Static Equilibrium and a wheel

In summary, to wheel over an obstacle of height h, the magnitude of force applied horizontally at the axle of the wheel is necessary to be equal to or greater than the torque due to the weight of the object.
  • #1
Litcyb
36
0

Homework Statement


what magnitude of force applied horizontally at the axle of the wheel is necessary to wheel over an obstacle of height h = 0.108 m? The wheel's radius is r = 0.698 m and its mass is m = 1.71 kg.

Homework Equations



I know that in order to have a static equilibrium we must have the total amount of forces and the total amount of torques(at any point) equal zero.
ƩF=0
Ʃτ=0

The Attempt at a Solution


I have no idea how to solve this problem since we have a circular object. do we apply angular momentum? I am really confused on how to solve this problem! :( please help.
 
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  • #2
The torque exerted by the applied force must overcome the torque due to the weight of the object. Hint: Take the top of the [STRIKE]object[/STRIKE] obstacle as your pivot point.

(Edit: Corrected typo. I meant obstacle, but wrote object!)
 
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  • #3
Why are we making the pivot point at the top of the circle, wouldn't it make more sense to put it at the corner since we don't know the friction giving by the corner? because if we put the pivot point at the corner we would just need to calculate the torques with respect to the corner.

We have two forces acting at the center of mass ; gravity( Mg) and F(the force being exerted)
now, we would just calculate the lever arm for each force acting upon the object in respect to the pivot point(corner).

Then we calculate the lever arm of Mg and F, in which they have to be perpendicular to the axis of the pivot point.

Am I doing the right steps here?
 
  • #4
Litcyb said:
Why are we making the pivot point at the top of the circle,
Who said anything about putting the pivot at the top of the circle? (Ah... when I said top of the "object" I meant to say top of the "obstacle". My bad! :redface:)
wouldn't it make more sense to put it at the corner since we don't know the friction giving by the corner?
Imagining the obstacle as a step with a corner (of the given height), then that is exactly where the pivot should be.

We have two forces acting at the center of mass ; gravity( Mg) and F(the force being exerted)
now, we would just calculate the lever arm for each force acting upon the object in respect to the pivot point(corner).
Sure.

Then we calculate the lever arm of Mg and F, in which they have to be perpendicular to the axis of the pivot point.
The lever arm for a particular force is the perpendicular distance between the line of the force and the pivot.

Am I doing the right steps here?
I think so.
 
  • #5
Thank you so much for reassuring me. I got the correct answer. the lever arm for force mg is (2rh-h^2))^1/2 and the lever arm for force F is (r-h)

Thus, the sum of all torques Ʃτ= τF-τmg= 0 => τF=τmg
= F(h-r)=mg(2hr-h^2)^1/2
F= [mg(2hr-h^2)^1/2]/(r-h)

Thanks!
 
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1. What is static equilibrium?

Static equilibrium is a state in which an object is at rest and all forces acting on it are balanced, resulting in a net force of zero.

2. How does a wheel achieve static equilibrium?

A wheel achieves static equilibrium when the forces acting on it are balanced around its center of rotation. This means that the forces pushing the wheel in one direction are equal in magnitude and opposite in direction to the forces pushing it in the other direction, resulting in a net force of zero.

3. What is the role of friction in static equilibrium?

Friction plays a crucial role in achieving and maintaining static equilibrium in a wheel. It helps to prevent the wheel from sliding or slipping out of place, ensuring that the forces acting on it remain balanced.

4. Can a wheel be in static equilibrium while in motion?

No, a wheel cannot be in static equilibrium while in motion. Static equilibrium refers to a state of rest, so if a wheel is in motion, it is not in static equilibrium. However, there are other types of equilibrium that apply to objects in motion, such as dynamic equilibrium.

5. How does the distribution of weight affect static equilibrium in a wheel?

The distribution of weight in a wheel can affect its static equilibrium. If the weight is evenly distributed, it will help to maintain balance. However, if the weight is unevenly distributed, it can cause the wheel to become unbalanced and result in a net force acting on it, causing it to move.

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