The uniform rod AB is suspended in air and its two ends are attached to supports by hinges.The rod has a weight of 15 lb and the spring is unstretched when θ = 0°. If θ = 30°, determine the stiffness k of the spring so that the rod is in equilibrium.
You need to describe the system better, since there is no figure. Where is the spring attached? In what direction does it pull or push? How is theta measured?
To determine the stiffness (k) of the spring so that the rod is in equilibrium when θ = 30°, we need to analyze the forces acting on the rod.
First, let's draw a free-body diagram to visualize the forces involved:
```
A B
--------|----------|--------
↓ ← F_spring ↓
```
Here, A and B represent the hinges, and F_spring is the force exerted by the spring. The weight of the rod (15 lb) acts at its center of gravity, which we'll mark as G.
Now, let's break down the forces acting on the rod:
1. Weight (W):
The weight, which acts through the center of gravity G, can be resolved into horizontal (x) and vertical (y) components:
W_x = 0 (since the rod is in equilibrium horizontally)
W_y = -15 lb
2. Tension at the hinges (T_A and T_B):
The tension in the hinges acts horizontally to provide the required support. Since the rod is uniform and in equilibrium, we can assume both hinge tensions are equal in magnitude.
T_A = T_B (let's call this common value T)
3. Force exerted by the spring (F_spring):
The force exerted by the spring can also be resolved into horizontal (F_spring_x) and vertical (F_spring_y) components.
To find the stiffness (k) of the spring, we need to consider the condition for equilibrium. In equilibrium, the sum of the horizontal and vertical forces acting on the rod is zero.
Vertical equilibrium:
T - 15 lb - F_spring_y = 0 (Equation 1)
Horizontal equilibrium:
F_spring_x = 0 (Equation 2)
Now let's determine the values of F_spring_x and F_spring_y:
Given that the spring is unstretched when θ = 0°, we can determine the displacement (x) of point A at θ = 30° using trigonometry.
x = l * sin(θ) (Equation 3)
where l is the length of the rod.
Since point A is displaced by x, point B will also be displaced by x. Therefore, the total displacement of the spring is 2x.
The force exerted by the spring is given by Hooke's Law:
F_spring = -k * x (Equation 4)
Substituting Equation 3 into Equation 4, we get:
F_spring = -k * l * sin(θ) (Equation 5)
Next, we can resolve the forces horizontally and vertically:
F_spring_x = -k * l * sin(θ) * cos(θ) (Equation 6)
F_spring_y = -k * l * sin²(θ) (Equation 7)
Now, substituting Equations 6 and 7 into Equations 1 and 2:
T - 15 lb - k * l * sin²(θ) = 0 (Equation 8)
-k * l * sin(θ) * cos(θ) = 0 (Equation 9)
Since the rod is in equilibrium, Equations 8 and 9 must hold true. From Equation 9, we can determine that either k = 0 (which implies no spring force) or sin(θ) * cos(θ) = 0.
To ensure there is a non-zero spring force, we must have sin(θ) * cos(θ) = 0. Since sin(θ) * cos(θ) = 1/2 * sin(2θ), we set 2θ = 90° (π/2) to satisfy this condition.
Therefore, θ = π/4 (45°).
Now, solving Equation 8 for k:
T - 15 lb - k * l * sin²(π/4) = 0
Given that T = 15 lb (as T_A = T_B), and l is the length of the rod (which is not provided in the question), we can solve for k:
k * l * 0.5 = 15 lb
k * l = 30 lb
The value of k depends on the length of the rod (l). Without the specified value of l, we cannot determine the exact value of k.