A 24.0 kg backpack is suspended midway between two trees by a light cord as in the figure (Figure 1) . A bear grabs the backpack and pulls vertically downward with a constant force, so that each section of cord makes an angle of 27 degrees below the horizontal. Initially, without the bear pulling, the angle was 17 degrees ; the tension in the cord with the bear pulling is double what it was when he was not
Well, it seems like the bear decided to take the backpack hostage! No need to panic though, I've got some physics humor to lighten the mood and help you with the problem:
Why did the bear get pulled into physics class?
Because he heard there were plenty of forces to bear!
Now, let's tackle the problem. We have a backpack hanging in the middle of two trees, suspended by a cord. Initially, the cord made an angle of 17 degrees below the horizontal. But then the bear decided to add its own force and make the cord angle increase to 27 degrees.
We're also told that the tension in the cord when the bear pulls is double what it was before. So, if we call the initial tension T, then the new tension with the bear is 2T.
Now, let's find a solution that isn't just "bear"ly accurate!
In order to solve this, we'll use the principle of equilibrium for the vertical and horizontal components of the forces acting on the backpack. This will give us two equations to work with.
In the vertical direction, the weight of the backpack (mg) is balanced by the vertical component of the tension in the cord (T sin θ), where θ is the angle.
So, for the initial case when the angle is 17 degrees:
mg = T sin(17)
When the bear starts pulling and the angle becomes 27 degrees:
mg = (2T) sin(27)
Now, let's unleash our calculating prowess and solve this equation!
To solve this problem step-by-step, we can follow these steps:
Step 1: Analyze the forces acting on the backpack
- In the absence of the bear pulling, the backpack is suspended midway between two trees.
- The weight of the backpack (mg) acts downward vertically.
- The tension in the cord (T) acts upward at an angle of 17 degrees with the horizontal.
- With the bear pulling, the tension in the cord becomes 2T and it acts upward at an angle of 27 degrees below the horizontal.
Step 2: Resolve the forces into vertical and horizontal components
- Let's resolve the tension force and the weight force into vertical and horizontal components.
- The vertical component of the tension force cancels out the weight of the backpack (mg).
- The horizontal components of the tension forces add up to balance each other.
Step 3: Calculate the vertical components of the forces
- The vertical component of the tension force without the bear pulling (T_v1) cancels out the weight of the backpack (mg).
- Vertically: T_v1 = mg
Step 4: Calculate the horizontal components of the forces
- The horizontal component of the tension force without the bear pulling (T_h1) balances each other out, so the value is zero.
- The horizontal component of the tension force with the bear pulling (T_h2) also balances each other out, so the value is zero.
Step 5: Calculate the vertical components of the forces with the bear pulling
- The total vertical force (F_v2) with the bear pulling is equal to the weight of the backpack plus the vertical component of the tension force (2T) downwards.
- Vertically: F_v2 = mg + 2T_v2
Step 6: Calculate the tension in the cord without the bear pulling
- Using the given information, we know that the tension in the cord with the bear pulling (2T) is double what it was without pulling.
- T = 2T_v1
Step 7: Calculate the vertical component of the tension force with the bear pulling
- From the previous step, we know that T = 2T_v1.
- Rearranging the equation, we get: T_v2 = T/2
Step 8: Substitute the values into the equation for F_v2
- Using the expressions for F_v2, mg, and T_v2, we can substitute the values into the equation from Step 5.
- F_v2 = mg + 2T_v2
- F_v2 = mg + 2(T/2) [Substituting T_v2 = T/2]
- F_v2 = mg + T
Step 9: Rearrange the equation for F_v2 to solve for T
- Rearranging the equation from Step 8, we get:
- F_v2 - mg = T
Step 10: Substitute the value of T from Step 6 into the equation from Step 9
- Substituting the value of T from Step 6 into the equation from Step 9, we get:
- F_v2 - mg = 2T_v1
Step 11: Rearrange the equation for T_v1 to find its value
- Rearranging the equation from Step 10, we get:
- T_v1 = (F_v2 - mg) / 2
Using these steps, you can calculate the tension in the cord without the bear pulling (T_v1) and the vertical component of the tension force with the bear pulling (T_v2).
To find the tension in the cord when the bear is pulling, we can use the concept of equilibrium. When the backpack is initially suspended without the bear pulling, the tension in the cord can be determined by considering the forces acting on the backpack.
First, we need to analyze the forces acting on the backpack when it is initially suspended:
1. Weight of the backpack (mg): The weight of the backpack is given as 24.0 kg. The acceleration due to gravity (g) is approximately 9.8 m/s^2. Therefore, the weight of the backpack is 24.0 kg × 9.8 m/s^2 = 235.2 N.
2. Tension in the cord (T1): We can break down the tension force into horizontal and vertical components. The angle between the cord and the horizontal is given as 17 degrees. So, the vertical component of tension is T1 × sin(17°) and the horizontal component is T1 × cos(17°).
When the backpack is in equilibrium, the sum of the forces in the vertical direction and horizontal direction should be zero:
Vertical equilibrium: T1 × sin(17°) = mg
Horizontal equilibrium: T1 × cos(17°) = 0 (since there are no horizontal forces acting on the backpack)
Now, let's find the value of T1:
T1 × sin(17°) = mg
T1 = mg / sin(17°)
T1 = 235.2 N / sin(17°)
T1 ≈ 703.65 N
So, the tension in the cord without the bear pulling (T1) is approximately 703.65 N.
Now, let's find the tension in the cord when the bear is pulling. It is mentioned that the tension in the cord with the bear pulling is double what it was when the bear was not pulling.
Let's denote the tension in the cord with the bear pulling as T2. We know that:
T2 = 2 × T1
T2 = 2 × 703.65 N
T2 ≈ 1407.3 N
Therefore, the tension in the cord when the bear is pulling (T2) is approximately 1407.3 N.