The net force on the cylinder is parallel to the a;-axis, by symmetry; taken positive 

 toward negative x, it is, per unit of length perpendicular to the flow, 



F = fp cos t ds 



where p is the pressure on the cylinder at any point, ds is an element of distance along S, 

 and < is the angle between the normal to ds and the aj-axis; see Figure 81. But, also, ds cos 

 t = dy where dy is the element of y corresponding to ds. Hence 



F = ;p 



[53k] 



Inserting p = - p ?^/2, also t^ = r^ ^ ^2 _ 2gT cos 6, and using [53h,i], the contribution 

 of the Bernoulli term to the pressure is found to be 



Fb-- y^p^^'- 



But 



J' '' / 2 sin 

 A'— 



d cos d sin^ d 



d6 



r " ■ 



sin 



2 f) sin^ d j 

 '^ d9= — +2 



--rr —tr 



C sin 6 cos 6 ,„ C 



sin (j cos 



d9 



Hence 



FB=-%pg^ 



2 < dd = -npgV^ 



The same result is obtained with p = -p lJ^/2. Hence the total force on S is the same as if 

 the pressure in the fluid were uniform throughout. 



Half of the symmetric diagram of streamlines outside of S is illustrated in Figure 82. 

 The excess of pressure P - p^ along the a;-axis and then over S is shown on an arbitrary 

 scale. 



Figure 82 — Streamlines past a cylinder S 

 of semi-infinite cross-section, only half of 

 which is shown, of width 2ng at infinity; 

 also, the distribution of pressure along the 

 plane of symmetry and over the cylinder. 

 Constructed with use of a line source at 0. 



■ 



\y 



— -~^^^^^:z::z:^ 



~^~^^^^^^^^II~~ 



-- — __ 



■ 



_ 



■-~~-_ 



— ■ — --„_ 



~--.-______ ~ ' 



y-iiy~' __ -- 



-^f\'~~~- — 



-s <; 



?^^^2:^= 



Plane of Symmetry 



7\ — . 



______P-P~ 



/' 



123 



