A uniform rod of length `1 m` mass `4 kg` is supports on two knife-edges placed `10 cm` from each end. A `60 N` weight is suspended at `30 cm` from one end. The reactions at the knife edges is.

Options

A. 60 N, 40 N
B. 75 N, 25 N
C. 65 N, 35 N
D. 55 N, 45 N

Explanation

To solve the problem, we need to find the reactions at the two knife edges supporting the uniform rod. We will follow these steps:

### Step 1: Understand the Setup
We have a uniform rod of length 1 m and mass 4 kg, which means its weight (W_rod) is:
\[ W_{rod} = mass \times g = 4 \, kg \times 9.81 \, m/s^2 = 39.24 \, N \]

The rod is supported on two knife edges placed 10 cm from each end. A 60 N weight is suspended at a distance of 30 cm from one end of the rod.

### Step 2: Identify Forces and Distances
- Let the left end of the rod be point A and the right end be point B.
- The knife edges are at points C (10 cm from A) and D (10 cm from B).
- The suspended weight (60 N) is at a distance of 30 cm from A, which means it is at point E (30 cm from A).

### Step 3: Calculate Total Weight
The total downward force acting on the rod is the sum of the weight of the rod and the suspended weight:
\[ F_{total} = W_{rod} + W_{suspended} = 39.24 \, N + 60 \, N = 99.24 \, N \]

### Step 4: Set Up Equations for Equilibrium
For the rod to be in equilibrium, the sum of vertical forces must be zero:
\[ R_1 + R_2 = 99.24 \, N \] (Equation 1)

Where \( R_1 \) is the reaction at knife edge C and \( R_2 \) is the reaction at knife edge D.

### Step 5: Calculate Moments about a Point
To find the reactions, we can take moments about one of the knife edges. Let's take moments about point C.

The moment due to the weight of the rod acts at its center of mass, which is at 50 cm from A (or 40 cm from C). The moment due to the suspended weight acts at 20 cm from C (30 cm from A - 10 cm from C).

The clockwise moments must equal the counterclockwise moments about point C:
\[ 60 \, N \times 20 \, cm + 39.24 \, N \times 40 \, cm = R_2 \times 80 \, cm \]

### Step 6: Substitute and Solve for \( R_2 \)
Convert cm to m for calculations:
\[ 60 \times 0.20 + 39.24 \times 0.40 = R_2 \times 0.80 \]
\[ 12 + 15.696 = R_2 \times 0.80 \]
\[ 27.696 = R_2 \times 0.80 \]
\[ R_2 = \frac{27.696}{0.80} = 34.62 \, N \]

### Step 7: Substitute \( R_2 \) back into Equation 1 to find \( R_1 \)
Using Equation 1:
\[ R_1 + 34.62 = 99.24 \]
\[ R_1 = 99.24 - 34.62 = 64.62 \, N \]

### Final Answer
The reactions at the knife edges are:
- \( R_1 \approx 64.62 \, N \)
- \( R_2 \approx 34.62 \, N \)