183_notes:examples:statics

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183_notes:examples:statics [2016/03/18 16:13] klinkos1183_notes:examples:statics [2016/03/25 15:58] (current) klinkos1
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-{{ 183_projects:screen_shot_2016-03-18_at_11.15.44_am.png }} (This picture clearly hates me) +====== Example: Statics====== 
- +{{183_notes:examples:imag0334.jpg}} 
-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 be 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 componentsand then solve the resulting system of equations. 
  
  
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   *Gravity works in the negative y direction, with $g=9.81m/s^{2}$   *Gravity works in the negative y direction, with $g=9.81m/s^{2}$
   *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?+
      
 === Lacking === === Lacking ===
  
-  *Either force of tension of the ropes  +  *The force of tension of both of the ropes  
 +  *Mass of the object 
 +  *Angles of the ropes
 === Approximations & Assumptions === === Approximations & Assumptions ===
  
    *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 zerosince the system is stationary
  
 === Representations === === Representations ===
  
-  *I have a drawing but I'm having a hard time adding pictures+{{183_notes:examples:screen_shot_2016-03-18_at_11.15.44_am.png}}
  
 ==== Solution ==== ==== Solution ====
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 For rope one, $$T_{1,y} = T_{1}\cos{\alpha}$$ For rope one, $$T_{1,y} = T_{1}\cos{\alpha}$$
 For rope 2, $$T_{2,y} = T_{2}\cos{\beta}.$$ For rope 2, $$T_{2,y} = T_{2}\cos{\beta}.$$
-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 y direction but 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?)
 +$$T_{2} = \dfrac{Mg}{(\dfrac{\sin{\beta}}{\tan{\alpha}}+\cos{\beta})}$$
 +
 +Now we have solutions for both tension forces, $T_{1}$ and $T_{2}$
 +
 +
 +
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