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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. ..
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. 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. ...
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 Fall '08
 SPRUNGER
 Physics, Force

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