Math Problem Statement

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?

Solution

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:

  • θ1=35.0\theta_1 = 35.0^\circ
  • θ2=48.0\theta_2 = 48.0^\circ
  • T1=142NT_1 = 142 \, \text{N}

Let T2T_2 be the tension in the second parachute, and let the mass of the crate be mm. The weight of the crate is W=mgW = mg, where g=9.81m/s2g = 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 T1T_1 is T1sinθ1T_1 \sin \theta_1
  • The horizontal component of T1T_1 is T1cosθ1T_1 \cos \theta_1
  • The vertical component of T2T_2 is T2sinθ2T_2 \sin \theta_2
  • The horizontal component of T2T_2 is T2cosθ2T_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: T1cosθ1=T2cosθ2T_1 \cos \theta_1 = T_2 \cos \theta_2 From this equation, we can solve for T2T_2: T2=T1cosθ1cosθ2T_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: T1sinθ1+T2sinθ2=mgT_1 \sin \theta_1 + T_2 \sin \theta_2 = mg

Step 3: Substitute T2T_2

Substitute the expression for T2T_2 into the vertical force equation: T1sinθ1+(T1cosθ1cosθ2)sinθ2=mgT_1 \sin \theta_1 + \left( \frac{T_1 \cos \theta_1}{\cos \theta_2} \right) \sin \theta_2 = mg

Now, solve for mm: m=T1sinθ1+T1cosθ1sinθ2cosθ2gm = \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=142Nsin(35)+142Ncos(35)sin(48)cos(48)9.81m/s2m = \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 21.47kg21.47 \, \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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Trigonometry
Force Resolution
Equilibrium

Formulas

T1 cos θ1 = T2 cos θ2
T1 sin θ1 + T2 sin θ2 = mg
m = (T1 sin θ1 + (T1 cos θ1 * sin θ2) / cos θ2) / g

Theorems

Newton's First Law of Motion
Trigonometric identities

Suitable Grade Level

High School - Physics (Grades 10-12)