24. (a) A uniform beam AB of mass 12 kg is 12 m long. The beam is placed horizontally on supports at P and Q on the beam. P is 2 m from A and Q is 9 m from A. If two masses 10 kg and 5 kg are suspended at points 3 m from A and 1 m from B respectively on the beam, calculate, correct to three significant figures the reactions at P and Q. (Take g= 10 ms²).

We are dealing with a uniform beam placed horizontally on two supports with two additional masses attached. The problem can be solved using the principles of static equilibrium: 

1. The **sum of all vertical forces** must equal zero.
2. The **sum of moments** about any point must also equal zero.

### Step 1: Identify forces

- **Weight of the beam**: The beam is uniform, so its weight acts at its center of mass, which is at the midpoint of the beam (6 m from A). The weight is \( W_{\text{beam}} = m_{\text{beam}} \cdot g = 12 \text{ kg} \cdot 10 \text{ m/s}^2 = 120 \text{ N} \).
  
- **Weight of the 10 kg mass**: This mass is placed 3 m from point A. Its weight is \( W_1 = 10 \text{ kg} \cdot 10 \text{ m/s}^2 = 100 \text{ N} \).
  
- **Weight of the 5 kg mass**: This mass is placed 1 m from point B. Since the total length of the beam is 12 m, this point is 11 m from point A. Its weight is \( W_2 = 5 \text{ kg} \cdot 10 \text{ m/s}^2 = 50 \text{ N} \).

- **Reactions at supports P and Q**: Let \( R_P \) and \( R_Q \) be the reactions at P and Q, respectively.

### Step 2: Apply equilibrium equations

#### 1. Sum of vertical forces:

The total force must equal zero:

\[
R_P + R_Q = W_{\text{beam}} + W_1 + W_2
\]

\[
R_P + R_Q = 120 + 100 + 50 = 270 \, \text{N}
\]

#### 2. Sum of moments about point A:

To eliminate \( R_P \) from the equation, we take moments about point A (clockwise moments are positive):

\[
\text{Moment at A} = 0
\]

Moments due to various forces:

- Moment due to \( R_Q \) at point Q (9 m from A): \( R_Q \cdot 9 \)
- Moment due to the beam's weight (120 N acting at the center, 6 m from A): \( 120 \cdot 6 \)
- Moment due to the 10 kg mass (100 N acting 3 m from A): \( 100 \cdot 3 \)
- Moment due to the 5 kg mass (50 N acting 11 m from A): \( 50 \cdot 11 \)

So the sum of moments about point A is:

\[
R_Q \cdot 9 - 120 \cdot 6 - 100 \cdot 3 - 50 \cdot 11 = 0
\]

Simplifying:

\[
9 R_Q = 720 + 300 + 550
\]

\[
9 R_Q = 1570
\]

\[
R_Q = \frac{1570}{9} = 174.4 \, \text{N} \, (\text{to three significant figures})
\]

#### 3. Find \( R_P \):

From the sum of vertical forces equation:

\[
R_P + 174.4 = 270
\]

\[
R_P = 270 - 174.4 = 95.6 \, \text{N}
\]

### Final Answers:

- Reaction at \( P \): \( R_P = 95.6 \, \text{N} \)
- Reaction at \( Q \): \( R_Q = 174.4 \, \text{N} \)

Would you like any additional details or clarifications on the process?

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Here are 5 related questions for further exploration:

1. What would happen to the reactions if the positions of the masses were swapped?
2. How would the reactions change if the beam were not uniform but had varying density?
3. How does the value of gravitational acceleration affect the results?
4. What would be the reactions if the mass at 1 m from B was doubled?
5. How do you calculate the moments when the beam is inclined?

**Tip:** Always choose a point where one unknown force acts to eliminate that force when summing moments.

A uniform beam AB of mass 12 kg is 12 m long. The beam is placed horizontally on supports at P and Q on the beam. P is 2 m from A and Q is 9 m from A. If two masses 10 kg and 5 kg are suspended at points 3 m from A and 1 m from B respectively on the beam, calculate, correct to three significant figures, the reactions at P and Q. (Take g = 10 m/s²).

Learn how to calculate the reactions at supports P and Q of a 12 m long uniform beam with suspended masses. This problem uses the principles of static equilibrium, including force balance and moments calculation. Suitable for students in Grades 11-12, covering concepts in statics and equilibrium.

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