Two like parallel force 20 N and 30 N act at the ands A and B of a rod 1.5 m long. The resultant of the forces will act at the point

Options

A. 90 cm from A
B. 75 cm from B
C. 20 cm from B
D. 85 cm from A

Answer

To solve the problem of finding the point where the resultant of two like parallel forces (20 N and 30 N) acts on a rod of length 1.5 m, we can follow these steps:

### Step 1: Understand the Forces and Their Positions
We have two forces acting on the ends of a rod:
- A force of 20 N at point A (left end).
- A force of 30 N at point B (right end).
The length of the rod is 1.5 m.

### Step 2: Define the Distance from Point A
Let the distance from point A to the point where the resultant force acts be denoted as \( x \). Consequently, the distance from point B to the same point will be \( 1.5 - x \).

### Step 3: Calculate the Torque About Point A
The torque due to the 20 N force about point A is:
\[ \tau_A = 20 \times x \]
The torque due to the 30 N force about point A is:
\[ \tau_B = 30 \times (1.5 - x) \]

### Step 4: Set Up the Torque Equation
For the system to be in equilibrium, the sum of the clockwise torques must equal the sum of the counterclockwise torques. Therefore, we can set up the equation:
\[ 20x = 30(1.5 - x) \]

### Step 5: Solve for \( x \)
Expanding the right side:
\[ 20x = 45 - 30x \]
Now, add \( 30x \) to both sides:
\[ 20x + 30x = 45 \]
\[ 50x = 45 \]
Now, divide both sides by 50:
\[ x = \frac{45}{50} = 0.9 \text{ m} \]

### Step 6: Convert to Centimeters
To convert meters to centimeters, multiply by 100:
\[ x = 0.9 \times 100 = 90 \text{ cm} \]

### Conclusion
The resultant force will act at a point 90 cm from point A.