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| 183_notes:examples:statics [2016/03/18 16:13] – klinkos1 | 183_notes:examples:statics [2016/03/24 22:29] – klinkos1 | ||
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| - | {{ 183_projects:screen_shot_2016-03-18_at_11.15.44_am.png | + | ====== Example: Statics====== |
| + | {{183_notes:examples: | ||
| If a sign were hung like the one above, what would 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 ==== | ||
| Line 10: | Line 10: | ||
| *Gravity works in the negative y direction, with $g=9.81m/ | *Gravity works in the negative y direction, with $g=9.81m/ | ||
| *There are two forces of tension, on for rope 1, and one for rope 2 | *There are two forces of tension, on for rope 1, and one for rope 2 | ||
| - | *Mass of the object? | + | |
| - | *Angles of the ropes? | + | |
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| === Lacking === | === Lacking === | ||
| *Either force of tension of the ropes | *Either force of tension of the ropes | ||
| + | *Mass of the object | ||
| + | *Angles of the ropes | ||
| === Approximations & Assumptions === | === Approximations & Assumptions === | ||
| Line 24: | Line 24: | ||
| === Representations === | === Representations === | ||
| - | *I have a drawing but I'm having a hard time adding pictures | + | {{183_notes: |
| ==== Solution ==== | ==== Solution ==== | ||
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| When finding the net force in the y direction, we cannot forget our assumption that gravity also works in the dimension in the opposite direction as our tension forces. The net force in the y direction is, | When finding the net force in the y direction, we cannot forget our assumption that gravity also works in the dimension in the opposite direction as our tension forces. The net force in the y direction is, | ||
| $$\sum F_{y} = T_{1}\cos{\alpha}+T_{2}\cos{\beta}-Mg = 0.$$ | $$\sum F_{y} = T_{1}\cos{\alpha}+T_{2}\cos{\beta}-Mg = 0.$$ | ||
| + | Now we have two unknowns (the tension of the two ropes) and two equations, so we can solve this as a system of equations. | ||
| + | The force of tension in the x direction can be rearranged to solve for one of the unknowns, in this case $T_{1}$ | ||
| + | $$T_{2}\sin{\beta} - T_{1}\sin{\alpha} = 0$$ | ||
| + | $$T_{1} = T_{2} \dfrac{\sin{\beta}}{\sin{\alpha}}$$ | ||
| + | Now we can plug this solution for $T_{1}$ into the equation we have for the tension force in the y direction | ||
| + | $$T_{1}\cos{\alpha}+T_{2}\cos{\beta}-Mg = 0$$ | ||
| + | $$T_{2} \dfrac{\sin{\beta}}{\sin{\alpha}}cos{\alpha}+T_{2}\cos{\beta}-Mg = 0$$ | ||
| + | And solve for $T_{2}$ (should I add more steps?) | ||
| + | $$T_{2} = \dfrac{Mg}{(\dfrac{\sin{\beta}}{\tan{\alpha}}+\cos{\beta})}$$ | ||
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| + | Now we have solutions for both tension forces, $T_{1}$ and $T_{2}$ | ||
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