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183_notes:examples:statics [2016/03/18 16:13] – klinkos1 | 183_notes:examples:statics [2016/03/21 17:11] – klinkos1 | ||
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+ | ====== Example: Statics====== | ||
{{ 183_projects: | {{ 183_projects: | ||
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*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 === | ||
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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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