Question

Three forces that act on a particle are given by F1 = 20i + 36j – 65k N, F2 = 2i + 36j N and
F3 = 10i – 65k N. Find their resultant and magnitude.

Answer 1

32i  + 72j  - 130k N & magnitude =  152.9 N  where F1 = 20i + 36j – 65k N, F2 = 2i + 36j N and F3 = 10i – 65k N.

Step-by-step explanation:

F1 = 20i + 36j – 65k N,

F2 = 2i + 36j N and

F3 = 10i – 65k N

F1 + F2 + F3  =   20i + 36j – 65k  + 2i + 36j  + 10i – 65k   N

=> F1 + F2 + F3  =    32i  + 72j  - 130k

Magnitude = √36² + 72² + 130²  = √23380 = 152.9

Learn more :

find the value of Vector (A + 2B).(2A-B)

https://brainly.in/question/11569264

what angle is the resultant incline to vector of magnitude 9

https://brainly.in/question/34219

Answer 2

Resultant =$32 i+72 j-130 k$ N

Magnitude=152

Step-by-step explanation:

$F_1=20 i+36 j-65 k N$

$F_2=2 i+36 j$N

$F_3=10 i-65 k$N

Resultant force=$F_1+F_2+F_3$

Substitute the values then we get

Resultant force=$20 i+36 j-65 k+2 i+36 j+10 i-65 k=32 i+72 j-130 k$ N

Magnitude =$\sqrt{x^2+y^2+z^2}$

Where x= Coefficient of unit vector i

y=Coefficient of unit vector j

z=Coefficient of unit vector k

Magnitude =$\sqrt{(32)^2+(72)^2+(-130)^2}$

Magnitude=152

#Learns more:

https://brainly.in/question/3320047