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

In summary, the conversation is about a problem involving a wheel with a radius of 6.00cm and a mass of 0.800 kg, trying to be raised over an obstacle of height 3.00cm. The question is, what magnitude of constant force applied horizontally at the axle is needed to raise the wheel? The equations used to solve this problem are the balance of forces and the balance of torques. The solution is found to be F=(mgr)/(r-h), which is equal to 15.7N. However, the answer in the text is 13.6N, suggesting a mistake in the calculation. Upon further discussion, it is realized that the moment arm of the gravitational force was mistakenly taken as
  • #1
Curieuse
51
1
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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  • #2
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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  • #3
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!
 

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

1. What is static equilibrium of a wheel?

Static equilibrium of a wheel refers to the state in which the wheel is stationary and not rotating. This means that the forces acting on the wheel are balanced, resulting in no net torque or rotational motion.

2. How is static equilibrium achieved in a wheel?

Static equilibrium in a wheel can be achieved by ensuring that the weight or mass of the wheel is evenly distributed around its axis of rotation. This ensures that the forces acting on the wheel are balanced and there is no net torque.

3. What are the factors that affect static equilibrium in a wheel?

The factors that affect static equilibrium in a wheel include the distribution of weight or mass, the shape and size of the wheel, and the surface on which the wheel is resting. Any imbalance in these factors can lead to a lack of static equilibrium.

4. Why is static equilibrium important in a wheel?

Static equilibrium is important in a wheel because it ensures that the wheel remains stationary and does not rotate unintentionally. This is crucial for the stability and safety of the wheel, especially in vehicles and machinery.

5. How can static equilibrium in a wheel be tested?

Static equilibrium in a wheel can be tested by using a variety of methods, such as measuring the weight distribution, conducting a spin test, or using a force plate to determine the balance of forces acting on the wheel. These tests can help identify any imbalances and allow for adjustments to be made to achieve static equilibrium.

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