Question
A uniform beam of mass 4kg is hinged to a wall at end A and held horizontally by a with joining B to a point C which is on the waal vertically above A. If a stone of mass 10kg is hinged at B and given that B has an inclination to the string 30°. Find the force in the wire connecting B to C
The force exerted by the beam on the hinge
The force exerted by the beam on the hinge
Answers
GPT-4o mini
To find the force in the wire connecting point B to point C, we need to analyze the forces and moments acting on the system.
### Step 1: Draw the Diagram
Let's represent the situation:
- Point A is the hinge on the wall.
- Point B is where the beam is hinged and where the stone is attached (at an angle of 30° with respect to the wire).
- Point C is the point on the wall directly above A.
### Step 2: Identify the Forces
1. **Weight of the Beam** (\( W_b \)): The weight of the beam acts downwards through its center of gravity (which is at its midpoint). \( W_b = m_b \cdot g = 4 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 39.24 \, \text{N} \).
2. **Weight of the Stone** (\( W_s \)): The weight of the stone also acts downwards at point B. \( W_s = m_s \cdot g = 10 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 98.1 \, \text{N} \).
3. **Tension in the Wire** (\( T \)): The wire connecting B to C exerts an upward force \( T \) at angle 30° to the vertical.
4. **Reaction Force at Hinge A**: This includes both horizontal and vertical components, but its exact value and direction will be determined after analyzing the equilibrium conditions.
### Step 3: Set Up Equilibrium Conditions
Since the beam is held horizontally in static equilibrium, we can set up the following equations for the vertical and horizontal forces:
#### Vertical Forces:
Considering upward forces as positive:
\[
T \cdot \cos(30°) - (W_b + W_s) = 0 \quad \text{(1)}
\]
The total weight acting downwards is the weight of the beam plus the weight of the stone:
\[
W_b + W_s = 39.24 \, \text{N} + 98.1 \, \text{N} = 137.34 \, \text{N}
\]
#### Horizontal Forces:
Since there is no horizontal motion, the horizontal component of the tension must balance any horizontal forces. In this case, only the force at A (reaction force components) acts horizontally.
### Step 4: Find the Tension
From the vertical forces equation (1):
\[
T \cdot \cos(30°) = 137.34 \, \text{N}
\]
Using \( \cos(30°) = \sqrt{3}/2 \):
\[
T \cdot \frac{\sqrt{3}}{2} = 137.34
\]
\[
T = \frac{137.34 \times 2}{\sqrt{3}} \approx \frac{274.68}{1.732} \approx 158.41 \, \text{N}
\]
### Step 5: Find the Force Exerted by the Beam on the Hinge
Next, we need to analyze the moments around point A. The distance from A to B is \( L \) and from A to C is the vertical distance \( L \cdot \sin(30°) \).
Setting up the moment about point A and considering moments due to the weights and tension:
\[
T \cdot L \cdot \sin(30°) = (W_b + W_s) \cdot \frac{L}{2}
\]
Calculating:
\[
158.41 \cdot L \cdot \frac{1}{2} = 137.34 \cdot \frac{L}{2}
\]
This shows that we are indeed in equilibrium; thus, we can express the reaction force at A, \( R_A \), will be:
\[
R_A = (W_b + W_s) + T \cdot \sin(30°)
\]
Substituting values:
\[
R_A = 137.34 + 158.41 \cdot \frac{1}{2} \approx 137.34 + 79.205 = 216.545 \, \text{N}
\]
### Conclusion
1. The force in the wire connecting B to C: **Approximately 158.41 N.**
2. The force exerted by the beam at the hinge: **Approximately 216.55 N.**
### Step 1: Draw the Diagram
Let's represent the situation:
- Point A is the hinge on the wall.
- Point B is where the beam is hinged and where the stone is attached (at an angle of 30° with respect to the wire).
- Point C is the point on the wall directly above A.
### Step 2: Identify the Forces
1. **Weight of the Beam** (\( W_b \)): The weight of the beam acts downwards through its center of gravity (which is at its midpoint). \( W_b = m_b \cdot g = 4 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 39.24 \, \text{N} \).
2. **Weight of the Stone** (\( W_s \)): The weight of the stone also acts downwards at point B. \( W_s = m_s \cdot g = 10 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 98.1 \, \text{N} \).
3. **Tension in the Wire** (\( T \)): The wire connecting B to C exerts an upward force \( T \) at angle 30° to the vertical.
4. **Reaction Force at Hinge A**: This includes both horizontal and vertical components, but its exact value and direction will be determined after analyzing the equilibrium conditions.
### Step 3: Set Up Equilibrium Conditions
Since the beam is held horizontally in static equilibrium, we can set up the following equations for the vertical and horizontal forces:
#### Vertical Forces:
Considering upward forces as positive:
\[
T \cdot \cos(30°) - (W_b + W_s) = 0 \quad \text{(1)}
\]
The total weight acting downwards is the weight of the beam plus the weight of the stone:
\[
W_b + W_s = 39.24 \, \text{N} + 98.1 \, \text{N} = 137.34 \, \text{N}
\]
#### Horizontal Forces:
Since there is no horizontal motion, the horizontal component of the tension must balance any horizontal forces. In this case, only the force at A (reaction force components) acts horizontally.
### Step 4: Find the Tension
From the vertical forces equation (1):
\[
T \cdot \cos(30°) = 137.34 \, \text{N}
\]
Using \( \cos(30°) = \sqrt{3}/2 \):
\[
T \cdot \frac{\sqrt{3}}{2} = 137.34
\]
\[
T = \frac{137.34 \times 2}{\sqrt{3}} \approx \frac{274.68}{1.732} \approx 158.41 \, \text{N}
\]
### Step 5: Find the Force Exerted by the Beam on the Hinge
Next, we need to analyze the moments around point A. The distance from A to B is \( L \) and from A to C is the vertical distance \( L \cdot \sin(30°) \).
Setting up the moment about point A and considering moments due to the weights and tension:
\[
T \cdot L \cdot \sin(30°) = (W_b + W_s) \cdot \frac{L}{2}
\]
Calculating:
\[
158.41 \cdot L \cdot \frac{1}{2} = 137.34 \cdot \frac{L}{2}
\]
This shows that we are indeed in equilibrium; thus, we can express the reaction force at A, \( R_A \), will be:
\[
R_A = (W_b + W_s) + T \cdot \sin(30°)
\]
Substituting values:
\[
R_A = 137.34 + 158.41 \cdot \frac{1}{2} \approx 137.34 + 79.205 = 216.545 \, \text{N}
\]
### Conclusion
1. The force in the wire connecting B to C: **Approximately 158.41 N.**
2. The force exerted by the beam at the hinge: **Approximately 216.55 N.**