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If the magnitude of F2 were increased by 1/3, what is the resultant moment at A? Round your numbers to the nearest whole number.

If the magnitude of F2 were increased by 1/3, what is the resultant moment at A? Round your numbers to the nearest whole number.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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If the magnitude of F2 were increased by 1/3, what is the resultant moment at A? Round your numbers to the nearest whole number.

During a gantry crane maneuver forces exerted on the frame F₁ = = -720k lb 1 F2 = (-300i + 700j + 190k)lb F3 = (450i 700j – 320k)lb Determine the resultant moment of the three forces about the base of the crane at A. F3 75 ft Z B A F2 10 ft 50 ft FBD F3 = (450i -700j -320k) lb 75 ft Z T2,3 F₂: = (-300i + 700j + 190k) lb B(0,0,75) '10 ft Figure 4. Gantry crane diagram and FBD. F₁=(-720k) lb 50 ft (0,0,0)A C(0,10,0) MRA y
Transcribed Image Text:During a gantry crane maneuver forces exerted on the frame F₁ = = -720k lb 1 F2 = (-300i + 700j + 190k)lb F3 = (450i 700j – 320k)lb Determine the resultant moment of the three forces about the base of the crane at A. F3 75 ft Z B A F2 10 ft 50 ft FBD F3 = (450i -700j -320k) lb 75 ft Z T2,3 F₂: = (-300i + 700j + 190k) lb B(0,0,75) '10 ft Figure 4. Gantry crane diagram and FBD. F₁=(-720k) lb 50 ft (0,0,0)A C(0,10,0) MRA y
P1 = ¸ = (0 − 0)i + (10 − 0)j + (0 – 0)k = (10j) ft * 2,3 = (0 - 0)i + (0 - 0)j + (75 −0)k = (75k) ft MRA = • Σ(˜ × F) = (F1 × F1) + (F2 × F2) + (T3 × F3) i = 0 0 j 10 0 k 0 + -720 j k i j k 0 75 + 0 0 75 -300 700 190 450 -700 -320 i 0 = [(−7, 200)i – Oj + 0k)] + [(0 − 52, 500)į – (0+ 22, 500)j + 0k)] + [(0 + 52, 500)ż - (0 – 33, 750)j + Ok] ft — lb MRA = (-7,200i + 11,250j) ft – lb
Transcribed Image Text:P1 = ¸ = (0 − 0)i + (10 − 0)j + (0 – 0)k = (10j) ft * 2,3 = (0 - 0)i + (0 - 0)j + (75 −0)k = (75k) ft MRA = • Σ(˜ × F) = (F1 × F1) + (F2 × F2) + (T3 × F3) i = 0 0 j 10 0 k 0 + -720 j k i j k 0 75 + 0 0 75 -300 700 190 450 -700 -320 i 0 = [(−7, 200)i – Oj + 0k)] + [(0 − 52, 500)į – (0+ 22, 500)j + 0k)] + [(0 + 52, 500)ż - (0 – 33, 750)j + Ok] ft — lb MRA = (-7,200i + 11,250j) ft – lb
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