In Fig. 13–30 the sun gear is the input, and it is driven clockwise at 100 rev/min. The ring gear is held stationary by being fastened to the frame. Find the rev/min and direction of rotation of the arm and gear 4.

Question

Accepted Answer

Designate [latex]n_{F}=n_{2}=-100 rev / min[/latex], and [latex] n_{L}=n_{5}=0[/latex]. Unlocking gear 5 and holding the arm stationary, in our imagination, we find

[latex]e=-\left(\frac{20}{30}\right)\left(\frac{30}{80}\right)=-0.25[/latex]

Substituting this value in Eq. (13–32) gives

[latex]e=\frac{n_{L}-n_{A}}{n_{F}-n_{A}}[/latex] (13–32)

[latex]-0.25=\frac{0-n_{A}}{(-100)-n_{A}}[/latex]

or

[latex]n_{A}=-20 rev / min[/latex]

To obtain the speed of gear 4, we follow the procedure outlined by Eqs. (b), (c), and (d). Thus

[latex]n_{23}=n_{2}-n_{3}[/latex] (b)

[latex]n_{53}=n_{5}-n_{3}[/latex] (c)

[latex]\frac{n_{53}}{n_{23}}=\frac{n_{5}-n_{3}}{n_{2}-n_{3}} [/latex] (d)

[latex]n_{43}=n_{4}-n_{3} \quad n_{23}=n_{2}-n_{3}[/latex]

and so

[latex]\frac{n_{43}}{n_{23}}=\frac{n_{4}-n_{3}}{n_{2}-n_{3}}[/latex] (1)

But

[latex]\frac{n_{43}}{n_{23}}=-\frac{20}{30}=-\frac{2}{3}[/latex] (2)

Substituting the known values in Eq. (1) gives

[latex]-\frac{2}{3}=\frac{n_{4}-(-20)}{(-100)-(-20)}[/latex]

Solving gives

[latex]n_{4}=33 \frac{1}{3} rev / min[/latex]

Question 13.6: In Fig. 13–30 the sun gear is the input, and it is driven cl...

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