Answer :
### Section B
#### (a) Explain why it is easy to open a nut by using a spanner with a long arm than that of a short arm. (3 marks)
Using a spanner with a long arm to open a nut is easier because of the principle of moments (or torque). The moment is the product of the force applied and the distance from the pivot point (in this case, the nut). By using a longer spanner, you increase the distance from the pivot, which increases the torque for the same amount of force applied. This means less force is required to produce the same turning effect, making it easier to open the nut.
#### (b) A metal beam AB of mass 50 kg is supported at its ends. The beam carries a mass of 200 kg at a distance of 0.75 m from end A. If the beam is 2 m long, determine the thrust at supports A and B. (6 marks)
To determine the thrust (reactions) at supports A and B, let's denote them as [tex]\( R_A \)[/tex] and [tex]\( R_B \)[/tex].
Given:
- Mass of beam, [tex]\( m_{\text{beam}} = 50 \, \text{kg} \)[/tex]
- Mass carried, [tex]\( m_{\text{load}} = 200 \, \text{kg} \)[/tex]
- Distance from A to where load is applied, [tex]\( d_{\text{load}} = 0.75 \, \text{m} \)[/tex]
- Length of the beam, [tex]\( L = 2 \, \text{m} \)[/tex]
- Acceleration due to gravity, [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex]
Step-by-Step Solution:
1. Calculate the Weight:
- Weight of the beam, [tex]\( W_{\text{beam}} = m_{\text{beam}} \times g = 50 \times 9.81 = 490.5 \, \text{N} \)[/tex]
- Weight of the load, [tex]\( W_{\text{load}} = m_{\text{load}} \times g = 200 \times 9.81 = 1962 \, \text{N} \)[/tex]
2. Taking Moments about Point A:
The moment due to the load and the beam about point A is given by:
[tex]\[ \text{Moment about A} = W_{\text{load}} \times d_{\text{load}} + W_{\text{beam}} \times \left(\frac{L}{2}\right) \][/tex]
Substituting the values:
[tex]\[ \text{Moment about A} = 1962 \times 0.75 + 490.5 \times 1 = 1471.5 + 490.5 = 1962 \, \text{N} \cdot \text{m} \][/tex]
3. Calculate Reaction at B ([tex]\( R_B \)[/tex]):
[tex]\[ R_B = \frac{\text{Moment about A}}{L} = \frac{1962}{2} = 981 \, \text{N} \][/tex]
4. Using Equilibrium of Vertical Forces:
The sum of vertical forces will be zero because the system is in equilibrium:
[tex]\[ R_A + R_B = W_{\text{beam}} + W_{\text{load}} \][/tex]
Therefore:
[tex]\[ R_A + 981 = 490.5 + 1962 \][/tex]
[tex]\[ R_A + 981 = 2452.5 \][/tex]
[tex]\[ R_A = 2452.5 - 981 = 1471.5 \, \text{N} \][/tex]
Final Answers:
- The reaction at support A ([tex]\( R_A \)[/tex]) is [tex]\( 1471.5 \, \text{N} \)[/tex].
- The reaction at support B ([tex]\( R_B \)[/tex]) is [tex]\( 981 \, \text{N} \)[/tex].
You can proceed with the other questions in your test following similar detailed steps, with explanations for each step to demonstrate the relevant physical principles and calculations involved.
#### (a) Explain why it is easy to open a nut by using a spanner with a long arm than that of a short arm. (3 marks)
Using a spanner with a long arm to open a nut is easier because of the principle of moments (or torque). The moment is the product of the force applied and the distance from the pivot point (in this case, the nut). By using a longer spanner, you increase the distance from the pivot, which increases the torque for the same amount of force applied. This means less force is required to produce the same turning effect, making it easier to open the nut.
#### (b) A metal beam AB of mass 50 kg is supported at its ends. The beam carries a mass of 200 kg at a distance of 0.75 m from end A. If the beam is 2 m long, determine the thrust at supports A and B. (6 marks)
To determine the thrust (reactions) at supports A and B, let's denote them as [tex]\( R_A \)[/tex] and [tex]\( R_B \)[/tex].
Given:
- Mass of beam, [tex]\( m_{\text{beam}} = 50 \, \text{kg} \)[/tex]
- Mass carried, [tex]\( m_{\text{load}} = 200 \, \text{kg} \)[/tex]
- Distance from A to where load is applied, [tex]\( d_{\text{load}} = 0.75 \, \text{m} \)[/tex]
- Length of the beam, [tex]\( L = 2 \, \text{m} \)[/tex]
- Acceleration due to gravity, [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex]
Step-by-Step Solution:
1. Calculate the Weight:
- Weight of the beam, [tex]\( W_{\text{beam}} = m_{\text{beam}} \times g = 50 \times 9.81 = 490.5 \, \text{N} \)[/tex]
- Weight of the load, [tex]\( W_{\text{load}} = m_{\text{load}} \times g = 200 \times 9.81 = 1962 \, \text{N} \)[/tex]
2. Taking Moments about Point A:
The moment due to the load and the beam about point A is given by:
[tex]\[ \text{Moment about A} = W_{\text{load}} \times d_{\text{load}} + W_{\text{beam}} \times \left(\frac{L}{2}\right) \][/tex]
Substituting the values:
[tex]\[ \text{Moment about A} = 1962 \times 0.75 + 490.5 \times 1 = 1471.5 + 490.5 = 1962 \, \text{N} \cdot \text{m} \][/tex]
3. Calculate Reaction at B ([tex]\( R_B \)[/tex]):
[tex]\[ R_B = \frac{\text{Moment about A}}{L} = \frac{1962}{2} = 981 \, \text{N} \][/tex]
4. Using Equilibrium of Vertical Forces:
The sum of vertical forces will be zero because the system is in equilibrium:
[tex]\[ R_A + R_B = W_{\text{beam}} + W_{\text{load}} \][/tex]
Therefore:
[tex]\[ R_A + 981 = 490.5 + 1962 \][/tex]
[tex]\[ R_A + 981 = 2452.5 \][/tex]
[tex]\[ R_A = 2452.5 - 981 = 1471.5 \, \text{N} \][/tex]
Final Answers:
- The reaction at support A ([tex]\( R_A \)[/tex]) is [tex]\( 1471.5 \, \text{N} \)[/tex].
- The reaction at support B ([tex]\( R_B \)[/tex]) is [tex]\( 981 \, \text{N} \)[/tex].
You can proceed with the other questions in your test following similar detailed steps, with explanations for each step to demonstrate the relevant physical principles and calculations involved.
