Question 13.9: In Fig. 13–38 an electric motor transmits 1-hp at 1800 rev/m...

From Eq. (13–19)

\cos\psi =\frac{\tan\phi _n}{\tan\phi _t}                              (13.19)

we find

\phi _t=\tan ^{-1}\frac{\tan \phi _n}{\cos \psi }=\tan ^{-1}\frac{\tan 20°}{\cos 30°}=22.8°

Also, P_t = P_n \cos \psi = 12 cos 30° = 10.39 teeth/in. Therefore the pitch diameter of the pinion is d_p = 18/10.39 = 1.732 in. The pitch-line velocity is

V=\frac{\pi dn}{12}=\frac{\pi(1.732)(1800)}{12}=816  ft/min

The transmitted load is

W_t=\frac{33  000H}{V}=\frac{(33  000)(1)}{816}=40.4  lbf

From Eq. (13–40) we find

W_r = W_t \tan \phi_t = (40.4) \tan 22.8° = 17.0  lbf

W_a = W_t tan \psi = (40.4) tan 30° = 23.3  lbf

W=\frac{W_t}{\cos\phi_n\cos\psi}=\frac{40.4}{\cos20°\cos30°}=49.6  lbf

These three forces, W_r =  17.0 lbf in the -y direction, W_a = 23.3 lbf in the -x direction, and W_t = 40.4 lbf in the +z direction, are shown acting at point C in Fig. 13–39.

We assume bearing reactions at A and B as shown. Then F^x_{A} = W_a = 23.3 lbf. Taking moments about the z axis,

-(17.0)(13) + (23.3) (0.866) + 10F^y_{B} = 0

or F^y_{B} = 20.1 lbf. Summing forces in the y direction then gives F^y_{A} = 3.1 lbf. Taking moments about the y axis, next

10F^z_{B} – (40.4)(13) = 0

or F^z_{B} = 52.5 lbf. Summing forces in the z direction and solving gives F^z_{A} = 12.1 lbf.
Also, the torque is T = W_td_p/2 = (40.4)(1.732y2) = 35 lbf . in.

For comparison, solve the problem again using vectors. The force at C is

W = -23.3i – 17.0j + 40.4k lbf

Position vectors to B and C from origin A are

R_B = 10i              R_C = 13i + 0.866j

Taking moments about A, we have

R_B \times F_B + T + R_C \times W = 0

Using the directions assumed in Fig. 13–39 and substituting values gives

10i \times (F^y_{B}j – F^z_{B}k) – Ti + (13i + 0.866j) \times (-23.3i – 17.0j + 40.4k) = 0

When the cross products are evaluated we get

(10F^y_{B}k + 10F^z_{B}j) – Ti + (35i – 525j – 201k) = 0

obtaining T = 35 lbf . in, F^y_{B} = 20.1 lbf, and F^z_{B} = 52.5 lbf.

Next,

F_A = -F_B – W, and so F_A = 23.3i – 3.1j + 12.1k lbf.