Question 9.2.4: Reconsider the single-stage gear train shown in Figure 1 and...

Goal Determine the mechanical advantage of the gear train in Figure 1.
Given The same situation as in Example 9.2.3, except that kinetic friction of μk = 0.25 acts on the 5 mm radii shafts in the journal bearings.
Assume Components are massless. Assume points B and D act as thrust bearings.
Draw Figure 2 shows the free-body diagrams of each shaft and gear.
Formulate Equations and Solve:
The following forces are zero by inspection F_{Az}  ,  F_{Bx}  ,  F_{Bz}  ,  F_{Cz}  ,  F_{Dx}  ,  F_{Dz} .
From the free-body diagram of Shaft 1, symmetry, and Equation (7.18) that describes journal bearing friction:

M_{O} =r\bullet \mu F                   (7.18)
M_{in} =(25  mm) (F_{in}) i
M_{Ax}= -F_{A  friction} r_{1} i = – (0.25)(5  mm)(F_{Ay} ) i
M_{Bx}= -F_{B  friction} r_{ 2} i = – (0.25)(5  mm)(F_{By} )i
\sum{F_{y}\left(\uparrow + \right) } =F_{Ay} +F_{By} – F_{in} – F_{12}=0 \Rightarrow F_{Ay} +F_{By} =\frac{\left(F_{in} – F_{12}\right) }{2}
\sum{M_{x} }\left(\curvearrowleft + \right) =(25  mm)(F_{in}) − (25  mm)(F_{12}) − (0.25) (5  mm)(F_{Ay} )-(0.25)(5  mm)(F_{By})=0
\Rightarrow F_{in}=\frac{26.25 }{23.75}F_{12}

From the free-body diagram of Shaft 2 and symmetry:

M_{out} = (35  mm)(F_{out}) i
M_{Cx} = (0.25)(5  mm)(F_{Cy}) i
M_{Dx} = (0.25)(5  mm)(F_{Dy}) i
\sum{F_{y}\left(\uparrow + \right) } =0=F_{Cy}+F_{Dy}+F_{out}+F_{21}\rightarrow F_{Cy}=F_{Dy}=\frac{- \left(F_{out}+F_{21}\right) }{2}
\sum{M_{x} }\left(\curvearrowleft + \right) =(35  mm)(F_{out}) − (35  mm)(F_{21}) + (0.25)(5  mm)(F_{Cy})+(0.25)(5  mm)(F_{Dy})= 0
\Rightarrow F_{out}=\frac{36.25 }{33.75}F_{21}

From Newton’s third law, we know that F_{12}=F_{21}.

\frac{M_{out}}{ M_{in}}=\frac{35  mm F_{out}}{25   mm F_{in}} =\frac{(35   mm)\frac{36.25}{33.75}F_{21} }{(25mm)\frac{26.25}{23.75}F_{12} } =1.36

The mechanical advantage of the gear train is 1.36

The efficiency of the gear train with bearing friction, relative to one without bearing friction is defined as:

Gear Train Efficiency =100\times \left[M_{out} / M_{in}\right]_{w/friction} ]/\left[M_{out} / M_{in}\right]_{wo/friction}]

Substituting in values found for the specific case of r_{1} = 25 mm, r_{2} = 35 mm, we find

Gear Train Efficiency = 100 × (1.36/1.40) = 97.1 percent

We interpret this to mean the gear train with friction is 97.1 percent as effective in converting the input moment into output moment; alternately we could say that there is a 2.9 percent “loss” in this conversion because of the presence of friction. What might this “lost” moment be converted into?
Check The friction on each shaft reduced the ratio by approximately the same amount. Check the moment balance for each shaft using alternate moment centers to ensure equilibrium.