Determine the maximum weight of the flowerpot that can be supported without exceeding a cable tension of 50 lb in either cable AB or AC.

#### Solution:

Show me the final answerโ

We will first draw a free body diagram, depicting the forces around ring A.

We can now write equations of equilibrium for the x and y-axis forces.

\rightarrow ^+\sum \text{F}_\text{x}\,=\,0
F_{AC}\sin30^0-F_{AB}\dfrac{3}{5}=0

(Isolate for F_{AC})

F_{AC}=1.2F_{AB} (eq.1)

**Notice how the force of F_{AC} would be 1.2 times the force of F_{AB}, which means F_{AC} will fail first.**

+\uparrow \sum \text{F}_\text{y}\,=\,0

F_{AB}\dfrac{4}{5}+F_{AC}\cos30^0-W=0 (eq.2)

We will now equate F_{AC}=50 lb into eq.1.

F_{AC}=1.2F_{AB}
50=1.2F_{AB}

F_{AB}=41.67 lb

Substitute this value into eq.2.

(41.67)\dfrac{4}{5}+50\cos30^0-W=0

W=76.64 lb

#### Final Answer:

Maximum weight = 76.64 lb

Why is FAB being subtracted in the first step?

Hi Gabriel. We are only subtracting the x-component of FAB because it’s pointing to the left. You will see that we assumed forces to the right, and forces going up to be positive. So for the x-axis, the x-component of FAB is going left, so it’s negative. Hope that helps ๐

What does it mean by Fac will fail first

Cable Fac will break first, like it will snap/tear.

Last one, how can we know which cable will fail?

Fac would be 1.2 times the force of Fab. In other words, Fac carries a much larger load, so its going to break first.

Oh… I get it…. Thank you very much for the clarification and help

Excellent! Glad to have helped. Best of luck with your studies ๐

I think when the summation of x its cos(30) and y is sin(30)

No, for F_AC, the x component is from sin and y is from cosine. Don’t always assume that x axis is cosine and y axis is sine.

Confused

How do you get FAC =50lb

It’s given in the question. We just plugging in 50 since that’s the max that can be supported.

Why didn’t we put Fab=50lb

Because FAC will fail first, so we test that first.