F- 200 N F1 = {300(4/5) j + 300 (3/5) k} N F2 = 600 N F, = {240 j + 180 k} N F1=300N 3m F2 = {- 600 i} N 4 m F. F3 = {- 900 k} N F3=V900 N F, = F4 (rg/ TB) %3D 200 N [(3i – 4 j+ 6 k)/(3² + 4² + 6²)½] {76.8 i – 102.4 j+ 153.6 k} N F;= { F,i – Fy j +F, k} N %3| - EXAMPLE #1 (continued) Equating the respective i, j, k components to zero, we have EF, = 76.8 – 600 + F, 0; solving gives F= 523.2 N = %3D = 240 EF y' 102.4 + F, = 0; solving gives Fy =- 137.6 N - = 180– 900 + 153.6 + F,= 0; solving gives F, = 566.4 N EF z Thus, F5= {523 i – 138 j+566 k} N %3D
F- 200 N F1 = {300(4/5) j + 300 (3/5) k} N F2 = 600 N F, = {240 j + 180 k} N F1=300N 3m F2 = {- 600 i} N 4 m F. F3 = {- 900 k} N F3=V900 N F, = F4 (rg/ TB) %3D 200 N [(3i – 4 j+ 6 k)/(3² + 4² + 6²)½] {76.8 i – 102.4 j+ 153.6 k} N F;= { F,i – Fy j +F, k} N %3| - EXAMPLE #1 (continued) Equating the respective i, j, k components to zero, we have EF, = 76.8 – 600 + F, 0; solving gives F= 523.2 N = %3D = 240 EF y' 102.4 + F, = 0; solving gives Fy =- 137.6 N - = 180– 900 + 153.6 + F,= 0; solving gives F, = 566.4 N EF z Thus, F5= {523 i – 138 j+566 k} N %3D
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Question
the variable underlined in red is negative then turns to positive. is it positive or negative?
![F = 200 N
F1 = {300(4/5) j + 300 (3/5) k} N
6 m
F2 = 600 N
F. = {240 j + 180 k} N
F1=300N
3m
F2 = {- 600 i} N
4 m
F3
: {- 900 k} N
F3=V 900 N
F4
F4 (rg/ TB)
= 200 N[(3i – 4 j+6 k/(3² + 4² + 6?)¾]
{76.8 i
102.4 j + 153.6 k} N
=
F;= { F,i – F, j +F, k} N
у
EXAMPLE #1
(continued)
Equating the respective i, j, k components to zero, we have
EF, = 76.8 – 600 + F,
0;
solving gives F,= 523.2 N
EF,
240
102.4 + F, = 0;
solving gives Fy
=-137.6 N
=
180 – 900 + 153.6 + F,= 0; solving gives F, = 566.4 N
EF z
%3D
%3|
Thus, F5= {523 i – 138 j+566 k} N
%3D
Using this force vector, you can determine the force's magnitude
and coordinate direction angles as needed.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa309386b-12aa-4359-937d-54f87907681c%2Fbb560291-c324-4f97-8f19-04e6608c63cf%2F4i96jp_processed.png&w=3840&q=75)
Transcribed Image Text:F = 200 N
F1 = {300(4/5) j + 300 (3/5) k} N
6 m
F2 = 600 N
F. = {240 j + 180 k} N
F1=300N
3m
F2 = {- 600 i} N
4 m
F3
: {- 900 k} N
F3=V 900 N
F4
F4 (rg/ TB)
= 200 N[(3i – 4 j+6 k/(3² + 4² + 6?)¾]
{76.8 i
102.4 j + 153.6 k} N
=
F;= { F,i – F, j +F, k} N
у
EXAMPLE #1
(continued)
Equating the respective i, j, k components to zero, we have
EF, = 76.8 – 600 + F,
0;
solving gives F,= 523.2 N
EF,
240
102.4 + F, = 0;
solving gives Fy
=-137.6 N
=
180 – 900 + 153.6 + F,= 0; solving gives F, = 566.4 N
EF z
%3D
%3|
Thus, F5= {523 i – 138 j+566 k} N
%3D
Using this force vector, you can determine the force's magnitude
and coordinate direction angles as needed.
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