To analyze the forces exerted at the hinges of the farm gate, we need to consider the principles of static equilibrium. Since the gate is at rest and not moving, the sum of the forces and the sum of the moments (or torques) acting on it must equal zero. Let's break this down step by step.
Understanding the Setup
We have a 20-lb gate that is 12 ft long, supported by two hinges that are 4 ft apart. The weight of the gate acts downward at its center of gravity, which for a uniform gate is located at its midpoint, 6 ft from either end. The upper hinge supports the majority of the weight, while the lower hinge provides additional support and stability.
Identifying Forces
Let’s denote the forces at the upper hinge as \( F_u \) and at the lower hinge as \( F_l \). The weight of the gate, \( W \), is 20 lbs, acting downward at the center of the gate. Since the gate is in static equilibrium, we can set up our equations based on the following conditions:
- The sum of vertical forces must equal zero.
- The sum of moments about any point must equal zero.
Calculating Vertical Forces
From the vertical force equilibrium, we have:
\( F_u + F_l - W = 0 \)
Substituting the weight of the gate:
\( F_u + F_l - 20 = 0 \)
This simplifies to:
\( F_u + F_l = 20 \)
Calculating Moments
Next, we can take moments about the lower hinge to eliminate \( F_l \) from our calculations. The moment due to the weight of the gate is calculated as follows:
The distance from the lower hinge to the center of the gate (where the weight acts) is 6 ft. The moment due to the weight is:
\( \text{Moment} = W \times \text{distance} = 20 \, \text{lbs} \times 6 \, \text{ft} = 120 \, \text{lb-ft} \)
The moment due to the force at the upper hinge (acting upward) is:
\( \text{Moment} = F_u \times \text{distance} = F_u \times 4 \, \text{ft} \)
Setting the sum of moments about the lower hinge to zero gives us:
\( F_u \times 4 - 120 = 0 \)
From this, we can solve for \( F_u \):
\( F_u \times 4 = 120 \)
\( F_u = \frac{120}{4} = 30 \, \text{lbs} \)
Finding the Force at the Lower Hinge
Now that we have \( F_u \), we can substitute it back into our vertical forces equation:
\( 30 + F_l = 20 \)
Solving for \( F_l \) gives:
\( F_l = 20 - 30 = -10 \, \text{lbs} \)
The negative value indicates that the lower hinge is not actually supporting any weight; instead, it is experiencing a reaction force in the opposite direction, which means it is effectively lifting the gate slightly. This is typical in scenarios where one hinge bears the entire weight.
Summary of Forces
In conclusion, the forces exerted at the hinges are:
- Upper hinge force (F_u): 30 lbs (upward)
- Lower hinge force (F_l): -10 lbs (downward, indicating a reaction force)
This analysis illustrates the importance of understanding static equilibrium in mechanical systems, particularly in structures like gates where weight distribution and support play critical roles.
