er vector MME246 FA15 Hw03 HAUMA-DIOMADIK C ezto.mheducation.com/hm.tpx value 7.
ID: 1333215 • Letter: E
Question
er vector MME246 FA15 Hw03 HAUMA-DIOMADIK C ezto.mheducation.com/hm.tpx value 7.14 points From the figure, consider P- 430 N and Q 330 N. Determine the m er p: 430 N and Q = 330 N Determine the magnitude and direction of the resultant of the two forces. From the figure, consid 50 20° The magnitude of the resultant of the two forces is The direction of the resutant of the two forces n the x-direction is The direction of the resutant of the twe forces in the y-direction is The direction of the resutant of the two forces in the z-directon is N (Round the final answer to the nearest whole number) (Round the final answer to a single decimal place ) Round the inal answer to a single decimal place) Round the final answer to a single decimal p Hints ReferencesBook & Resources e
Explanation / Answer
First of all we have to resolve the force P in the Y direction and in the XZ plane
Along Y direction = PSin50 = 430Sin50 = 329.399 N
IN XZ plane = PCos50 = 276.398 N
Now in X and Z direction
Along X = PCos50*Cos20 = 259.729 N
Along Z = -PCos50Sin20 =- 94.5339 N
NOw similarly for Q
Along Y direction = QSin30 = 330Sin30 = 165 N
Along X =- QCos30SIn15 = - 73.967 N
Along Z direction = QCos30Cos15 = 276.05 N
Net force
Along Y = 329.399 + 165 = 494.399 N
Along X = 259.729 - 73.967 = 185.762 N
Along Z = 276.05 - 94.5339 = 181.516 N
Net Resultant = (X2 + Y2 + Z2)1/2 = 558.467 N
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