A bench seat is attached to a wall by hinges and supported from above by two ropes, as shown in the figure below. The seat weighs 139 N and is designed to support a person weighing up to 1300 N, sitting on the edge. How much tension must each rope be able to withstand? (Let θ = 31.5°.)
To solve this problem, we can analyze the forces acting on the bench seat using the principles of equilibrium.
Step 1: Draw a free body diagram of the bench seat.
We will draw a diagram showing all the forces acting on the bench seat.
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- There are two ropes supporting the bench seat from above.
- The weight of the bench seat acts downward.
- The tension in the ropes acts upward at an angle θ.
Step 2: Break down the forces into their components.
We need to break down the forces into their vertical and horizontal components, since the ropes make an angle with the vertical.
- The weight of the bench seat (139 N) can be broken down into a vertical component (Wv = 139 N * cosθ) and a horizontal component (Wh = 139 N * sinθ).
- The tension in the ropes can also be broken down into a vertical component (Tv = 2T * cosθ) and a horizontal component (Th = 2T * sinθ).
Step 3: Set up the equilibrium equations.
In order for the bench seat to be in equilibrium, the sum of the forces in the vertical direction and the sum of the forces in the horizontal direction must both equal zero.
Vertical equilibrium equation:
Tv + Wv = 0
Horizontal equilibrium equation:
Th = Wh
Step 4: Solve the equations to find the tension in the ropes.
From the vertical equilibrium equation, we can solve for Tv:
Tv = -Wv
2T * cosθ = -W * cosθ
2T = -W
T = -W/2
T = -(139 N * cosθ)/2
From the horizontal equilibrium equation, we can solve for Th:
Th = Wh
2T * sinθ = W * sinθ
2T = W
T = W/2
T = (139 N * sinθ)/2
So, the tension in each rope must be able to withstand:
T = -(139 N * cosθ)/2
T = (139 N * sinθ)/2
Given that θ = 31.5°, we can substitute the value of θ into the equations to find the tension values.
To find the tension in each rope, we will analyze the forces acting on the bench seat and use Newton's laws of motion.
Let's consider the bench seat as a system and draw a free-body diagram.
1. Start by drawing the bench seat as a rectangle, with the weight (139 N) acting at its center, pointing straight downwards.
<img src="https://i.imgur.com/qVgBHTG.png" alt="Free-body diagram of bench seat">
2. Now, let's resolve the weight force (139 N) into two components: one parallel to the ropes and one perpendicular to them.
<img src="https://i.imgur.com/JHfzHj7.png" alt="Resolved forces">
The component parallel to the ropes (mg*sin(theta)) will create tension in the ropes, and the component perpendicular to the ropes (mg*cos(theta)) will exert a force on the wall.
3. Apply Newton's second law of motion in the vertical direction (y-axis):
∑Fy = T₁ + T₂ - mg*sin(theta) = 0
Since the bench seat is in equilibrium, the sum of the vertical forces must be zero. Here, T₁ and T₂ are the tensions in the ropes, and mg*sin(theta) is the vertical component of the weight.
4. Apply Newton's second law of motion in the horizontal direction (x-axis):
∑Fx = -mg*cos(theta) = 0
Again, since the bench seat is in equilibrium, the sum of the horizontal forces must be zero. Here, mg*cos(theta) is the horizontal component of the weight. Note that this force is balanced by a normal force from the wall.
5. Now let's solve the equations. Rearrange the equation in the y-direction:
T₁ + T₂ = mg*sin(theta)
Substituting the given values:
T₁ + T₂ = (139 N + 1300 N)*sin(31.5°)
6. With the horizontal component in equilibrium (∑Fx = 0), we have:
-mg*cos(theta) = 0
Which is already satisfied by the given values, as there is no horizontal acceleration.
7. Solving the equation from step 5:
T₁ + T₂ = (1439 N)*sin(31.5°)
T₁ + T₂ = 736.6 N
8. Since the bench seat is symmetric, we can assume the tension in each rope is equal. So, we divide the total tension equally between the two ropes:
T₁ = T₂ = 736.6 N / 2
T₁ = T₂ = 368.3 N
Therefore, each rope must be able to withstand a tension of 368.3 N.