The device in Fig. 6.41 is used to compress air in a cylinder by applying a couple M to the arm AB. The pressure p in the cylinder and the net force F exerted on the piston by pressure are P = Patm (v0/v), F = Apatm (V0/V - 1) where A = 0.02 m² is the cross-sectional area of the piston, patm =

Question

The device in Fig. 6.41 is used to compress air in a cylinder by applying a couple M to the arm AB. The pressure p in the cylinder and the net force F exerted on the piston by pressure are

[latex] p = p_{atm} \left(\frac{V_{0}}{V} \right)[/latex],

[latex] p = Ap_{atm} \left(\frac{V_{0}}{V} - 1 \right)[/latex],

where A = 0.02 m² is the cross-sectional area of the piston, [latex]p_{atm}= 10^{5} Pa[/latex] (Pascals, or N/m²) is atmospheric pressure, V is the volume of air in the cylinder, and [latex]V_{0} [/latex] is the value of V when α = 0. The dimensions R = 150 mm, b = 350 mm, d = 150 mm, and L = 1050 mm. If M and α are initially zero and M is slowly increased until its value is 40 N-m, what are the resulting values of α and p?

Strategy
By expressing the volume of air in the cylinder in terms of α, we will determine the force exerted on the cylinder by pressure in terms of α. From the free-body diagram of the piston we will determine the axial force in the twoforce member BC in terms of the pressure force on the cylinder. Then from the free-body diagram of the arm AB we will obtain a relation between M and α.

Accepted Answer

From the geometry of the arms AB and BC (Fig. a), the volume of air in the cylinder is

[latex]V = A \left(L - d - \sqrt{b^{2} - R^{2} \sin ^{2}α }+ R \cos α ight) [/latex]

When α = 0, the volume is

[latex]V_{0} = A(L - d - b + R) [/latex]

Therefore the force exerted on the piston by pressure is

[latex] F = Ap_{atm} \left(\frac{V_{0}}{V} - 1 ight) = Ap_{atm}\left(\frac{L - d - b + R}{L - d - \sqrt{b^{2} - R^{2} \sin ^{2} α } + R \cos α } - 1 ight) [/latex],

We draw the free-body diagrams of the piston and the arm AB in Figs, b and c, where N is the force exerted on the piston by the cylinder (friction is neglected), Q is the axial force in the two-force member BC, and [latex]A_{x} and A_{y}[/latex] and Ay are the reactions due to the pin support A. From Fig. b, we obtain the equilibrium equation

[latex]\sum{F_{x} } = F - Q \cos \beta = 0,[/latex]

where

[latex]\beta = arctan \left(\frac{R \sin \alpha }{\sqrt{b^{2} - R^{2} \sin ^{2} \alpha } } ight).[/latex]

The force exerted on the arm Afi at B is

[latex] Q \cos \beta i + Q \sin \beta j [/latex]

The moment of this force about A is

[latex]r_{AB} imes\left(Q \cos \beta i + Q \sin \beta j ight) = \left|\begin{matrix}i & j & k \ R \cos \alpha & R \sin \alpha & 0 \ Q\cos \beta & Q \sin \beta & 0 \end{matrix} ight| = Q R \left(\cos \alpha \sin \beta - \sin \alpha \cos \beta ight)K . [/latex]Using this result, the sum of the moments about A is

[latex]\sum{M_{(point A)} } = M + Q R\left(\cos \alpha \sin \beta - \sin \alpha \cos \beta ight) = 0[/latex]

If we choose a value of a, we can sequentially calculate V, F, β, Q, and M. Computing M as a function of a, we obtain the graph shown in Fig. 6.42. The moment M = 40 N-m at approximately α = 80°. By examining computed results near 80° (see table), we estimate that α = 79.61° when M = 40 N-m. Once we know α, we can calculate V and then p. obtaining

[latex]p = 1.148 p_{atm} = 1.148 imes 10^{5} Pa.[/latex]

α	M (N-m)
79.59º	39.9601
79.60º	39.9769
79.61º	39.9937
79.62º	40.0105
79.63º	40.0272

Question 6.11: The device in Fig. 6.41 is used to compress air in a cylinde...

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