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Chapter 10: Problem 24

You have your bicycle upside-down for repairs. The front wheel is free to rotate and is perfectly balanced except for the \(25-g\) valve stem. If the valve stem is \(32 \mathrm{cm}\) from the rotation axis and at \(24^{\circ}\) below the horizontal, what's the resulting torque about the wheel's axis?

Short Answer

Expert verified
The resulting torque about the wheel's axis due to the valve stem is \(0.04 Nm\).

Step by step solution

01

Identify and Express Given Values

The mass of the valve stem is \(25 g\) or \(0.025 kg\). The distance from the valve stem to the axis of rotation is \(32 cm\) or \(0.32 m\). The standard acceleration due to gravity is \(g = 9.8 m/s^2\). We have also been given that the valve stem is below the horizontal by \(24^{\circ}\).
02

Calculate Force Due to Gravity

The force due to gravity (weight) on the valve stem can be calculated using the formula \(F = m \cdot g\), where \(m\) is the mass and \(g\) is the acceleration due to gravity. Substituting the given values, we get \(F = 0.025 kg \cdot 9.8 m/s^2 = 0.245 N\).
03

Calculate Torque

The torque (\(\tau\)) due to the gravitational force on the valve stem with respect to the wheel's axis can be calculated as \(\tau = F \cdot r \cdot \sin(\theta)\) where \(F\) is the force, \(r\) is the distance from the axis of rotation and \(\theta\) is the angle below the horizontal. Substituting the given values, we get \(\tau = 0.245 N \cdot 0.32 m \cdot \sin(24^{\circ}) = 0.03979 Nm = 0.04 Nm\), after rounding to two decimal places.

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