Force vector F with magnitude F = 5 kN is acting on the tip of the inclined pole

Question

Force vector F with magnitude F = 5 kN is acting on the tip of the inclined pole at B with the direction and sense shown.
Force vector F with magnitude F = 5 kN is acting on the tip of the inclined pole at B with the direction and sense shown. (a) Express F in Cartesian vector form. (b) Determine the magnitude of the component of F that runs parallel to the inclined pole (i.e., F? to BA) (c) Express the parallel component in (b) in Cartesian vector form (i.e., F? to BA) (d) Determine the component of F that runs perpendicular to the inclined pole in Cartesian vector form (i.e., F? to BA). (e) Determine the magnitude of the perpendicular component in (d) (i.e., F? to BA) (f) Calculate the angle ? between F and the inclined pole BA. (a) Express F in Cartesian vector form. (b) Determine the magnitude of the component of F that runs parallel to the inclined pole (i.e., F? to BA) (c) Express the parallel component in (b) in Cartesian vector form (i.e., F? to BA) (d) Determine the component of F that runs perpendicular to the inclined pole in Cartesian vector form (i.e., F? to BA). (e) Determine the magnitude of the perpendicular component in (d) (i.e., F? to BA) (f) Calculate the angle ? between F and the inclined pole BA. C (3,0,4) m B (5, 6, 1) m.

Explanation / Answer

a) F in Cartesian vector form.

=> F = 5 * (2 i + 6 j - 3 k) kN

b) the magnitude of the component of F that runs parallel   = 5 /sqrt(49) = 0.714 kN

c)   the parallel component = 5 * (2 i + 6 j - 3 k)/7   kN

d) the component of F that runs perpendicular = 5 * (2 i + 6 j - 3 k)/6 kN

e)   the magnitude of the perpendicular component = 0.8 kN

f)   the angle between F and the inclined pole BA = tan-1(0.8/0.714)

                                                                            = 48.25 degree

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