P15_007

# P15_007 - 7(a The pressure diﬀerence results in forces...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 7. (a) The pressure diﬀerence results in forces applied as shown in the ﬁgure. We consider a team of horses pulling to the right. To pull the sphere apart, the team must exert a force at least as great as the horizontal component of the total force determined by “summing” (actually, integrating) these force vectors. .. . . .. .. . .. .. .. . .. .. .. . .. .. .. . .. .. .. ........... .......... .. . ... .... ..... ..... ..... ..... .... ...... .... ..... . .... .... ....... .. .. ..... .... .... . .. .... . . ..... .... . ... . . .. .... .. . .. . ....... . .. ........ . .. ..... . . . ....... ...... .. .. . .. .. . . . . . . . .. . . .. . . . . . . . . . . . .. . .. . . . .. . . .. . . . .... ... ... ... . ... ... ...................................... .... .... .... . .... .... ...................................... . . .. . .. .. .. . . . . . . . . . . . . . .. ... ... ........ ........ .. . ..... .. . ....... .... .. .... ... .... .. .. .... .... ... ... .... .... ... ... . .... . . .... ....... ... . .... ......... .. . .. ... . .... ... ....... .. ....... .. ..... ..... .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. . We consider a force vector at angle θ. Its leftward component is ∆p cos θdA, where dA is the area element for where the force is applied. We make use of the symmetry of the problem and let dA be that of a ring of constant θ on the surface. The radius of the ring is r = R sin θ, where R is the radius of the sphere. If the angular width of the ring is dθ, in radians, then its width is R dθ and its area is dA = 2πR2 sin θ dθ. Thus the net horizontal component of the force of the air is given by r • θ R π /2 Fh = 2πR2 ∆p sin θ cos θ dθ 0 = πR2 ∆p sin2 θ π /2 = πR2 ∆p . 0 (b) We use 1 atm = 1.01 × 105 Pa to show that ∆p = 0.90 atm = 9.09 × 104 Pa. The sphere radius is R = 0.30 m, so Fh = π (0.30 m)2 (9.09 × 104 Pa) = 2.6 × 104 N. (c) One team of horses could be used if one half of the sphere is attached to a sturdy wall. The force of the wall on the sphere would balance the force of the horses. ...
View Full Document

## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online