter.
- Torque (τ_rod) = Weight of the rod × Distance = 98.1 N × 1 m = 98.1 N·m (clockwise).

- Torque due to the weight placed at the end (counterclockwise):
- Distance from the pivot (1 meter from the left end to the right end) = 3 meters.
- Torque (τ_weight) = W × 3 m.

### Step 4: Set Up the Equation
For the rod to be in equilibrium (horizontal), the clockwise torque must equal the counterclockwise torque:

\[ \tau_{\text{rod}} = \tau_{\text{weight}} \]

Substituting the values we have:

\[ 98.1 N·m = W × 3 m \]

### Step 5: Solve for W
Now, we can solve for W:

\[ W = \frac{98.1 N·m}{3 m} = 32.7 N \]

### Step 6: Convert Weight to Mass
To find the mass equivalent of this weight, we can use the relation \( W = mg \):

\[ m = \frac{W}{g} = \frac{32.7 N}{9.81 m/s²} \approx 3.33 kg \]

### Final Answer
The weight that must be placed at the end of the rod to keep it horizontal is approximately **32.7 N** or **3.33 kg**.

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