fisica1-youn-e-freedman-exercicios-resolvidos

−4
−4
2
11.64: a) ∆w = −σ
( ∆l
l)
w 0
= −(0.23)(9.0
× 10 ) 4(0.30 × 10 m ) π = 1.3 µ m.
b)
F
⊥
∆l
1 ∆w
= AY = AY
l σ w
11
(2.1×
10 Pa) ( π (2.0×
10
=
0.42
−2
m)
2
)
0.10×
10
−
2.0 × 10
−3
2
m
= 3.1×
10
m
6
N,
where the Young’s modulus for nickel has been used.
11.65: a) The tension in the horizontal part of the wire will be 240 N. Taking torques
about the center of the disk, ( 240 N)(0.250 m) − w (1.00 m)) = 0, or w = 60 N.
b) Balancing torques about the center of the disk in this case,
( 240 N) (0.250 m) − ((60 N)(1.00 m) + (20 N)(2.00 m)) cos θ = 0, so θ = 53. 1°
.
11.66: a) Taking torques about the right end of the stick, the friction force is half the
w
weight of the stick, f = 2
⋅ Taking torques about the point where the cord is attached to
the wall (the tension in the cord and the friction force exert no torque about this
point),and noting that the moment arm of the normal force is
w
f
l tan θ , n tan θ = ⋅ Then, = tan θ < 0.40, so θ<
arctan (0.40) = 22°
.
2
n
b) Taking torques as in part (a), and denoting the length of the meter stick as l ,
l
l
fl = w + w(
l − x)
and nl
tan θ = w + wx.
2
2
In terms of the coefficient of friction µ
s
,
l
f + ( l − x)
2
3l
− 2x
µ
s
> = tanθ
= tanθ.
l
n + x l + 2x
2
Solving for x,
l 3tanθ
− µ
s
x >
= 30.
2 cm.
µ + tanθ
2
s
c) In the above expression, setting x = 10 cm and solving for µ
s
gives
(3 − 20 l) tanθ
µ
s
>
= 0.625.
1+
20 l