3.A motor-lorry frame can be considered as a beam 4.2 m long and simply supported at two points, A and B, 3 m apart. The frame is subjected to concentrated loads of 5 kN, 10 kN, 15 kN and 10 kN 											                                                                                   (a)  the value of the support reactions at A and B
(b)  the position at which the 15 kN load must be applied to make the support reactions equal.

Let's analyze the problem step-by-step to determine the support reactions and the position where the 15 kN load must be applied to make the support reactions equal.

### Step 1: Determine the Support Reactions at A and B

1. **Length of the beam (L)**: 4.2 m
2. **Distance between supports A and B**: 3 m
3. **Loads applied**:
   - \( F_1 = 5 \text{ kN} \)
   - \( F_2 = 10 \text{ kN} \)
   - \( F_3 = 15 \text{ kN} \)
   - \( F_4 = 10 \text{ kN} \)

Let's denote:
- \( R_A \) as the reaction at support A
- \( R_B \) as the reaction at support B

For equilibrium, the sum of the vertical forces and the sum of moments about any point must be zero.

#### Vertical Force Equilibrium:
\[ R_A + R_B = F_1 + F_2 + F_3 + F_4 \]
\[ R_A + R_B = 5 + 10 + 15 + 10 \]
\[ R_A + R_B = 40 \text{ kN} \]

#### Moment Equilibrium about A:
Let's denote the distances of the loads from point A as \( d_1, d_2, d_3, \) and \( d_4 \).

For simplicity, let's assume the positions of the loads along the beam.

\[ \text{Sum of moments about A:} \]
\[ R_B \times 3 - (5 \times d_1) - (10 \times d_2) - (15 \times d_3) - (10 \times d_4) = 0 \]

### Step 2: Determine the Position to Make Reactions Equal
To make the reactions equal (\( R_A = R_B \)):

\[ R_A = R_B \]
\[ R_A = R_B = \frac{40 \text{ kN}}{2} \]
\[ R_A = R_B = 20 \text{ kN} \]

The moment equilibrium around point A will help us find the correct position for \( F_3 \) (15 kN load).

Since the reactions are equal:
\[ 20 \times 3 - (5 \times d_1) - (10 \times d_2) - (15 \times d_3) - (10 \times d_4) = 0 \]

### Specific Load Positions:
Let’s consider the distances as follows for our calculation:
- \( d_1 = 0.5 \text{ m} \)
- \( d_2 = 1.5 \text{ m} \)
- \( d_3 = x \text{ m} \) (unknown, where we place 15 kN to balance)
- \( d_4 = 3.5 \text{ m} \)

\[ 20 \times 3 - (5 \times 0.5) - (10 \times 1.5) - (15 \times x) - (10 \times 3.5) = 0 \]
\[ 60 - 2.5 - 15 - 15x - 35 = 0 \]
\[ 60 - 52.5 - 15x = 0 \]
\[ 7.5 = 15x \]
\[ x = \frac{7.5}{15} \]
\[ x = 0.5 \text{ m} \]

Therefore, to make the support reactions equal, the 15 kN load must be applied 0.5 m from point A.

### Summary:
(a) The support reactions \( R_A \) and \( R_B \) are 20 kN each when the 15 kN load is applied at 0.5 m from point A.

(b) The 15 kN load must be applied 0.5 m from point A to make the support reactions equal.

Would you like further details or have any questions?

### Related Questions:
1. How do you calculate the moment of a force about a point?
2. What is the significance of having equal reactions at supports in a beam?
3. How do different load positions affect the reactions at the supports?
4. Can we have unequal loads at different positions and still achieve equal reactions?
5. What is the effect of the beam length on the support reactions?
6. How do you verify the accuracy of calculated support reactions?
7. How would the support reactions change if one of the loads is removed?
8. What role does the distance between supports play in determining the reactions?

### Tip:
Always ensure the sum of moments around a support is zero to maintain equilibrium in beam problems.

Explore how to calculate support reactions and determine the position for a load on a simply supported beam. This problem involves statics and mechanics of materials concepts, including the application of the sum of forces and sum of moments principles. Suitable for college-level engineering students.

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