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PROBLEM 05. Refer to problem SP203. Change P=25 KN up to the left at 30 degrees w/ x-axis, Q=30 KN, W=35 KN, T=18 KN directed to the right.

PROBLEM 05. Refer to problem SP203. Change P=25 KN up to the left at 30 degrees w/ x-axis, Q=30 KN, W=35 KN, T=18 KN directed to the right.

Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter1: Tension, Compression, And Shear
Section: Chapter Questions
Problem 1.8.22P: A cargo ship is tied down to marine boll arts at a number of points along its length while its cargo...
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Please refer to PROBLEM SP203 to answer PROBLEM #5.

OBLEM SP203. From the gven set of concurrent forces shown in the figure, determine the Resultant force which could represents them. P=I OKN Q=20KN Figure, 60° 3. Solubion: T=I OKN a). METHOD I (Using the components of the resultant equations), Rx =£x = P,x +Qr=Ty =10KN(cos 60° )+20KN(3/5)–10KN =7 KN Ry = £Y=- P, +Q, -W, = -10KN(sin 60° )+20KN(4/5)–30KN =- 22.66KN w=30KN %3D Ry (RF +(R,} =/(7) +(22.66) = 23.72KN 0, = tan RY 22.66 = 72.83° 7 -1 = tan downwarc to the nght, DR
Transcribed Image Text:OBLEM SP203. From the gven set of concurrent forces shown in the figure, determine the Resultant force which could represents them. P=I OKN Q=20KN Figure, 60° 3. Solubion: T=I OKN a). METHOD I (Using the components of the resultant equations), Rx =£x = P,x +Qr=Ty =10KN(cos 60° )+20KN(3/5)–10KN =7 KN Ry = £Y=- P, +Q, -W, = -10KN(sin 60° )+20KN(4/5)–30KN =- 22.66KN w=30KN %3D Ry (RF +(R,} =/(7) +(22.66) = 23.72KN 0, = tan RY 22.66 = 72.83° 7 -1 = tan downwarc to the nght, DR
PROBLEM 05. Refer to problem SP203. Change P=25 KN up to the left at 30 degrees w/ x-axis, Q=30 KN, W=35 KN, T=18 KN directed to the right.
Transcribed Image Text:PROBLEM 05. Refer to problem SP203. Change P=25 KN up to the left at 30 degrees w/ x-axis, Q=30 KN, W=35 KN, T=18 KN directed to the right.
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