PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
8th Edition
ISBN: 9781337904285
Author: Larson
Publisher: CENGAGE L
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Chapter 10.2, Problem 68E
To determine

To find: the tension in each of the supporting cables shown in the given figure.

Expert Solution & Answer
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Answer to Problem 68E

The tension in each of the supporting cables is: 159 newtons, 202 newtons, 229 newtons.

Explanation of Solution

Given information: The weight of a crate 500 newtons. The given figure is shown

below.

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 10.2, Problem 68E

Calculation:

From the figure:

First need to find the coordinates of each point.

D = (45, -65, 0)

C = (-60, 0, 0)

B = (0, 70, 0)

A = (0, 0, -115)

Next, need to label the 4 vectors. They are the 3 from the supporting cables and 1 from the weight of the crate that goes straight down the z axis.

u = cable from A to D

v = cable from A to C

z = cable from A to B

w = weight of the crate.

Then need to find vectors u, v, and z based on their coordinates.

A to D = <45-0, -65-0, 0-115> = <45, -65, 115>

A to C = <-60-0, 0-0, 0-115> = <60, 0, 115>

A to B = <0-0, 70-0, 0-115> = <0, 70, 115>

Find the force vectors, u, v, and z

  u=‖u‖AD¯‖AD‖=‖u‖<45,−65,115>452+(−65)2+1152=‖u‖<45,−65,115>19475=‖u‖<0.321,−0.464,0.821>v=‖v‖AC¯‖AC‖=‖v‖<−60,0,115>(−60)2+(0)2+1152=‖v‖<−60,0,115>168255=‖v‖<0.462,0,0.885>z=‖z‖AB¯‖AB‖=‖z‖<0,70,115>02+(70)2+1152=‖z‖<0,70,115>18125=‖z‖<0,0.518,0.851>

Find the force vector for w . It says in the problem that the weight of the crate is 500N. The 500 is only along the z axis so x and y are 0.

w = <0, 0, 500>

Use u, v, z, and w to write a system of equations. U will be the first time in each equation. V will be the second term in each equation. Z will be the third term in each equation. W will be the answer for each equation.

321||u|| + -.462||v|| + 0||z|| = 0

-.464||u|| + 0||v|| + .518||z|| = 0

.821||u|| + .885||v|| + .851||z|| = 500

Solve the system of equations.

.321||u|| + - .462||v|| + 0||z|| = 0

.211||u|| = .462||v||

||u|| = 1.44||v||

-.464||u|| + 0llvll + .518||z|| = 0

.518||z|| = .474||u||

||z|| = .900||u||

||z|| = .900||u|| = .900(1.44||v||)

||z|| = 1.27||v||

.82||u|| + .885||v|| + .851||z|| = 500

.821(1.44||v||) + .885||v|| + .851(1.27||v||) = 500

1.18||v|| + .885||v|| + 1.08||v|| = 500

3.15||v|| = 500

||v|| = 159

||z|| = 1.27||v|| = 1.27(159) = 202

||u|| = 1.44||v|| = 1.44(159) = 229

Therefore, the tension in each of the supporting cables is: 159 newtons, 202 newtons, 229 newtons.

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Chapter 10.2, Problem 67E
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Chapter 10 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

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