Question

We have 3 gears meshed with the following teeth. A – 20; B – 50; C – 30. Another set have the following teeth. X – 20; Y – 25; Z – 30. Assuming gears A and X are running same rpm

Answer 1

The answer is    $N_{C}=N_{z}=\frac{2}{3} N _{p m}$

Given, that Geor A and year x running at some speed (in rpm).

$\text { So, } N_{A}=N_{x}=N \text { (in rpm) }$

Now

1st case - Let Geor A, B and C are in Mass.

So,

N T= Compton

for gears in mesh.

$\text { So, } N_{B}=-N_{A}\left(\frac{T_{A}}{T_{B}}\right)\\=-N \times\left(\frac{20}{50}\right)=-\frac{2}{5} N$

$N_{C}=-N_{B}\left(\frac{T_{B}}{T_{C}}\right)\\=-\left(-\frac{2}{5}\right)\left(\frac{50}{30}\right)=\frac{2}{3} N$

So,

$N_{A}=N \\\quad N_{B}=-\frac{2}{5} N \\N_{C}=\frac{2}{3} N$

Case 2

Now consider gears x, y and z are in mesh together.

$\begin{aligned}&N_{x}=N^{1} \\&N_{y}=-N_{x}\left(\frac{T_{x}}{T_{y}}\right)\end{aligned}$

$\begin{aligned}&=-N\left(\frac{20}{25}\right)=-\frac{4}{5} N \\N_{z} &=-N_{y}\left(\frac{T_{y}}{T_{z}}\right)\end{aligned}$

$=-\left(\frac{-4}{5} N\right)\left(\frac{25}{30}\right)\\=\frac{4}{5} N \times \frac{5}{6}=\frac{2}{3} N$

So,

$N_{x}=N \\\quad N_{y}=-\frac{4}{5} N \quad \\N_{z}=\frac{2}{3} N$

$\rightarrow$ year C and year z will run with

some rpm, as,$N_{C}=N_{z}=\frac{2}{3} N _{p m}$

What is gears ?

• A gear is a spinning circular machine element with cut teeth, or inserted teeth in the case of a cogwheel or gearwheel, that mesh with another toothed part to impart torque and speed. The basic concept underlying gear operation is identical to the basic principle behind lever operation.
• Gears are employed in mechanical devices to convey motion and torque between machine components. Gears can alter the direction of movement and/or increase the output speed or torque depending on the design and construction of the gear pair used.

To learn more about  gear https://brainly.com/question/14455728

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