Someone help me to do this?
1 answer:
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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
