The three cables are used to support the 40-kg flowerpot. Determine the force developed in each cable for equilibrium.

The image of the 3 cables used to support the flower pot is not given so i have attached it.

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

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6