You are attempting to raise a wheel of radius R and mass m over a step of height

Question

You are attempting to raise a wheel of radius R and mass m over a step of height h by applying a horizontal force F. What is the minimum magnitude force F that will accomplish this task if the force is applied at the center of the wheel? At the top of the wheel?

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. A solution can be found either following a geometrical approach (in terms of r and h), or else by trigonometrical means (in terms of the angle between r and the horizontal -or vertical. Much more often than not, those who post Physics problems in Y!A know the answer beforehand. I have no way to know if the answer you got -I'm sure you do- is in a trigonometrical, rather than geometrical, format. However, since h is mentioned in the problem statement, the latter alternative seems to me more likely. 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 v[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 v[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 45°. On the other hand, in the second case F 1. The resulting fraction will in every case be less than unity.

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