# Homework Help: Weird problem

1. Dec 17, 2003

### formulajoe

what minimum horizontal force is needed to pull a wheel of radius R and mass M over a step of height H. force is supplied at center of wheel.
???

2. Dec 17, 2003

### Njorl

Draw yourself a force diagram. Just to get you started right, make the step height smaller than R, otherwise no force will be sufficient. The verticle component of the force exerted by the step upon the wheel must counteract gravity, so that it lifts off of the ground.

You might also want to show that the process "runs away". As the wheel climbs over the step, less force is needed. This ensures that you don't have to worry about climbing partly up the step and rolling back down.

Njorl

3. Dec 17, 2003

### himanshu121

Given force is horizontal

4. Dec 17, 2003

### himanshu121

We should look for conservation of Energy

5. Dec 17, 2003

### formulajoe

torque = F * R, but where do i factor in the H?

6. Dec 17, 2003

### himanshu121

nope torque is not FR here

7. Dec 17, 2003

### formulajoe

is torque even used in this problem?

8. Dec 17, 2003

### himanshu121

$$Fsin \theta R=I \alpha$$

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9. Dec 17, 2003

### Njorl

Yes, but reactive forces are not.

The applied forces are a horizontal force upon the axis, and gravity. The reactive forces are from the ground, and the step. The wheel will roll over the step when the reactive force from the step has a verticle component equal and opposite to the gravitaional effect upon the wheel. This will mean the ground supplies no force, because the wheel will stop touching the ground.

Njorl

10. Dec 17, 2003

### himanshu121

How that is going to help to solve the prob,

11. Dec 17, 2003

### formulajoe

im stuck at F*sintheta = m * a now.
how do i factor in the H?

12. Dec 17, 2003

### himanshu121

$$Fsin\theta r = i \frac{d\omega}{dt}$$
$$FRsin\theta d\theta=\omega I d\omega$$

solving this from fig uget

$$\frac{I\omega^2}{2}=\frac{FRH}{\sqrt{R^2+(R-H)^2}}$$

13. Dec 17, 2003

### himanshu121

Apply Conservation of Energy

$$F=\frac{mg \sqrt{R^2+(R-H)^2}}{R}$$

14. Dec 17, 2003

### NateTG

Do the torque calculations about the edge of the step. Torque due to the normal force is zero. That leaves torque due to gravity, and torque due to the applied force which must be larger.
I get
$$\tau_F=-(r-h)F$$
(negative because it is clockwise)
$$\tau_G=mg\sqrt{R^2-(R-H)^2}$$
and
$$\tau_F>-\tau_G$$
so
$$F > \frac{mg \sqrt{2RH-H^2}}{R-H}}$$

15. Dec 17, 2003

### himanshu121

Sorry i lost NateTG is correct