The three cables are used to support the 40-kg flowerpot. Determine the force developed in each cable for equilibrium.
1 answer:
The image of the 3 cables used to support the flower pot is not given so i have attached it.
Answer:
F_ab = 523.2 N
F_ac = 392.4 N
F_ad ≈ 762.7 N
Explanation:
First of all, let's Calculate the sum of forces around the z-axis.
But before then, we need to resolve the distances along the 3 axis to get;
1.5/√(1.5² + 2² + 1.5²) = 0.5145
Thus;
ΣF_z = 0;
(F_ad × 0.5145) - (40 × 9.81) = 0
Where;
(40 × 9.81) is the weight of the flower pot.
F_ad is the force in cable AD
Thus;
0.5145F_ad - 392.4 = 0
F_ad = 392.4/0.5145
F_ad ≈ 762.7 N
Now, let's do the same for the x-axis;
ΣF_z = 0;
Since x = 1.5, just like z was the same 1.5, we will adopt same resolved distance.
F_ac - F_ad(0.5145) = 0
Where F_ac is force in cable AC
F_ac = 762.7 × 0.5145
F_ac = 392.4 N
Lastly, for the y-axis;
ΣF_y = 0;
y is 2 and so, we need to recalculate the resolved distance.
F_ab - F_ad(2/√(1.5² + 2² + 1.5²)) = 0
Where F_ab is the force in cable AB
F_ab = 762.7 × 0.686
F_ab = 523.2 N
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