My attempt: At first I thought it was 0N but that was wrong. Then I thought it w

Question

My attempt: At first I thought it was 0N but that was wrong. Then I thought it was essentially 2 * .143kg * 9.8....but that's wrong apparantly?

A tire balance weight is a piece of lead with a mass of 143 grams. The total diameter of the wheel, including the tire, is 29.8 inches. The weight is mounted on the rim, and the tire wall is 3.5 inches tall. After the weight is installed, the owner drives the car on the highway at 65 miles per hour. What is the magnitude of the difference between the two net forces experienced by the weight, | F1 - F2| ? Force F1 is the net force on the weight when it is passing through its lowest position on the wheel, while force F2 is the net force on the weight when it is passing through its highest position on the wheel rim. In your calculation, use 1 inch = 2.54 cm. The magnitude of the difference | F1 - F2 | is N. To get a feeling for the magnitude of the force imbalance caused by (or balanced by) a few-oz piece of lead, you can compare this force differential to the weight of a typical wheel [rim + tire], which is of the order of 50 lbs.

Explanation / Answer

Radial position of weight = 29.8-3.5 = 26.3 in = 0.668 m

Vel of car = 65 mph = 28.89 m/s

assuming upward direction as positive

hence, F1 = - m(g + v^2/r )

F2 = m( -g + v^2/r )

|F1-F2| = 2m v^2/r = 2*0.143 * 28.89^2/0.668 = 357.34 N

|F1-F2|/weight of tyre = |F1-F2|/(50*0.453*9.81) = 1.6 times

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