Solving Rotational Dynamics Problem: Finding Magnitude of Force

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To solve the rotational dynamics problem of a bicycle wheel against a step, focus on the torques about the point of contact. The applied horizontal force F creates a torque that can be expressed as F(r-h), while the gravitational force's torque depends on the geometry of the wheel and the step height. When the torque from the applied force exceeds the torque from gravity, the wheel will begin to rise. It's essential to relate the distances and angles involved to find the correct expressions for these torques. Understanding these relationships will help determine the magnitude of the force F when the wheel loses contact with the ground.
zachattackback
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I have a quiz tommorrow and i am reviewing the harder problems and i can't get started on this one

A bicycle Wheel is resting against a small step whose height is h=0.120 m. The weight and radius of the whell are W=25.0N and r=0.340 m. A horizontal force F is applied to the axle of the whell. As the magnitude of F increases, there comes a time when the wheel just begins to rise up and loses contact with the ground. What is the magnitude of the force when this happenes?

I don't kinwo where to start i knwo the formulas
t=FL
and that the sum of the forces =0 but i don't know where to start

Please help me get started

thanks
 
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When the wheel starts to move, it will be rotating about the point of contact with the step. You will want to investigate torques and moment of inertia about that point.
 
do u ahve to make the force perpendicular to the point f contact so one equation would be Frcos45=t

then i don't knwo where to go i am trying to make stuff up to trigger my mind
 
zachattackback said:
do u ahve to make the force perpendicular to the point f contact so one equation would be Frcos45=t

then i don't knwo where to go i am trying to make stuff up to trigger my mind
What is the torque about the point of contact from the force acting at the point of contact? The only other forces are the applied horizontal force on the axle and gravity. The torques related to those forces will depend on the height of the step and the radius of the wheel. The torque for the horizontal force is easy to find in terms of F. The one for gravity is a bit tougher, but you can get it from the geometry of the problem.
 
can u maybe help me with an equation i am confused on what u are saying
 
zachattackback said:
can u maybe help me with an equation i am confused on what u are saying

Draw a circle to represent the wheel and a horizontal tangent below the wheel to represent the ground. Draw the step of height h with the corner touching the wheel. Mg acts downward at the center of the wheel. F acts horizontally toward the step. The distance between the line of F and the contact point is r-h. The torque is F(r-h). Draw a radius from the center of the wheel straight down and another from the center to the point of contact. Call the angle between these radii θ. r-h can be written in terms of r and θ, so θ can be determined from r and h. The distance between the line of mg and the contact point can be written in terms of r and θ, and the torque from gravity is this distanve times mg. When F is big enough to make the F torque greater than the mg torque, the wheel will leave the ground.
 
Say I am standing on the ground. As we know, gravity is acting on me. But why am I not sinking into the ground? Because there is an opposite force acting on me, this force is called the Normal force. If I want to lift off the ground, what do I have to do?
 

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