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A system of concurrent forces P,Q,& R has following magnitudes and passing through origin and the indicated points are
P=280N(12,6,-4)
Q=520N(-3,-4,12)
R=270N(6,-3,-6)

Profile image of dave nirav
14 Years agoGrade
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2 Answers

Profile image of Sourabh Singh
10 Years ago
What is the question that you want to ask. Repost and we will reach over to you ASAP
Profile image of SHIVAM KHAJURIA
5 Years ago
394N
P= 280 ( 12i + 6j - 4k)/((12)^2 + (6)^2 + (-4)^2)^(1/2)
= 280 (12i +6j -4k )/14
= 20(12i+6j-4k)-------------------------(1
Q= 520 (-3i -4j +12k)/((-3)^2 +(-4)^2 +(12)^2)^(1/2)
Q=520 (-3i-4j+12k)/13
Q=40(-3i-4j+12k)---------------------------(2
 
R = 270(6i-3j-6k)/((6)^2+(-3)^2+(-6)^2))^(1/2)
R= 270(6i-3j-6k)/9
R= 30(6i-3j-6k)-------------------(3
 
Add 1 + 2 +3 
R=20(12i+6j-4k) +40(-3j-4j+12k)+30(6i-3j-6k)
R= 300i -130j +220k
Magnitude =((300)^2+(-130)^2+(220)^2)^(1/2)

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