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| 183_notes:examples:statics [2016/03/21 17:18] – [Example: Statics] klinkos1 | 183_notes:examples:statics [2016/03/25 15:58] (current) – klinkos1 | ||
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| ====== Example: Statics====== | ====== Example: Statics====== | ||
| - | {{ 183_notes: | + | {{183_notes: |
| - | + | If a sign were hung like the one above, what would be the tension forces acting on both of the ropes? | |
| - | If a sign were hung like the one above, what would the tension forces acting on both of the ropes? | + | |
| ==== Setup ==== | ==== Setup ==== | ||
| - | To solve for the force of tension in both rope 1 and 2, both forces have to broken down into their x and y components and then solve the resulting system of equations. | + | To solve for the force of tension in both rope 1 and 2, both forces have to broken down into their x and y components, and then solve the resulting system of equations. |
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| === Lacking === | === Lacking === | ||
| - | *Either | + | *The force of tension |
| *Mass of the object | *Mass of the object | ||
| *Angles of the ropes | *Angles of the ropes | ||
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| *The lengths of the two ropes is irrelevant, only the angle matters to solve for the two forces | *The lengths of the two ropes is irrelevant, only the angle matters to solve for the two forces | ||
| - | *The net force is zero since the system is stationary | + | *The net force is zero, since the system is stationary |
| === Representations === | === Representations === | ||
| - | *I have a drawing but I'm having a hard time adding pictures | + | {{183_notes: |
| ==== Solution ==== | ==== Solution ==== | ||
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| For rope one, T1,y=T1cosα | For rope one, T1,y=T1cosα | ||
| For rope 2, T2,y=T2cosβ. | For rope 2, T2,y=T2cosβ. | ||
| - | When finding the net force in the y direction, we cannot forget our assumption that gravity also works in the dimension | + | When finding the net force in the y direction, we cannot forget our assumption that gravity also works in the y direction but in the opposite direction as our tension forces. The net force in the y direction is, |
| ∑Fy=T1cosα+T2cosβ−Mg=0. | ∑Fy=T1cosα+T2cosβ−Mg=0. | ||
| Now we have two unknowns (the tension of the two ropes) and two equations, so we can solve this as a system of equations. | Now we have two unknowns (the tension of the two ropes) and two equations, so we can solve this as a system of equations. | ||