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| Both sides previous revision Previous revision Next revision | Previous revision | ||
| 183_notes:examples:statictorque [2016/03/21 15:06] – klinkos1 | 183_notes:examples:statictorque [2016/03/25 16:04] (current) – klinkos1 | ||
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| ====== Example: Statics with Torque ====== | ====== Example: Statics with Torque ====== | ||
| + | {{183_notes: | ||
| Find the force a person would have to apply support their friend doing a keg stand | Find the force a person would have to apply support their friend doing a keg stand | ||
| ==== Setup ==== | ==== Setup ==== | ||
| Line 7: | Line 8: | ||
| === Facts ==== | === Facts ==== | ||
| - | * Gravity will apply a negative acceleration of $9.82 m/s^2$ | + | * Gravity will apply a negative acceleration of $9.81 m/s^2$ |
| === Lacking === | === Lacking === | ||
| Line 18: | Line 19: | ||
| * The pivot will be at the keg | * The pivot will be at the keg | ||
| === Representations === | === Representations === | ||
| - | + | {{183_notes: | |
| - | * | + | |
| ==== Solution ==== | ==== Solution ==== | ||
| Since the system is stationary, we know that both the net force and the net torque are equal to zero | Since the system is stationary, we know that both the net force and the net torque are equal to zero | ||
| Line 30: | Line 29: | ||
| 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. | + | Where r (show by h in this problem) |
| 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 | + | 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 radii of the three torques in our system. Torque due to the keg ($\tau_{k}$), |
| $$\tau_{net} = \tau_{k}+\tau_{p}-\tau_{g} = 0$$ | $$\tau_{net} = \tau_{k}+\tau_{p}-\tau_{g} = 0$$ | ||
| The radius of the torque due to the keg is zero, since we established that the keg is at the pivot point. The radius of the torque due to the support of the person is the full length of the person being supported, since they are supported at their feet. The radius of the torque due to gravity is at half of the length of the supported person, since this is approximately their center of mass. | The radius of the torque due to the keg is zero, since we established that the keg is at the pivot point. The radius of the torque due to the support of the person is the full length of the person being supported, since they are supported at their feet. The radius of the torque due to gravity is at half of the length of the supported person, since this is approximately their center of mass. | ||