Question

a compound gear train is composed of four gears: a, b, c, and d. gear a has 120 teeth. gear b has 12 teeth and shares a shaft with gear c, which has 24 teeth. gear c is meshed with gear d, the output gear. power is supplied at gear a with 200 ft. lb of torque and is traveling at 400 rpm. with a torque output of 1600 ft lb, how many teeth should gear d have?

Explanation:

Step1: Gear ratio and torque relationship

The torque ratio is equal to the product of the gear ratios. The torque ratio \(T_{ratio}=\frac{T_{output}}{T_{input}}=\frac{1600}{200} = 8\). The gear ratio for gears \(A - B\) is \(r_{AB}=\frac{N_{A}}{N_{B}}\), and for gears \(C - D\) is \(r_{CD}=\frac{N_{D}}{N_{C}}\). Since \(B\) and \(C\) are on the same shaft, the overall gear ratio \(r = r_{AB}\times r_{CD}=\frac{N_{A}}{N_{B}}\times\frac{N_{D}}{N_{C}}\)

Step2: Substitute values

We know \(N_{A} = 120\), \(N_{B}=12\), \(N_{C} = 24\). And \(r = 8\) (from torque ratio). Substitute into the formula \(8=\frac{120}{12}\times\frac{N_{D}}{24}\)

First, simplify \(\frac{120}{12}=10\). Then the equation becomes \(8 = 10\times\frac{N_{D}}{24}\)

Step3: Solve for \(N_{D}\)

Rearrange the equation \(8=\frac{10N_{D}}{24}\). Cross - multiply: \(10N_{D}=8\times24\). So \(10N_{D}=192\). Then \(N_{D}=\frac{192\times10}{10}\) (Wait, no, correct cross - multiply: \(N_{D}=\frac{8\times24\times12}{120}\) (original formula \(r=\frac{N_{A}}{N_{B}}\times\frac{N_{D}}{N_{C}}\), \(8=\frac{120}{12}\times\frac{N_{D}}{24}\), \(N_{D}=\frac{8\times12\times24}{120}\))

\(N_{D}=\frac{8\times12\times24}{120}=\frac{8\times24}{10}= \frac{192}{10}\) (Wrong, re - do)

Correct: \(r=\frac{N_{A}}{N_{B}}\times\frac{N_{D}}{N_{C}}\), \(8=\frac{120}{12}\times\frac{N_{D}}{24}\), \(8 = 10\times\frac{N_{D}}{24}\), \(N_{D}=\frac{8\times24}{10}=19.2\) (No, wrong approach. Use power transmission (assuming no power loss \(P = T\omega\), \(\omega=\frac{2\pi n}{60}\), \(P_{input}=P_{output}\), \(T_{input}\omega_{input}=T_{output}\omega_{output}\), \(\frac{\omega_{input}}{\omega_{output}}=\frac{T_{output}}{T_{input}}\). Also \(\frac{\omega_{A}}{\omega_{D}}=\frac{N_{B}}{N_{A}}\times\frac{N_{C}}{N_{D}}\) (gear ratio rules: \(\frac{\omega_{1}}{\omega_{2}}=\frac{N_{2}}{N_{1}}\) for meshed gears). Since \(\frac{T_{output}}{T_{input}}=\frac{\omega_{input}}{\omega_{output}}\) (from \(T\omega = constant\)), \(\frac{1600}{200}=\frac{N_{B}}{N_{A}}\times\frac{N_{D}}{N_{C}}\) (signs ignored as we are dealing with magnitudes). So \(8=\frac{12}{120}\times\frac{N_{D}}{24}\), \(N_{D}=\frac{8\times120\times24}{12}\)

Step4: Calculate \(N_{D}\)

\(N_{D}=\frac{8\times120\times24}{12}\)
\(N_{D}=8\times10\times24\)
\(N_{D}=1920\)

Answer:

1920