# Forces Equillibrium Problem

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• March 3rd, 2008, 10:16 AM
myoplex11
Forces Equillibrium Problem
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. Note that the 2 cables make an angle 60 and 53.1 respectively with the horizontal

Fxnet=TABcos60-TACcos53.1 =0
Fynet-TABsin60+TACsin53.1-W=0
I dont know how to figure out if 50lb is the tension in cable AB or in AC
does the more vertical cable have the higher tension why?
• March 3rd, 2008, 12:17 PM
organic god
can we have a diagram please
• March 3rd, 2008, 01:24 PM
bit4bit
If the forces are in equillibrium, then it looks to be a setup where A is the position of the plantpot, which is supended by two ropes to points B and C.

Labelling AB as T<sub>1</sub>, and AC as T<sub>2</sub>:

x: 0=T<sub>1</sub>cos120 + T<sub>2</sub>cos53.1
y: W=T<sub>1</sub>sin120 + T<sub>2</sub>sin53.1

Calculate the angles, and solve simultaneously for (say) T<sub>1</sub>. Then plug in 50lb for T<sub>1</sub>, and calculate W. Do the same for T<sub>2</sub>, and calculate W again. The lowest W value should be the maximum weight of the plantpot.
• March 4th, 2008, 07:02 AM
Harold14370
Re: Forces Equillibrium Problem
Quote:

Originally Posted by myoplex11
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. Note that the 2 cables make an angle 60 and 53.1 respectively with the horizontal

Fxnet=TABcos60-TACcos53.1 =0
Fynet-TABsin60+TACsin53.1-W=0
I dont know how to figure out if 50lb is the tension in cable AB or in AC
does the more vertical cable have the higher tension why?

If you rearrange your first equation you find that the ratio of TAB to TAC is cos53.1/cos60, which means that TAB is 1.2 times TAC. Therefore AB has to be the one with 50 lb tension, and this is the more vertical cable.