Unformatted text preview: r the wire beneath the limb is as follows:
Taking upward and to the right as the positive
directions, we find for the x components of the
forces that +y
T 447 N ΣFx = Tx − ( 447 N ) cos 14.0° = 0
Tx = ( 447 N ) cos 14.0° = 434 N
For the y components of the forces we have θ
Tx 14.0º Ty +x 151 N ΣFy = Ty + ( 447 N ) sin14.0° − 151 N = 0
Ty = − ( 447 N ) sin14.0° + 151 N = 43 N
The magnitude of the tension force is
T = Tx2 + Ty2 = ( 434 N )2 + ( 43 N )2 = 436 N Since the components of the tension force and the angle θ are related by tan θ = Ty / Tx , we
find that Ty −1 43 N = tan = 5.7° Tx 434 N θ = tan −1 Chapter 4 Problems 235 111. SSM REASONING The shortest time to pull the person from the cave corresponds to
the maximum acceleration, a y , that the rope can withstand. We first determine this
acceleration and then use kinematic Equation 3.5b ( y = v0 y t + 1 a y t 2 ) to find the time t.
2
SOLUTION As the person is being pulled from the cave, there are two forces that act on
him; they are the tension T in the rope that points vertically upward, and the weight of the
person mg that points vertically downward. Thus, if we take upward as the positive
direction, Newton's second law gives ∑ Fy = T – mg = ma y . Solving for a y , we have
ay = T
T
569 N
–g=
–g=
− 9.80 m/s 2 = 0.92 m/s 2
2
2
m
W /g
(5.20 ×10 N)/(9.80 m/s ) Therefore, from Equation 3.5b with v0 y = 0 m/s, we have y = 1 a y t 2 . Solving for t, we find
2 t= 2y
2(35.1 m)
=
= 8.7 s
ay
0.92 m/s 2 ____________________________________________________________________________________________ 112. REASONING The freebody diagrams for the large cube (mass = M) and the small cube
(mass = m) are shown in the following drawings. In the case of the large cube, we have
omitted the weight and the normal force from the surface, since the play no role in the
solution (although they do balance).
fsMAX – FN
P FN
mg In these diagrams, note that the two blocks exert a normal force on each other; the large
block exerts the force FN on the smaller block, while the smaller block exerts the force –FN
on the larger block. In accord with Newton’s third law these forces have opposite directions
and equal magnitudes FN . Under the influence of the forces shown, the two blocks have the
same acceleration a. We begin our solution by applying Newton’s second law to each one. SOLUTION According to Newton’s second law, we have ΣF = P – FN = Ma
144 2444
4
3 FN = ma
1 24
43 Large block Small block 236 FORCES AND NEWTON'S LAWS OF MOTION Substituting FN = ma into the largeblock expression and solving for P gives
P = (M + m) a
For the smaller block to remain in place against the larger block, the static frictional force
must balance the weight of the smaller block, so that fsMAX = mg. But fsMAX is given by
fsMAX = µsFN , where, from the Newton’s second law, we know that FN = ma. Thus, we
have µsma = mg or a = g/µs . Using this result in the expression for P gives P = ( M + m) ( M + m ) g = ( 25 kg + 4.0 kg ) ( 9.80 m/s
a=
µs 2 0.71 )= 4.0 × 102 N ____________________________________________________________________________________________ 113. REASONING According to Newton’s second law, the acceleration of the probe is
a = ΣF/m. Using this value for the acceleration in Equation 2.8 and noting that the probe
starts from rest (v0 = 0 m/s), we can write the distance traveled by the probe as
1 ΣF
x = v0t + 1 at 2 = 2
2 m 2
t This equation is the basis for our solution.
SOLUTION Since each engine produces the same amount of force or thrust T, the net force
is ΣF = 2T when the engines apply their forces in the same direction and ΣF = T 2 + T 2 = 2T when they apply their forces perpendicularly. Thus, we write the
distances traveled in the two situations as follows:
1 2T 2
x= t
2
14 m 3
4
244 and Engines fired in
same direction 1 2T 2
x= t⊥ 2 m 244
144
3
Engines fired
perpendicularly Since the distances are the same, we have
1 2T 2 m 2 1 2T 2 t⊥
t = 2 m or 2
2 t 2 = t⊥ The firing time when the engines apply their forces perpendicularly is, then, t⊥ = ( 4 2 ) t = ( 4 2 ) ( 28 s ) = 33 s ____________________________________________________________________________________________ Chapter 4 Problems 237 114. REASONING The drawing shows the point between the earth and the moon where the
gravitational force exerted on the spacecraft by the earth balances that exerted by the moon.
The magnitude of the gravitational force exerted on the spacecraft by the earth is Fearth = G mearth mspacecraft
r2 while that exerted on the spacecraft by the moon is Fmoon = G mmoon mspacecraft ( rearthmoon − r )2 By setting these two expressions equal to each other (since the gravitational forces balance),
we will be able to find the distance r.
Point where the
gravitational forces
balance Earth
Moon
r
rearthmoon SOLUTION Setting Fearth equal to Fmoon, we have G mearth mspacecraft
r2 =G mmoon mspacecraft ( rearthmoon − r )2 Solving this expression for r gives ( rearthmoon )
r= mearth
mmoon mearth
1+
mmoon = ( 3.85 ×108 m )
1 + 81.4 81.4 = 3.47 ×108 m ______________________________________________________________________________ 238 FORCES AND NEWTON'S LAWS OF MOTION 115....
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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