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kozerog [31]
3 years ago
7

The resultant force is directed along the positive x axis and has a magnitude of 1330 N.

Engineering
1 answer:
Andrew [12]3 years ago
3 0

Answer:

the magnitude of F_A is 752 N

the direction theta of F_A is 57.9°

Explanations:

Given that,

Resultant force = 1330 N in x direction

∑Fx = R

from the diagram of the question which i uploaded along with this answer

FB = 800 N

FAsin∅ + FBcos30 = 1330 N

FAsin∅ = 1330 - (800 × cos30)

FA = 637.18 / sin∅

Now ∑Fx = 0

FAcos∅ - FBsin30 = 0

we substitute for FA

(637.18 / sin∅)cos∅ = 800 × sin30

637.18 / 800 × sin30 = sin∅/cos∅

and we know that { sin∅/cos∅ = tan∅)

so tan∅ = 1.59295

∅ = 57.88° ≈ 57.9°

THEREFORE FROM THE EQUATION

FA = 637.18 / sin∅

we substitute ∅

so FA = 637.18 / sin57.88

FA = 752 N

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