A uniform beam of length 20 m and mass 100 kg is supported by pylons on either end. A 200-kg box is centered at 12 m from the right end of the beam. What is the downward force on the left most pylon?

To solve this problem, we can use the principles of static equilibrium. The beam and the box are in static equilibrium, which means the sum of forces and the sum of torques (moments) must both equal zero.

### Step 1: Define the forces

- Let \( F_L \) be the force on the leftmost pylon (what we need to find).
- Let \( F_R \) be the force on the rightmost pylon.
- The downward forces acting on the system are:
  - The weight of the beam, \( W_{\text{beam}} = mg = 100 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 980 \, \text{N} \).
  - The weight of the box, \( W_{\text{box}} = 200 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 1960 \, \text{N} \).

### Step 2: Torque about the rightmost pylon

We will sum the torques about the rightmost pylon to eliminate the unknown \( F_R \). Since the beam is in static equilibrium, the total torque about any point must be zero.

Taking torques about the rightmost pylon:

1. The beam’s center of mass is at its midpoint (10 m from either end). The perpendicular distance from the center of mass to the right pylon is \( 20 \, \text{m} - 10 \, \text{m} = 10 \, \text{m} \).
   - Torque due to the beam’s weight: \( \tau_{\text{beam}} = 980 \, \text{N} \times 10 \, \text{m} = 9800 \, \text{N} \cdot \text{m} \).

2. The box is placed 12 m from the right end. The perpendicular distance to the right pylon is \( 12 \, \text{m} \).
   - Torque due to the box’s weight: \( \tau_{\text{box}} = 1960 \, \text{N} \times 12 \, \text{m} = 23,520 \, \text{N} \cdot \text{m} \).

Now, the total torque around the right pylon must be balanced by the torque due to the force on the left pylon, which is exerted at a distance of 20 m from the right pylon:
\[
F_L \times 20 \, \text{m} = 9800 \, \text{N} \cdot \text{m} + 23,520 \, \text{N} \cdot \text{m}
\]
\[
F_L \times 20 \, \text{m} = 33,320 \, \text{N} \cdot \text{m}
\]
\[
F_L = \frac{33,320 \, \text{N} \cdot \text{m}}{20 \, \text{m}} = 1,666 \, \text{N}
\]

### Step 3: Verify the forces

The total downward force is the sum of the beam and the box:
\[
F_{\text{total}} = 980 \, \text{N} + 1960 \, \text{N} = 2940 \, \text{N}
\]
Since the system is in equilibrium, the sum of the forces must also equal zero. Therefore:
\[
F_L + F_R = 2940 \, \text{N}
\]
\[
1,666 \, \text{N} + F_R = 2940 \, \text{N}
\]
\[
F_R = 2940 \, \text{N} - 1,666 \, \text{N} = 1274 \, \text{N}
\]

Thus, the force on the right pylon is \( 1274 \, \text{N} \) and the force on the leftmost pylon is \( \boxed{1,666 \, \text{N}} \).

Would you like further details or have any questions?

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**Related Questions:**

1. How would the forces change if the box were moved to the center of the beam?
2. What if the beam were not uniform but instead had more mass concentrated towards one end?
3. How would the problem change if the box was located 5 m from the leftmost pylon?
4. How does the concept of torque help in balancing objects in equilibrium?
5. What is the effect of increasing the length of the beam while keeping the mass and positions constant?

**Tip:** When calculating torques, always choose a pivot point that simplifies the math, often one where you can eliminate unknowns.

Learn how to solve a statics problem involving a uniform beam and a 200-kg box placed 12 m from the right end. This detailed solution uses principles of static equilibrium and torque calculations to find the force on the leftmost pylon. Suitable for high school physics students in Grades 11-12.

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