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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)

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)

Grade:

2 Answers

Sourabh Singh IIT Patna
askIITians Faculty 2104 Points
5 years ago
What is the question that you want to ask. Repost and we will reach over to you ASAP
SHIVAM KHAJURIA
14 Points
one month 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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