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| 183_notes:examples:statics [2016/03/18 16:13] – klinkos1 | 183_notes:examples:statics [2016/03/18 16:25] – klinkos1 | ||
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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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