A supply crate is dropped from an airplane to the survivors of a devastating hurricane. For safety purposes, the crate has two parachutes which deploy from and keep the crate falling at a constant and vertical velocity. The parachutes make upward angles θ1=35.0°θ1​=35.0° and θ2=48.0°θ2​=48.0° with the horizontal. If the tension in the parachute making angle θ1θ1​ is T1=142  NT1​=142N, what must the mass of the crate be?

The problem describes a crate falling vertically with two parachutes, each forming an angle with the horizontal and contributing to the overall tension. The tension forces can be resolved into their horizontal and vertical components. Since the crate is falling vertically with a constant velocity, the horizontal components of the tensions cancel out, and the vertical components balance the weight of the crate. Here's how we solve for the mass of the crate.

### Given Data:
- \( \theta_1 = 35.0^\circ \)
- \( \theta_2 = 48.0^\circ \)
- \( T_1 = 142 \, \text{N} \)

Let \( T_2 \) be the tension in the second parachute, and let the mass of the crate be \( m \). The weight of the crate is \( W = mg \), where \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity.

### Step 1: Resolve Tensions
For each parachute, we can resolve the tension into horizontal and vertical components.

- The vertical component of \( T_1 \) is \( T_1 \sin \theta_1 \)
- The horizontal component of \( T_1 \) is \( T_1 \cos \theta_1 \)
- The vertical component of \( T_2 \) is \( T_2 \sin \theta_2 \)
- The horizontal component of \( T_2 \) is \( T_2 \cos \theta_2 \)

### Step 2: Balance of Forces
#### Horizontal Force Balance:
Since the crate is falling vertically, the horizontal components must cancel each other:
\[
T_1 \cos \theta_1 = T_2 \cos \theta_2
\]
From this equation, we can solve for \( T_2 \):
\[
T_2 = \frac{T_1 \cos \theta_1}{\cos \theta_2}
\]

#### Vertical Force Balance:
The vertical components of the tensions must balance the weight of the crate:
\[
T_1 \sin \theta_1 + T_2 \sin \theta_2 = mg
\]

### Step 3: Substitute \( T_2 \)
Substitute the expression for \( T_2 \) into the vertical force equation:
\[
T_1 \sin \theta_1 + \left( \frac{T_1 \cos \theta_1}{\cos \theta_2} \right) \sin \theta_2 = mg
\]

Now, solve for \( m \):
\[
m = \frac{T_1 \sin \theta_1 + \frac{T_1 \cos \theta_1 \sin \theta_2}{\cos \theta_2}}{g}
\]

### Step 4: Plug in Values
\[
m = \frac{142 \, \text{N} \cdot \sin(35^\circ) + \frac{142 \, \text{N} \cdot \cos(35^\circ) \cdot \sin(48^\circ)}{\cos(48^\circ)}}{9.81 \, \text{m/s}^2}
\]

Let me calculate that for you.The mass of the crate is approximately \( 21.47 \, \text{kg} \).

Would you like more details on the calculations or any other clarification?

Here are 5 related questions to explore further:
1. How would the mass change if the tension in both parachutes was increased?
2. What if the angles of the parachutes were smaller? How would that affect the mass?
3. What would happen if the crate were accelerating downwards instead of falling at a constant velocity?
4. How would the forces change if the parachute cords had some elasticity?
5. What is the role of air resistance in this type of problem?

**Tip:** When resolving forces into components, always check if the system is in equilibrium to simplify the problem-solving process.

A supply crate is dropped from an airplane to the survivors of a devastating hurricane. For safety purposes, the crate has two parachutes which deploy from and keep the crate falling at a constant and vertical velocity. The parachutes make upward angles θ1=35.0° and θ2=48.0° with the horizontal. If the tension in the parachute making angle θ1 is T1=142 N, what must the mass of the crate be?

Find out how to calculate the mass of a supply crate dropped with two parachutes by resolving forces using trigonometry and Newton's laws. Given the angles and tension in one parachute, learn the step-by-step solution for this real-world physics problem. Suitable for high school students.

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