You have your bicycle upside down for repairs. The front wheelis free to rotate

Question

You have your bicycle upside down for repairs. The front wheelis free to rotate and is perfectly balanced except for the 25gvalve stem. If the valve stem is 32cm from the rotation axis and islocated 24o below the horizontal, what is the resultingtorque about the wheel's axis?     You have your bicycle upside down for repairs. The front wheelis free to rotate and is perfectly balanced except for the 25gvalve stem. If the valve stem is 32cm from the rotation axis and islocated 24o below the horizontal, what is the resultingtorque about the wheel's axis?    

Explanation / Answer

The weight of the valve stem is (25g)*(9.8m/s2). The torque about a point is the perpendicular distance from thatpoint to the line of action of the force. The valve stem is24 degrees below the horizontal. To find the horizontaldistance to the force's line of action, we take the cosine: T = r x F = (32cm)*cos(24)*(25g)*(9.8m/s2) = (7162cm-g-m/s2)*(0.01m/cm)*(0.001kg/g) = 0.07162N-m

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