A 3 ft by 4 ft plate moves at 44 ft/sec in still air at an angle of 12° with the horizontal. Using a coefficient of drag CD = 0.17 and a coefficient of lift CL = 0.72, determine (a) the resultant force exerted by the air on the plate, (b) the frictional force, and (c) the horsepower required to

Question

A 3 ft by 4 ft plate moves at 44 ft/sec in still air at an angle of 12° with the horizontal. Using a coefficient of drag [latex]C_{D}=0.17 ext { and a coefficient of lift } C_{L}=0.72[/latex], determine (a) the resultant force exerted by the air on the plate, (b) the frictional force, and (c) the horsepower required to keep the plate moving. (Use γ = 0.0752 lb/ft³.)

Accepted Answer

[latex] ext { (a) Drag force }=C_{D}\left(\frac{\gamma}{g} ight) A \frac{V^{2}}{2}=(0.17)\left(\frac{0.0752}{32.2} ight)(12) \frac{(44)^{2}}{2}=4.61 lb [/latex].

[latex] ext { Lift force }=C_{L}\left(\frac{\gamma}{g} ight) A \frac{V^{2}}{2}=(0.72)\left(\frac{0.0752}{32.2} ight)(12) \frac{(44)^{2}}{2}=19.5 lb [/latex].

Referring to Fig. 13-13, the resultant of the drag and lift components is

[latex]R=\sqrt{(4.61)^{2}+(19.5)^{2}}=20.0 lb \quad ext { acting on the plate at } heta_{x}= an ^{-1}(19.5 / 4.61)=76^{\circ} 42^{\prime}[/latex].

(b) The resultant force might also have been resolved into a normal component and a frictional component (shown dotted in the figure). From the vector triangle,

[latex] ext { frictional component }=R \cos \left( heta_{x}+12^{\circ} ight)=(20.0)(0.0227)=0.45 lb[/latex].

(c) Horsepower = (force in direction of motion × velocity)/550 = (4.61 × 44)/550 = 0.369.

Question 13.28: A 3 ft by 4 ft plate moves at 44 ft/sec in still air at an a...

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