Solving a Wheel on Step Question - Torque Calculation and Solution Attempt

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SUMMARY

The discussion focuses on calculating torque in a physics problem involving a wheel on a step. The key equation used is torque (\tau = \overline{r} \times \overline{F}), emphasizing the importance of the moment arm, which is defined as the radius (R) minus the height (h) of the step. The solution approach involves determining the force (F_{min}) acting perpendicularly to the lever arm to counteract the forces on the wheel's axle. The discussion also suggests applying the same torque calculation strategy to subsequent parts of the problem.

PREREQUISITES
  • Understanding of torque and its calculation using the equation \tau = \overline{r} \times \overline{F}
  • Familiarity with the concepts of moment arm and force vectors
  • Knowledge of basic physics principles related to rotational motion
  • Ability to interpret and analyze physics problems involving forces and torques
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  • Study the application of torque in rotational dynamics using examples
  • Learn about the relationship between torque, angular acceleration, and moment of inertia (Mgr = I * alpha)
  • Explore advanced torque problems involving multiple forces and moments
  • Investigate the effects of varying the radius and height on torque calculations
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Students studying physics, particularly those focusing on mechanics, as well as educators and tutors looking for effective strategies to teach torque and rotational dynamics.

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Homework Statement



Question is here. http://i.imgur.com/WOy0w.png

WOy0w.png


Homework Equations


Torque=FR


The Attempt at a Solution



1)
1)I thought it is radius. But radius is not one of the option
2)I thought this is radius too.
3)Not sure about this question,maybe get torque get acceleration and than use Mgr=I*alpha
Thanks!
 

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Greetings! Remember that torque is [tex]\tau = \overline{r} \times \overline{F}[/tex], which is different than [tex]\tau = rF[/tex]. In other words, torque is the radius at which a force acts perpendicularly to the lever arm, times the magnitude of the force. Thus, for part A, note that the corner of the step will be exerting a force [tex]F_{min}[/tex] directly to the left in order to counteract the rightward force on the wheel's axle. If we extend this leftward [tex]F_{min}[/tex], we see that it acts perpendicularly to the wheel at a radius of [tex]R - h[/tex] (the moment arm is conveniently indicated by the vertical dotted line).

Therefore, for part A, [tex]\tau = \overline{r} \times \overline{F} = (R-h)F_{min}[/tex]. Try applying this same strategy to part B.
 

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