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snow_lady [41]
1 year ago
9

Determine the magnitudes of the components of f acting along and perpendicular to segment bc of the pipe assembly

Chemistry
1 answer:
Burka [1]1 year ago
3 0

The magnitude of the component perpendicular to segment BC is 68.0 lb.

Let's take the vectors for points B and C.

rB = [ 3, 4, 0] ft

rC = [ 7, 6, −4] ft

The unit vector going from B to C is then ˆuBC = rC − rB / |rC − rB| = [4, 2, −4] / √(4)^2 + (2)^2 + (−4)^2

Now the dot product of F with ˆuBC to find the component of the force along segment BC of the pipe.

F║ = F · ˆuBC = [30, −45, 50] · [4, 2, −4] / √(4)^2 + (2)^2 + (−4)^2 lb = − 85 / 3 lb

Thus, the magnitude of the component along segment BC is

|F║| = 85/3 lb ≈ 28.3 lb.

Subtract the parallel component of F from F we can get the perpendicular component of F.

F = F − F║ = [30, −45, 50] lb − F║ˆuBC

= [30, −45, 50] lb + 85/3 . [4, 2, −4] / √(4)^2 + (2)^2 + (−4)^2 lb

= [ 440/9, −320/9, 280/9] lb

|F| = √(440/9)^2 + (−320/9)^2 + (280/9)^2 lb ≈ 68.0 lb.

Thus, the magnitude of the component perpendicular to segment BC is  68.0 lb.

If you need to learn more about magnitude click here

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