A three-legged structure is shown below. It is attached to the ground at points

Question

A three-legged structure is shown below. It is attached to the ground at points B = (4.52, 0, 0) and C = (?4.52, 0, 0) and connected to the ceiling at point A = (?5.60, 3.76, 5.92). The three legs connect at point D = (0, 3.10, 4.00), where two forces, F and P, act. Force F is given by F=7.80 i?14.0 j+24.4 k N; P has magnitude 40.0 N and direction angles ?=140.0?, ?=54.5?, and ?=74.0? for the x, y, and z axes, respectively.

I NEED A PART D!

Part D - Finding the component of a force perpendicular to a direction

Given F=7.80 i ?14.0 j +24.4 kN, find the component of F that acts perpendicular to member DAsuch that the vector addition of the perpendicular and parallel components of F (F=F?+F?) with respect to DA equals F. Express your answer in component form.

Express your answers, separated by commas, to three significant figures.

Explanation / Answer

vector DA = A - D = (-5.60, 3.76, 5.92) - ( 0, 3.10, 4.00)

vector DA = -5.60i + 0.66j 1.92k

|DA| = sqrt(5.60^2 + 0.66^2 + 1.92^2) = 5.96

component of F along DA

magnitude of F|| = F. DA / |DA|

vector F|| = ( F.DA / |DA|) ( DA / |DA| )

= [(7.80i - 14j + 24.4k).(-5.60i + 0.66j 1.92k ) / 5.96 ] ( -5.60i + 0.66j 1.92k)/5.96

= - 1.09i + 0.13j + 0.375k N

and F = F|| + F_

(7.80i - 14j + 24.4k) = (- 1.09i + 0.13j + 0.375k ) + F_

F_ = 8.89i - 14.13j + 24.02k N

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