Solução_Calculo_Stewart_6e

F.
136 ¤ PROBLEMS PLUS
TX.10
5. (a) When θ = θ s, the block is not moving, so the sum of the forces on the block
must be 0, thus N + F + W = 0. This relationship is illustrated
geometrically in the figure. Since the vectors form a right triangle, we have
tan(θ s)= |F|
|N| = μ sn
n = μ s.
(b) We place the block at the origin and sketch the force vectors acting on the block, including the additional horizontal force
H, with initial points at the origin. We then rotate this system so that F lies along the positive x-axis and the inclined plane
is parallel to the x-axis.
|F| is maximal, so |F| = μ s n for θ>θ s. Then the vectors, in terms of components parallel and perpendicular to the
inclined plane, are
N = n j
W =(−mg sin θ) i +(−mg cos θ) j
F =(μ s n) i
H =(h min cos θ) i +(−h min sin θ) j
Equating components, we have
μ s n − mg sin θ + h min cos θ =0 ⇒ h min cos θ + μ s n = mg sin θ (1)
n − mg cos θ − h min sin θ =0 ⇒ h min sin θ + mg cos θ = n (2)
(c) Since (2) is solved for n, we substitute into (1):
h min cos θ + μ s (h min sin θ + mg cos θ)=mg sin θ
⇒
h min cos θ + h min μ s sin θ = mg sin θ − mgμ s cos θ
⇒
sin θ − μs cos θ tan θ − μs
h min = mg
= mg
cos θ + μ s sin θ 1+μ s tan θ
tan θ − tan θs
From part (a) we know μ s =tanθ s , so this becomes h min = mg
and using a trigonometric identity,
1+tanθ s tan θ
this is mg tan(θ − θ s ) as desired.
Note for θ = θ s, h min = mg tan 0 = 0, which makes sense since the block is at rest for θ s, thus no additional force H
is necessary to prevent it from moving. As θ increases, the factor tan(θ − θ s), and hence the value of h min,increases
slowly for small values of θ − θ s but much more rapidly as θ − θ s becomes significant. This seems reasonable, as the