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- - **statics class, ended up with 2 equations and 2 variables, rules to isolate variables**
(*http://mymathforum.com/algebra/46305-statics-class-ended-up-2-equations-2-variables-rules-isolate-variables.html*)

statics class, ended up with 2 equations and 2 variables, rules to isolate variablesHey so i have this excersise in my statics class where im going to solve tension of 2 ropes which are holding a load, in addition there is a force pulling the load in the X directions. When apllied law of statics that states if a load is not moving all forces needs to equal each other out so i ended up with these 2 expressions and cant seem to solve them. guess it is to long since math class so please fill me in with rules of isolating those 2 expressions. I have been trying to do the same thing on both sides but cant seem to get it right :/ 1: F=cos51,3*AB+cos26,6*AC | where F equals 420 2: G=sin51,3*AB+sin26,6*AC | where G equals 245 So I know i have to get an expression for either AB or AC and subsditute it into the other equation. As far as i remember I have to perform the same stebs on left hand as i did on right side so i tried dividing F with cos51,3*AB and dividing G with sin26,6 * AC. Then im left with expressions for AB and AC? it does not feel right so im here to ask. |

What you are doing seems to make a certain amount of sense, but it's hard to tell because you equations aren't clear. Can you use brackets to make it clear what the arguments of the trigonometric functions are and what they are multiplied by? |

Well there are 2 wires which are making tansions due to whe load hangin verticly and the force F working horizontal, the opposit way from the cables. The 2 cables hae a D=2m from the load on the X axsis and one of the wires have H=1m and the second H=2.5m in Y directions. So i used tan inverse O/A whitch gives men Tan inverse of 2,5/2 and Tan inverse of 1/2 So i found that the wire T1 are making 51,3 degrees from the X axsis and the T2 giving me 26.6 deg from the x axsis Im saying that the sum of the forces has to equal 0 which gives me for forces in X direction: F workin completely horizontal has to equal cos(angle1)*T1 + cos(angle2)*T2 420=(cos51,3*T1)+(cos26,6*T2) and i did the same for the Y directions, the weith are making a force of 245N down on Y axsis so the forces up the Y axsis has to equal that force giving me: 245=(sin51,3*T1)+(sin26,6*T2) When im trying to solve this im always ending up with this 2 expression which seems way off: Im using cosine and sinus to know how much force going in each directions $\displaystyle F=T1x+T2x|||| G=T1y+T2y|||| F=cos51,3*T1+cos26,6*T2|||| G=sin51,3*T1+sin26,6*T2|||| T1=(420-cos26,6*T2)/cos51,3||| T2=(245,3-sin〖51,3*T1〗)/sin26,6 $ |

1 Attachment(s) Do you mean something like this? |

Excactly, F and G working from the green point |

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