You are trying to raise a bicycle wheel of mass m and radius R up over a curb of

Question

You are trying to raise a bicycle wheel of mass m and radius R up over a curb of height h. To do this, you apply a horizontal force F.

Part A

What is the smallest magnitude of the force F that will succeed in raising the wheel onto the curb when the force is applied at the center of the wheel?

Express your answer in terms of m , R , h , and g .

Part B

What is the smallest magnitude of the force F that will succeed in raising the wheel onto the curb when the force is applied at the top of the wheel?

Express your answer in terms of m , R , h , and g .

Part C

In which case is less force required?

Explanation / Answer

a) Both applied force F and weight W exert opposing torques, both pivoting in the point where the wheel touches the edge of the curb. Briefly stated, force F will succeed to lift the wheel if its associated torque is greater than the torque due to weight. In mathematical terms,

L = r × F + r × W.

All terms in the equation above are vectors. L is the resulting torque; r × F is considered positive. F will lift the wheel whenever L > 0.

As much can be said without resorting to a figure. Please follow these directions:
1. Draw a circle.
2. Draw a cross-sectional view of the curb, on the right side of wheel. Curb height h should be smaller than wheel radius r.
3. Draw a line from the center of circle to the point where the circle touches curb edge. Label this line "r".
4. Both F and W are supposed to act on the center of the circle. You may tentalively draw both, F as a horizontal line, directed to the right; W pointing down from the center.

Since F and W are perpendicular, one horizontal and the other vertical, resolve r into its horizontal and vertical components. These will be the effective "lever-arm" for each of theseforces. The vertical component is readily found in terms of r and h: it is just (r - h). The horizontal component can be found by Pythagoras' Theorem. We have (in scalar format):

L = F(r - h) - W [r² - (r - h)²]

Minimum value for F can be found letting L = 0 and solving for F. The resulting algebraic expression is too cumbersome to be included here, and I think it is hardly necessary to do so. Besides, there are several algebraic manipulations possible. Likely, I would end with a different -albeit equivalent- expression.


b) When F is applied to the top of the wheel, the effective lever-arm for F increases to 2r - h. Refer to figure, or, rather, draw a new one. Join the uppermost point on the wheel to the contact point between wheel and curb edge. Now, persuade yourself that the vertical component of this line equals wheel diameter - curb height, or 2r - h.

Torque due to W remains as before. Consequently,

F(2r - h) = W [r² - (r - h)²].

Least F is found solving for F.


c) Clearly, with increased leverage, it would be easier to lift the wheel onto the curb in the second case. This is not apparent from the formulas, however.

For this purpose, the solution in trigonometric form offers much better insight. Here they are:

1st case: F = W / tan , where is the (clockwise) angle r makes with the horizontal.
2nd case: F = W cos /(1 + sin )

Note that for < 45°, F > W, in the first case. Also, F = W if = 45°, and F < W if > 45°. On the other hand, in the second case F < W is always met, since both sin and cos lie in the range 0 to 1. Thus, the numerator will be typically < 1, while the denominator is guaranteed to be > 1. The resulting fraction will in every case be less than unity.

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