To solve this problem, we need to apply the principle of moments (or the principle of torque) about the fulcrum C to ensure the lever remains horizontal. This principle states that for a lever in equilibrium, the sum of clockwise moments about any point is equal to the sum of counterclockwise moments about that same point.
Here's the step-by-step solution:
### Step 1: Identify the Forces and Distances
- Length of the lever AE = 7 meters.
- Fulcrum C is located 3 meters from point A.
- Force at point A ( \( F_A \) ) = 20 Newtons (downward).
- Force at point B ( \( F_B \) ) = 10 Newtons (downward), 2 meters from point A.
- Force at point E ( \( F_E \) ) = 30 Newtons (downward).
- Point D is 1 meter from point E, thus it is 6 meters from point A ( \( AD \) ).
### Step 2: Calculate the Distances from the Fulcrum to Each Force
- Distance from A to C ( \( d_{AC} \) ) = 3 meters.
- Distance from B to C ( \( d_{BC} \) ) = \( d_{AC} - d_{AB} = 3 - 2 = 1 \) meter.
- Distance from E to C ( \( d_{EC} \) ) = \( length_{AE} - d_{AC} = 7 - 3 = 4 \) meters.
- Distance from D to C ( \( d_{DC} \) ) = \( length_{AE} - d_{DE} - d_{AC} = 7 - 1 - 3 = 3 \) meters.
### Step 3: Moment Calculation
A moment (or torque) is the product of a force and its distance from the fulcrum.
First, let's calculate the clockwise moments:
1. Force at A:
[tex]\[ Moment_{A} = F_{A} \times d_{AC} = 20 \times 3 = 60 \, \text{Nm (Clockwise)} \][/tex]
2. Force at B:
[tex]\[ Moment_{B} = F_{B} \times d_{BC} = 10 \times 1 = 10 \, \text{Nm (Clockwise)} \][/tex]
3. Force at E:
[tex]\[ Moment_{E} = F_{E} \times d_{EC} = 30 \times (-4) = -120 \, \text{Nm (Clockwise)} \][/tex]
Note: Negative sign indicates that it acts in the opposite direction.
Now, sum these clockwise moments:
[tex]\[ Moment_{clockwise} = Moment_{A} + Moment_{B} + Moment_{E} \][/tex]
[tex]\[ Moment_{clockwise} = 60 + 10 - 120 = -50 \, \text{Nm} \][/tex]
### Step 4: Determine the Counter-clockwise Moment
Since we need to balance the moments to keep the lever horizontal, let's establish the position of force \( F_D \):
[tex]\[ Moment_{counter\_clockwise} = F_D \times d_{DC} \][/tex]
Setting the clockwise and counterclockwise moments equal:
[tex]\[ |Moment_{clockwise}| = F_D \times d_{DC} \][/tex]
[tex]\[ |-50| = F_D \times 1 \][/tex]
[tex]\[ F_D = \frac{50}{1} \][/tex]
[tex]\[ F_D = -50 \, \text{N} \][/tex]
### Step 5: Direction of the Force
Since we calculated \( F_D \) as a negative value, this indicates the force \( F_D \) must be applied upward to maintain balance.
Therefore, the force \( F_D \) equals 50 Newtons and acts in an upward direction.
### Final Answer:
[tex]\[ \boxed{(a) \, \text{A vertical force of 50 N must be applied at point D, acting upward, in order for the lever AE to remain horizontal.}} \][/tex]
