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Using a vector triangle, cosine law, and sine law, determine the resultant of the forces shown in the Figure below. 40 KN ت 50° 10° 30 kN A. The resultant of the two forces, R, is given by 402+302-2(40) (30) cos 130° kN and the direction of the resultant is 30 kN - sin 130° 60°-sin R counterclockwise from the positive x-axis OB. The resultant of the two forces, R, is given by 402 +302-2(40) (30) cos 50° kN and the direction of the resultant is 30 kN - sin 130° 60°-sin R counterclockwise from the positive x-axis OC. The resultant of the two forces, R, is given by 402 +302-2(40)(30) cos 60° kN and the direction of the resultant is 30 kN. sin 60° 60°-sin R counterclockwise from the positive x-axis OD. The resultant of the two forces, R, is given by 40²+30² KN and the direction of the resultant is sin 30 kN - sin 130° R anticounterclockwise from the positive x-axis ○ E. The resultant of the two forces, R, is given by R 30 kN sin 120° sin 50° and the direction of the resultant is 60°, clockwise from the positive x-axis.

Using a vector triangle, cosine law, and sine law, determine the resultant of the forces shown in the Figure below. 40 KN ت 50° 10° 30 kN A. The resultant of the two forces, R, is given by 402+302-2(40) (30) cos 130° kN and the direction of the resultant is 30 kN - sin 130° 60°-sin R counterclockwise from the positive x-axis OB. The resultant of the two forces, R, is given by 402 +302-2(40) (30) cos 50° kN and the direction of the resultant is 30 kN - sin 130° 60°-sin R counterclockwise from the positive x-axis OC. The resultant of the two forces, R, is given by 402 +302-2(40)(30) cos 60° kN and the direction of the resultant is 30 kN. sin 60° 60°-sin R counterclockwise from the positive x-axis OD. The resultant of the two forces, R, is given by 40²+30² KN and the direction of the resultant is sin 30 kN - sin 130° R anticounterclockwise from the positive x-axis ○ E. The resultant of the two forces, R, is given by R 30 kN sin 120° sin 50° and the direction of the resultant is 60°, clockwise from the positive x-axis.

Practical Management Science
Practical Management Science
6th Edition
ISBN:9781337406659
Author:WINSTON, Wayne L.
Publisher:WINSTON, Wayne L.
Chapter11: Simulation Models
Section: Chapter Questions
Problem 54P
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Using a vector triangle, cosine law, and sine law, determine the resultant of the forces shown in the Figure below. 40 KN ت 50° 10° 30 kN A. The resultant of the two forces, R, is given by 402+302-2(40) (30) cos 130° kN and the direction of the resultant is 30 kN - sin 130° 60°-sin R counterclockwise from the positive x-axis OB. The resultant of the two forces, R, is given by 402 +302-2(40) (30) cos 50° kN and the direction of the resultant is 30 kN - sin 130° 60°-sin R counterclockwise from the positive x-axis OC. The resultant of the two forces, R, is given by 402 +302-2(40)(30) cos 60° kN and the direction of the resultant is 30 kN. sin 60° 60°-sin R counterclockwise from the positive x-axis OD. The resultant of the two forces, R, is given by 40²+30² KN and the direction of the resultant is sin 30 kN - sin 130° R anticounterclockwise from the positive x-axis ○ E. The resultant of the two forces, R, is given by R 30 kN sin 120° sin 50° and the direction of the resultant is 60°, clockwise from the positive x-axis.
Transcribed Image Text:Using a vector triangle, cosine law, and sine law, determine the resultant of the forces shown in the Figure below. 40 KN ت 50° 10° 30 kN A. The resultant of the two forces, R, is given by 402+302-2(40) (30) cos 130° kN and the direction of the resultant is 30 kN - sin 130° 60°-sin R counterclockwise from the positive x-axis OB. The resultant of the two forces, R, is given by 402 +302-2(40) (30) cos 50° kN and the direction of the resultant is 30 kN - sin 130° 60°-sin R counterclockwise from the positive x-axis OC. The resultant of the two forces, R, is given by 402 +302-2(40)(30) cos 60° kN and the direction of the resultant is 30 kN. sin 60° 60°-sin R counterclockwise from the positive x-axis OD. The resultant of the two forces, R, is given by 40²+30² KN and the direction of the resultant is sin 30 kN - sin 130° R anticounterclockwise from the positive x-axis ○ E. The resultant of the two forces, R, is given by R 30 kN sin 120° sin 50° and the direction of the resultant is 60°, clockwise from the positive x-axis.
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