183_notes:examples:statictorque

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183_notes:examples:statictorque [2016/03/24 22:26] – [Setup] klinkos1183_notes:examples:statictorque [2016/03/25 16:03] klinkos1
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 All of these forces apply a torque on the person performing a keg stand. The torque equation is All of these forces apply a torque on the person performing a keg stand. The torque equation is
 $$\tau = rF\sin{\theta}.$$ $$\tau = rF\sin{\theta}.$$
-Where r is the distance between where the force is applied and the pivot of the system (the point it is rotating around), F is the amount of force, and $\theta$ is the angle at which this force is applied. (Figure here??) +Where r (show by h in this problem) is the distance between where the force is applied and the pivot of the system (the point it is rotating around), F is the amount of force, and $\theta$ is the angle at which this force is applied. 
 Now we can find the net torque of the system. Now we can find the net torque of the system.
 Based on the diagram, we can determine that the pivot point of the system is at the keg, and the person turns around this. We can use this information to determine the radiuses of the three torques in our system. Torque due to the keg ($\tau_{k}$), torque due to the support of the person ($\tau_{p}$), and torque due to gravity ($\tau_{g}$). The torque due to gravity will be negative since it acts in the opposite direction of the other two forces.  Based on the diagram, we can determine that the pivot point of the system is at the keg, and the person turns around this. We can use this information to determine the radiuses of the three torques in our system. Torque due to the keg ($\tau_{k}$), torque due to the support of the person ($\tau_{p}$), and torque due to gravity ($\tau_{g}$). The torque due to gravity will be negative since it acts in the opposite direction of the other two forces. 
  • 183_notes/examples/statictorque.txt
  • Last modified: 2016/03/25 16:04
  • by klinkos1