PROBLEM 3: Given are concurrent forces having the following magnitudes and passing through the origin and the indicated points: A = 200 N@ P1(+4,+3,+5) B = 400N@P2(+6,-3,-5) C = 300N@P3(-3,+6,-4) a. Compute the x- component of the resultant b. Compute the y-component of the resultant c. Compute the z-component of the resultant d. Resultant force

Structural Analysis
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Chapter2: Loads On Structures
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PROBLEM 3: Given are concurrent
forces having the following magnitudes
and passing through the origin and the
indicated points:
A = 200 N@ P1(+4,+3,+5)
B = 400N@P2(+6,-3,-5)
C= 300N@P3(-3,+6,-4)
a. Compute the x- component of the
resultant
b. Compute the y-component of the
resultant
c. Compute the z-component of the
resultant
d. Resultant force
Transcribed Image Text:PROBLEM 3: Given are concurrent forces having the following magnitudes and passing through the origin and the indicated points: A = 200 N@ P1(+4,+3,+5) B = 400N@P2(+6,-3,-5) C= 300N@P3(-3,+6,-4) a. Compute the x- component of the resultant b. Compute the y-component of the resultant c. Compute the z-component of the resultant d. Resultant force
RESULTANT OF
CONCURRENT FORCES IN
SPACE
Principle: Components ofForce Fare:
Fx = F cosex = F (x/d) – fomula 1
Fy = F cosey = F (y/d) – fomula 2
Fz = F cos8z = F (z/d) – fomula 3
d= Jx2 +y2 +z2 → fomula 4
Transcribed Image Text:RESULTANT OF CONCURRENT FORCES IN SPACE Principle: Components ofForce Fare: Fx = F cosex = F (x/d) – fomula 1 Fy = F cosey = F (y/d) – fomula 2 Fz = F cos8z = F (z/d) – fomula 3 d= Jx2 +y2 +z2 → fomula 4
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