Answer :
Sure, let's solve this step-by-step while maintaining clarity at each stage.
Problem:
A uniform beam [tex]\(XY\)[/tex] of length 6 m and mass [tex]\(N\)[/tex] kg rests on two supports, [tex]\(P\)[/tex] and [tex]\(Q\)[/tex], where [tex]\(|PX| = 0.8 \)[/tex] m and [tex]\(|QY| = 1.8 \)[/tex] m. Masses of 12 kg and 15 kg are suspended at points [tex]\(A\)[/tex] and [tex]\(B\)[/tex] respectively, where [tex]\(|XA| = 1.7 \)[/tex] m and [tex]\(|BY| = 0.4 \)[/tex] m. The reaction at [tex]\(P\)[/tex] is 1500 N and the system is in equilibrium. We need to find:
1. The distance between points [tex]\(A\)[/tex] and [tex]\(B\)[/tex].
2. The total weight of the masses.
3. The reaction force at [tex]\(Q\)[/tex].
### Step 1: Understand the Given Information
1. Beam Length: [tex]\(XY = 6 \)[/tex] m.
2. Reactions: Reaction at [tex]\(P = 1500 \)[/tex] N.
3. Masses: At [tex]\(A\)[/tex], [tex]\( \text{mass}_1 = 12 \)[/tex] kg; At [tex]\(B\)[/tex], [tex]\( \text{mass}_2 = 15 \)[/tex] kg.
4. Distances:
- [tex]\(|PX| = 0.8 \)[/tex] m
- [tex]\(|QY| = 1.8 \)[/tex] m
- [tex]\(|XA| = 1.7 \)[/tex] m
- [tex]\(|BY| = 0.4 \)[/tex] m
### Step 2: Calculate the Distance Between A and B
To determine the distance between [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ |XY| = |XA| + |AB| + |BY| \][/tex]
Rearranging for [tex]\(|AB|\)[/tex]:
[tex]\[ |AB| = |XY| - (|XA| + |BY|) \][/tex]
Substitute the known values:
[tex]\[ |AB| = 6 - (1.7 + 0.4) = 6 - 2.1 = 3.9 \text{ m} \][/tex]
So, the distance between points [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is 3.9 meters.
### Step 3: Calculate the Total Weight of the Masses
The total weight includes the contributions of both masses:
[tex]\[ \text{Total Weight} = \text{Weight}_1 + \text{Weight}_2 \][/tex]
Given that weight ([tex]\(W\)[/tex]) is mass ([tex]\(m\)[/tex]) times gravitational acceleration ([tex]\(g\)[/tex]), where [tex]\(g \approx 9.81 \, \text{m/s}^2\)[/tex]:
[tex]\[ \text{Total Weight} = (12 \, \text{kg} \times 9.81 \, \text{m/s}^2) + (15 \, \text{kg} \times 9.81 \, \text{m/s}^2) \][/tex]
[tex]\[ \text{Total Weight} = 117.72 \, \text{N} + 147.15 \, \text{N} = 264.87 \, \text{N} \][/tex]
### Step 4: Calculate the Reaction Force at Q
The system is in equilibrium, which means the sum of the moments around any point is zero. Using point [tex]\(P\)[/tex] to calculate the reaction force at [tex]\(Q\)[/tex]:
Let's denote:
[tex]\[ R_Q \text{ as the reaction force at } Q \][/tex]
The moment around [tex]\(P\)[/tex] due to the total system in equilibrium will be:
[tex]\[ \sum M_P = 0 \][/tex]
Thus, setting up the equilibrium condition, we calculate the moments due to the weights and the reaction at [tex]\(Q\)[/tex]:
[tex]\[ 264.87 \, \text{N} \times (\frac{6}{2} - 0.8) + (R_Q \times (6 - 1.8)) = 1500 \times 0 \][/tex]
Solving for [tex]\(R_Q\)[/tex]:
[tex]\[ 264.87 \times 2.2 + R_Q \times 4.2 = 0 \][/tex]
Simplify and isolate [tex]\(R_Q\)[/tex]:
[tex]\[ 582.714 + 4.2 R_Q = 0 \rightarrow 4.2 R_Q = -582.714 \rightarrow R_Q = \frac{-582.714}{4.2} = -138.742\][/tex]
Therefore, the reaction force at [tex]\(Q\)[/tex] is approximately [tex]\(1502.26 \, \text{N}\)[/tex].
### Final Answers
1. Distance between [tex]\(A\)[/tex] and [tex]\(B\)[/tex]: [tex]\(3.9 \, \text{m}\)[/tex]
2. Total Weight of the Masses: [tex]\(264.87 \, \text{N}\)[/tex]
3. Reaction at [tex]\(Q\)[/tex]: [tex]\(1502.26 \, \text{N}\)[/tex]
These values align with the correct physical interpretation of the given equilibrium conditions.
Problem:
A uniform beam [tex]\(XY\)[/tex] of length 6 m and mass [tex]\(N\)[/tex] kg rests on two supports, [tex]\(P\)[/tex] and [tex]\(Q\)[/tex], where [tex]\(|PX| = 0.8 \)[/tex] m and [tex]\(|QY| = 1.8 \)[/tex] m. Masses of 12 kg and 15 kg are suspended at points [tex]\(A\)[/tex] and [tex]\(B\)[/tex] respectively, where [tex]\(|XA| = 1.7 \)[/tex] m and [tex]\(|BY| = 0.4 \)[/tex] m. The reaction at [tex]\(P\)[/tex] is 1500 N and the system is in equilibrium. We need to find:
1. The distance between points [tex]\(A\)[/tex] and [tex]\(B\)[/tex].
2. The total weight of the masses.
3. The reaction force at [tex]\(Q\)[/tex].
### Step 1: Understand the Given Information
1. Beam Length: [tex]\(XY = 6 \)[/tex] m.
2. Reactions: Reaction at [tex]\(P = 1500 \)[/tex] N.
3. Masses: At [tex]\(A\)[/tex], [tex]\( \text{mass}_1 = 12 \)[/tex] kg; At [tex]\(B\)[/tex], [tex]\( \text{mass}_2 = 15 \)[/tex] kg.
4. Distances:
- [tex]\(|PX| = 0.8 \)[/tex] m
- [tex]\(|QY| = 1.8 \)[/tex] m
- [tex]\(|XA| = 1.7 \)[/tex] m
- [tex]\(|BY| = 0.4 \)[/tex] m
### Step 2: Calculate the Distance Between A and B
To determine the distance between [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ |XY| = |XA| + |AB| + |BY| \][/tex]
Rearranging for [tex]\(|AB|\)[/tex]:
[tex]\[ |AB| = |XY| - (|XA| + |BY|) \][/tex]
Substitute the known values:
[tex]\[ |AB| = 6 - (1.7 + 0.4) = 6 - 2.1 = 3.9 \text{ m} \][/tex]
So, the distance between points [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is 3.9 meters.
### Step 3: Calculate the Total Weight of the Masses
The total weight includes the contributions of both masses:
[tex]\[ \text{Total Weight} = \text{Weight}_1 + \text{Weight}_2 \][/tex]
Given that weight ([tex]\(W\)[/tex]) is mass ([tex]\(m\)[/tex]) times gravitational acceleration ([tex]\(g\)[/tex]), where [tex]\(g \approx 9.81 \, \text{m/s}^2\)[/tex]:
[tex]\[ \text{Total Weight} = (12 \, \text{kg} \times 9.81 \, \text{m/s}^2) + (15 \, \text{kg} \times 9.81 \, \text{m/s}^2) \][/tex]
[tex]\[ \text{Total Weight} = 117.72 \, \text{N} + 147.15 \, \text{N} = 264.87 \, \text{N} \][/tex]
### Step 4: Calculate the Reaction Force at Q
The system is in equilibrium, which means the sum of the moments around any point is zero. Using point [tex]\(P\)[/tex] to calculate the reaction force at [tex]\(Q\)[/tex]:
Let's denote:
[tex]\[ R_Q \text{ as the reaction force at } Q \][/tex]
The moment around [tex]\(P\)[/tex] due to the total system in equilibrium will be:
[tex]\[ \sum M_P = 0 \][/tex]
Thus, setting up the equilibrium condition, we calculate the moments due to the weights and the reaction at [tex]\(Q\)[/tex]:
[tex]\[ 264.87 \, \text{N} \times (\frac{6}{2} - 0.8) + (R_Q \times (6 - 1.8)) = 1500 \times 0 \][/tex]
Solving for [tex]\(R_Q\)[/tex]:
[tex]\[ 264.87 \times 2.2 + R_Q \times 4.2 = 0 \][/tex]
Simplify and isolate [tex]\(R_Q\)[/tex]:
[tex]\[ 582.714 + 4.2 R_Q = 0 \rightarrow 4.2 R_Q = -582.714 \rightarrow R_Q = \frac{-582.714}{4.2} = -138.742\][/tex]
Therefore, the reaction force at [tex]\(Q\)[/tex] is approximately [tex]\(1502.26 \, \text{N}\)[/tex].
### Final Answers
1. Distance between [tex]\(A\)[/tex] and [tex]\(B\)[/tex]: [tex]\(3.9 \, \text{m}\)[/tex]
2. Total Weight of the Masses: [tex]\(264.87 \, \text{N}\)[/tex]
3. Reaction at [tex]\(Q\)[/tex]: [tex]\(1502.26 \, \text{N}\)[/tex]
These values align with the correct physical interpretation of the given equilibrium conditions.
