Download Ch 08-151

15. What is the tension T exerted by the hamstring muscles in the back of the thigh and the compressive force Fc
in the knee joint due to the application of a horizontal force of 100 N to the ankle as shown in Figure P8.15?
8.15
Require that Στ = 0 about an axis through the knee
joint and perpendicular to the page. This gives
100 N
Στ = Fc ( 0 ) − T ( 4.00 cm ) + ( 100 N )( 42.0 cm ) = 0 ,
or the tension in the muscle is
T = 1.05 × 103 N = 1.05 kN
Then, ΣFx = − Fc − 100 N + 1.05 × 103 N=0 gives the
compression force as Fc = 950 N
42.0 cm
T
4.00 cm
Fc
36. A cylindrical 5.00-kg reel with a radius of 0.600 m and a frictionless axle, starts from rest and speeds up
uniformly as a 3.00-kg bucket falls into a well, making a light rope unwind from the reel (Fig. P8.36). The bucket
starts from rest and falls for 4.00 s. (a) What is the linear acceleration of the falling bucket? (b) How far does it
drop? (c) What is the angular acceleration of the reel?
FIGURE P8.36
8.36
The moment of inertia of the reel is
R
1
1
2
I = MR 2 = ( 5.00 kg ) ( 0.600 m ) = 0.900 kg ⋅ m 2
2
2
T
Applying Newton’s second law to the falling bucket gives
T
29.4 N − T = ( 3.00 kg ) a t
3.00 kg
Then, Newton’s second law for the reel gives
⎛ at
⎝R
τ = T R = Iα = I ⎜
or
T=
⎞
⎟,
⎠
I a t ( 0.900 kg ⋅ m
=
R2
( 0.600 m )2
at
(1)
29.4 N
2
)a =
t
( 2.50 kg ) a t .
(a) Solving equations (1) and (2) simultaneously gives
2
a t = 5.35 m s downward
1
1
2
(b) Δy = v i t + a t t 2 = 0 + ( 5.35 m s 2 ) ( 4.00 s ) = 42.8 m
2
2
(2)
(c) α =
a t 5.35 m s 2
=
= 8.91 rad s 2
R
0.600 m
60. A 12.0-kg object is attached to a cord that is wrapped around a wheel of radius r = 10.0 cm (Fig. P8.60). The
acceleration of the object down the frictionless incline is measured to be 2.00 m/s2. Assuming the axle of the
wheel to be frictionless, determine (a) the tension in the rope, (b) the moment of inertia of the wheel, and (c) the
angular speed of the wheel 2.00 s after it begins rotating, starting from rest.
FIGURE P8.60
8.60
(a) Consider the free-body diagram of the block given at
the right. If the +x-axis is directed down the incline,
ΣFx = max gives
mg sin 37.0° − T = m at , or T = m ( g sin 37.0° − at )
n
T
at
12.0 kg
+x
T = ( 12.0 kg ) ⎡⎣( 9.80 m s 2 )sin 37.0° − 2.00 m s 2 ⎤⎦
= 46.8 N .
(b)
Now, consider the free-body diagram of the pulley.
Choose an axis perpendicular to the page and
passing through the center of the pulley,
⎛ at
Στ = Iα gives T ⋅ r = I ⎜
⎝ r
⎞
⎟ , or
⎠
T ⋅ r 2 ( 46.8 N )( 0.100 m )
I=
=
=
at
2.00 m s 2
2
0.234 kg ⋅ m 2
⎛a
(c) ω = ωi + α t = 0 + ⎜ t
⎝r
2
⎞ ⎛ 2.00 m s ⎞
=
t
⎟( 2.00 s ) = 40.0 rad s
⎟ ⎜
⎠ ⎝ 0.100 m ⎠
mg 37.0°
65. In Figure P8.65 the sliding block has a mass of 0.850 kg, the counterweight has a mass of 0.420 kg, and the
pulley is a uniform solid cylinder with a mass of 0.350 kg and an outer radius of 0.0300 m. The coefficient of
kinetic friction between the block and the horizontal surface is 0.250. The pulley turns without friction on its
axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of 0.820 m/s toward
the pulley when it passes through a photogate. (a) Use energy methods to predict its speed after it has moved to a
second photogate, 0.700 m away. (b) Find the angular speed of the pulley at the same moment.
FIGURE P8.65
8.65
Let mp be the mass of the pulley, m1 be the mass of the sliding block, and m2 be
the mass of the counterweight.
1
(a) The moment of inertia of the pulley is I = m p R p2 and its angular velocity at
2
v
, where v is the linear speed of the other masses. The
any time is ω =
Rp
friction force retarding the sliding block is fk = μ k n = μk ( m1 g ) .
Choose PEg = 0 at the level of the counterweight when the sliding mass
reaches the second photogate. Then, from the work-kinetic energy theorem,
(
Wnc = KEtrans + KErot + PEg
) − ( KE
f
trans
+ KErot + PEg
)
i
− fk ⋅ s =
⎛ v2 ⎞
1
1 1
( m1 + m2 )v 2f + ⎛⎜ mp Rp2 ⎞⎟⎜⎜ f2 ⎟⎟ + 0
2
2⎝ 2
⎠⎝ R p ⎠
−
or
⎛ v2 ⎞
1
1 1
( m1 + m2 )v i2 − ⎜⎛ mp Rp2 ⎟⎞⎜⎜ i2 ⎟⎟ − m2 gs ,
2
2⎝ 2
⎠⎝ R p ⎠
1⎛
1 ⎞ 2 1⎛
1 ⎞ 2
⎜ m1 + m2 + m p ⎟v f = ⎜ m1 + m2 + m p ⎟v i + m2 gs − μ k ( m1 g ) ⋅ s .
2⎝
2 ⎠
2⎝
2 ⎠
This reduces to v f = v i2 +
2 ( m2 − μk m1 ) g s
,
1
m1 + m2 + mp
2
and yields
2
2
m ⎞ 2 ( 0.208 kg )( 9.80 m s )( 0.700 m )
⎛
v f = ⎜ 0.820
= 1.63 m s .
⎟ +
s ⎠
1.45 kg
⎝
(b) ω f =
vf
Rp
=
1.63 m s
= 54.2 rad s
0.0300 m