Question

part 9 - mechanical comprehension question 2 of 15 the mechanical advantage of the wheel - and - belt assembly shown in the illustration is $\frac{1}{3}$. what is the diameter of the driver wheel? a 2 inches b 3 inches c 4 inches d 6 inches click the button or type the letter next to your answer.

Explanation:

Step1: Recall Mechanical Advantage Formula

For wheel - and - belt (or pulley - belt) systems, the mechanical advantage (MA) is given by the ratio of the diameter of the driven wheel (\(d_{driven}\)) to the diameter of the driver wheel (\(d_{driver}\)). So, \(MA=\frac{d_{driven}}{d_{driver}}\)

Step2: Substitute Known Values

We know that \(MA = \frac{1}{3}\) and \(d_{driven}=12\) inches. Substituting these values into the formula \(\frac{1}{3}=\frac{12}{d_{driver}}\)

Step3: Solve for \(d_{driver}\)

Cross - multiply to get \(d_{driver}\times1 = 12\times3\), so \(d_{driver}=36\)? Wait, no, wait. Wait, actually, the mechanical advantage for a wheel - belt system (where the belt connects two wheels) can also be thought of as the ratio of the speed of the driver to the speed of the driven, but the ratio of diameters is inverse to the ratio of speeds. Wait, maybe I mixed up. Let's correct.

The relationship between the diameters and the mechanical advantage: If the driver wheel has diameter \(d_1\) and the driven wheel has diameter \(d_2\), the mechanical advantage \(MA=\frac{d_2}{d_1}\) when the driven wheel is the one doing the work. Wait, no, actually, mechanical advantage in terms of force: the force is related to the torque. Torque \(\tau = F\times r\) (or \(F\times\frac{d}{2}\)). For the same belt tension \(T\), the torque on the driver is \(T\times\frac{d_1}{2}\) and on the driven is \(T\times\frac{d_2}{2}\). The mechanical advantage (force advantage) is \(\frac{\text{Output Force}}{\text{Input Force}}=\frac{\tau_{driven}/(\frac{d_2}{2})}{\tau_{driver}/(\frac{d_1}{2})}=\frac{\tau_{driven}}{\tau_{driver}}\times\frac{d_1}{d_2}\). But if the wheels are connected by a belt, the linear speed at the rim is the same, so \(v = \pi d_1n_1=\pi d_2n_2\), where \(n_1\) is the rotational speed of the driver and \(n_2\) is the rotational speed of the driven. The mechanical advantage (in terms of force) is \(\frac{F_2}{F_1}=\frac{d_2}{d_1}\) (because \(F_1\times\frac{d_1}{2}=F_2\times\frac{d_2}{2}\) for equal torque, wait no, if \(F_1\) is the input force on the driver and \(F_2\) is the output force on the driven, then torque on driver: \(F_1\times\frac{d_1}{2}\), torque on driven: \(F_2\times\frac{d_2}{2}\). For a belt - driven system, the torques are related by the fact that the tension provides the torque, and if we assume no slip, the torques are such that \(F_1\times\frac{d_1}{2}=F_2\times\frac{d_2}{2}\) only if it's a fixed - axis, but actually, the mechanical advantage for force is \(\frac{F_2}{F_1}=\frac{d_1}{d_2}\) when the driven wheel is larger? Wait, I think I had it reversed earlier. Let's start over.

The problem says the mechanical advantage is \(\frac{1}{3}\). Let's assume that the mechanical advantage is defined as \(\frac{\text{Force on Driven}}{\text{Force on Driver}}=\frac{1}{3}\). So \(\frac{F_{driven}}{F_{driver}}=\frac{1}{3}\). Since torque \(\tau = F\times r\), and for the belt - driven wheels, the tension \(T\) is the same, so the torque on the driver is \(T\times\frac{d_{driver}}{2}\) and on the driven is \(T\times\frac{d_{driven}}{2}\). The force on the driver \(F_{driver}\) is related to its torque: \(\tau_{driver}=F_{driver}\times\frac{d_{driver}}{2}\), and force on the driven \(F_{driven}\) is related to its torque: \(\tau_{driven}=F_{driven}\times\frac{d_{driven}}{2}\). Since the torques are related by the fact that the belt doesn't slip, the power is conserved (ignoring losses), so \(\tau_{driver}n_{driver}=\tau_{driven}n_{driven}\), but also \(v = \pi d_{driver}n_{driver}=\pi d_{driven}n_{driven}\), so \(d_{driver}n_{driver}=d_{driven}n_{driven}\), or \(n_{driven}=\frac{d_{driver}}{d_{driven}}n_{driver}\).

But going back to the mechanical advantage formula for wheel - belt systems: the mechanical advantage (MA) is equal to the ratio of the diameter of the driven wheel to the diameter of the driver wheel when MA is defined as the ratio of the distance moved by the load (driven) to the distance moved by the effort (driver). Wait, no, distance moved: the distance a point on the wheel moves is \(s=\pi d\times n\) (number of revolutions). Since the belt length is the same, the distance moved by the driver wheel's rim and the driven wheel's rim is the same. So \(\pi d_{driver}n_{driver}=\pi d_{driven}n_{driven}\), so \(d_{driver}n_{driver}=d_{driven}n_{driven}\), or \(\frac{n_{driven}}{n_{driver}}=\frac{d_{driver}}{d_{driven}}\).

The mechanical advantage in terms of force: If we apply a force \(F_{driver}\) to the driver wheel, the torque is \(F_{driver}\times\frac{d_{driver}}{2}\). The torque on the driven wheel is the same (because the belt tension provides the torque, and for no slip, the torques are related by the belt tension), so \(F_{driver}\times\frac{d_{driver}}{2}=F_{driven}\times\frac{d_{driven}}{2}\), so \(F_{driven}=F_{driver}\times\frac{d_{driver}}{d_{driven}}\). So the mechanical advantage \(MA = \frac{F_{driven}}{F_{driver}}=\frac{d_{driver}}{d_{driven}}\). Wait, now I'm confused. Let's check the problem statement again.

The problem says "the mechanical advantage of the wheel - and - belt assembly... is \(\frac{1}{3}\)". Let's assume that the driven wheel has diameter 12 inches, and we need to find the driver's diameter. Let's use the formula \(MA=\frac{d_{driven}}{d_{driver}}\) (maybe the problem defines MA as \(\frac{d_{driven}}{d_{driver}}\)). So if \(MA=\frac{1}{3}=\frac{12}{d_{driver}}\), then solving for \(d_{driver}\) gives \(d_{driver}=12\times3 = 36\)? No, that can't be, because the options are 2, 3, 4, 6. So I must have the formula reversed.

Ah! Wait, maybe the mechanical advantage is the ratio of the driver's diameter to the driven's diameter, but inverted. Wait, no, let's think about speed. If the driver wheel is smaller, it has to rotate more times to move the belt the same distance as the larger driven wheel. So the speed of the driven wheel is slower than the driver. The mechanical advantage (force advantage) is higher when the driven wheel is larger. Wait, maybe the formula is \(MA=\frac{d_{driver}}{d_{driven}}\). So if \(MA = \frac{1}{3}\), then \(\frac{d_{driver}}{d_{driven}}=\frac{1}{3}\), so \(d_{driver}=\frac{d_{driven}}{3}\). Since \(d_{driven}=12\) inches, then \(d_{driver}=\frac{12}{3}=4\) inches. Yes! That makes sense with the options. So the correct formula is \(MA=\frac{d_{driver}}{d_{driven}}\) (maybe the problem defines MA as the ratio of the driver's diameter to the driven's, or maybe I had the force advantage reversed). Let's verify: if the driver has diameter 4 inches and the driven has 12 inches, then the ratio of diameters is \(\frac{4}{12}=\frac{1}{3}\), which matches the given mechanical advantage. So that must be it.

So, using \(MA=\frac{d_{driver}}{d_{driven}}\), we have \(\frac{1}{3}=\frac{d_{driver}}{12}\), so \(d_{driver}=12\times\frac{1}{3}=4\) inches.

Answer:

C. 4 inches