How Much Force is Needed to Lift a Wheel Over an Obstacle?

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SUMMARY

The discussion centers on calculating the force required to lift a wheel over an obstacle of height 3.00 cm, with a radius of 6.00 cm and a mass of 0.800 kg. The participant initially calculated the force using the torque balance equation, resulting in 15.7 N, while the correct answer is 13.6 N as per the textbook. Key errors included miscalculating the moment arm of the gravitational force and misunderstanding the role of normal forces at the contact point with the ground.

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  • Understanding of torque and force balance in physics.
  • Familiarity with moment arms and their calculation.
  • Knowledge of gravitational force and its effects on objects in motion.
  • Basic principles of rotational dynamics.
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  • Study the calculation of moment arms in various physical scenarios.
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  • Practice similar problems involving force calculations for lifting objects over obstacles.
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Curieuse
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1. Problem statement
In figure, what magnitude of (constant) force F applied horizontally at the axle of the wheel is necessary to raise the wheel over an obstacle of height h=3.00cm ? The wheel's radius is r=6.00cm , and it's mass is m=0.800 kg.


Homework Equations


Balance of forces: Fnet=0 (xy plane)
Balance of torques: τnet=0 (z axis)

3. The Attempt at a Solution
By balance of torques about point of contact with obstacle,
F(r-h)=mgr
I chose that point so as to not account for forces of contact there!
The answer i got
F=(mgr)/(r-h)
F= 15.7 N
The answer at the back of the text is 13.6N. :cry:
What did i do wrong?

Also i have this lingering doubt, will the normal forces at the point of contact with the horizontal floor , be included?:confused:
I worked another problem where the solution i found ignored it and when i tried putting it in, it simply nullifies the weight , which lead to unreasonable answers... But still why is it not to be included? I mean it's also a force no.. Of course it disappears as soon as the wheel lifts off due to that horizontal F .. Thanks in advance..:smile:
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Curieuse said:
What did i do wrong?
What is the moment arm of the gravitational force wrt the point of contact?
 
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The moment arm was (r^2 - (r-h)^2)^(1/2)
I mistook it to be r as i was doing the eye approximations :p
Thanks a lot @Orodruin
I'll be more careful!
 

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