Skip to main content
Ch. 7 - Applications of Trigonometry and Vectors
Lial - Trigonometry 12th Edition
Lial12th EditionTrigonometryISBN: 9780136552161Not the one you use?Change textbook
All textbooksLial 12th EditionCh. 7 - Applications of Trigonometry and VectorsProblem 44
Chapter 8, Problem 44

A crate is supported by two ropes. One rope makes an angle of 46° 20′ with the horizontal and has a tension of 89.6 lb on it. The other rope is horizontal. Find the weight of the crate and the tension in the horizontal rope.

Verified step by step guidance
1
Identify the forces acting on the crate: the weight (W) acting vertically downward, the tension in the inclined rope (T_1 = 89.6 lb) at an angle of 46° 20′ with the horizontal, and the tension in the horizontal rope (T_2) acting horizontally.
Resolve the tension in the inclined rope into its horizontal and vertical components using trigonometric functions: the horizontal component is \(T_1 \cos(46^\circ 20')\) and the vertical component is \(T_1 \sin(46^\circ 20')\).
Since the crate is in equilibrium, apply the condition that the sum of forces in the horizontal direction is zero: \(T_2 = T_1 \cos(46^\circ 20')\).
Apply the equilibrium condition in the vertical direction: the upward vertical component of the inclined rope's tension balances the weight of the crate, so \(W = T_1 \sin(46^\circ 20')\).
Use the above equations to express the weight of the crate and the tension in the horizontal rope in terms of the given tension and angle, then calculate their numerical values if needed.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Resolving Forces into Components

Forces acting at an angle can be broken down into horizontal and vertical components using trigonometric functions sine and cosine. This allows analysis of each direction separately, which is essential for solving equilibrium problems involving multiple forces.
Recommended video:
03:55
Position Vectors & Component Form

Equilibrium of Forces

An object in equilibrium has no net force acting on it, meaning the sum of forces in both horizontal and vertical directions must be zero. This principle helps set up equations to find unknown tensions and weights in static systems.
Recommended video:
04:39
45-45-90 Triangles

Trigonometric Functions and Angle Conversion

Understanding how to use sine and cosine with angles given in degrees and minutes is crucial. Converting 46° 20′ into decimal degrees allows accurate calculation of force components, ensuring precise results in tension and weight determination.
Recommended video:
6:04
Introduction to Trigonometric Functions
Related Practice
Textbook Question

Given vectors u and v, find: v - 3u. 

u = 〈-1, 2〉, v = 〈3, 0〉

Textbook Question

A ship leaves port on a bearing of 34.0° and travels 10.4 mi. The ship then turns due east and travels 4.6 mi. How far is the ship from port, and what is its bearing from port?

Textbook Question

A luxury liner leaves port on a bearing of 110.0° and travels 8.8 mi. It then turns due west and travels 2.4 mi. How far is the liner from port, and what is its bearing from port?

6
views
Textbook Question

Given vectors u and v, find: 2u + 3v. 

u = 〈-1, 2〉, v = 〈3, 0〉

Textbook Question

Write each vector in the form a i + b j.

〈6, -3〉

1
views
Textbook Question

Given vectors u and v, find: 2u. 

u = 〈-1, 2〉, v = 〈3, 0〉