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This is a problem about balancing forces on a uniform beam. Here's how we can solve it step-by-step.

### Problem Setup:
- A uniform beam is 1 meter long, and three forces act on it: two downward forces of 3 N at 60 cm and 90 cm from the left end, and two unknown forces, \( X \) and \( Y \), at 20 cm and 50 cm, respectively.
- We assume the beam is in equilibrium, meaning both the sum of forces and the sum of torques (moments) around any point must be zero.

### Step 1: Equilibrium of Forces
For equilibrium in the vertical direction, the total upward forces must equal the total downward forces.

\[
X + Y = 3 \, \text{N} + 3 \, \text{N} = 6 \, \text{N}
\]

### Step 2: Equilibrium of Moments
To ensure rotational equilibrium, we calculate the torque about a convenient point. Let’s choose the left end of the beam (at 0 cm) as the pivot point. The torque is calculated by multiplying the force by the perpendicular distance from the pivot.

- The torque due to \( X \) at 20 cm: \( X \times 0.2 \, \text{m} \)
- The torque due to \( Y \) at 50 cm: \( Y \times 0.5 \, \text{m} \)
- The torque due to the 3 N force at 60 cm: \( 3 \times 0.6 \, \text{m} = 1.8 \, \text{Nm} \)
- The torque due to the 3 N force at 90 cm: \( 3 \times 0.9 \, \text{m} = 2.7 \, \text{Nm} \)

In equilibrium, the sum of torques must be zero. Assuming clockwise torques as positive:

\[
X \times 0.2 + Y \times 0.5 = 1.8 + 2.7
\]

\[
0.2X + 0.5Y = 4.5
\]

### Step 3: Solve the Equations
Now we solve the system of equations:
1. \( X + Y = 6 \)
2. \( 0.2X + 0.5Y = 4.5 \)

Multiply the second equation by 10 to eliminate decimals:

\[
2X + 5Y = 45
\]

Now solve the system:
- From equation 1: \( X = 6 - Y \)
- Substitute into the second equation:

\[
2(6 - Y) + 5Y = 45
\]
\[
12 - 2Y + 5Y = 45
\]
\[
3Y = 33
\]
\[
Y = 11
\]

Substitute \( Y = 11 \) into \( X + Y = 6 \):

\[
X + 11 = 6 \quad \Rightarrow \quad X = -5
\]

It seems there is an inconsistency here because forces cannot be negative. Let me check again.

A 1 m long uniform beam is balanced with forces X and Y. Two downward forces of 3N are applied at 60 cm and 90 cm. What are the forces X and Y?

Learn how to calculate the forces required to balance a uniform beam. This problem involves concepts of equilibrium, torque, and force balance. Suitable for high school students in Grades 9-12.

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