P= 120 lb30°70°0 = 100 lbUsing direction cosines, the vector P in the diagram can be expressed asO P = (120lb cos 30°) i + (120lb sin 60°) jO P = (120lb cos 60°) i + (120lb cos 30°) jO P = (120lb cos 30°) i + (120lb cos 60°) j

Given: The magnitude of the P is 120 lb. The angle of the P is 30o with positive x-axis.

P= 120 lb 30° 70° 0 = 100 lb Using direction cosines, the vector P in the diagram can be expressed as * O OP = (120lb cos 30°) i + (120lb sin 60*) j O P = (120lb cos 60°) i + (120lb cos 30°) j OP = (120lb cos 30°) i + (120lb cos 60°) j

P= 120 lb 30° 70° 0 = 100 lb Using direction cosines, the vector P in the diagram can be expressed as * O OP = (120lb cos 30°) i + (120lb sin 60*) j O P = (120lb cos 60°) i + (120lb cos 30°) j OP = (120lb cos 30°) i + (120lb cos 60°) j

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

P= 120 lb
30°
70°
0 = 100 lb
Using direction cosines, the vector P in the diagram can be expressed as
O P = (120lb cos 30°) i + (120lb sin 60°) j
O P = (120lb cos 60°) i + (120lb cos 30°) j
O P = (120lb cos 30°) i + (120lb cos 60°) j

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