Unformatted text preview: uation (1) for the airplane’s acceleration:
ax = µs mg cos θ
T cos θ
=
M
M ( cos θ − µs sin θ ) (6) SOLUTION We apply Equation (6) to calculate the acceleration of the airplane: ( ) ( 0.77 )( 85 kg ) 9.80 m/s cos 9.0
µs mg cos θ
=
= 6.7 × 10−3 m/s 2
ax =
o
o
M ( cos θ − µs sin θ ) (109 000 kg ) cos 9.0 − 0.77 sin 9.0 ( 2 o ) Chapter 4 Problems 87. REASONING As the freebody diagram shows, there are two forces
acting on the fireman as he slides down the pole: his weight W and the
kinetic frictional force fk. The kinetic frictional force opposes the
motion of the fireman, so it points upward in the +y direction. In accord
with Newton’s second law, the net force, which is the sum of these two
forces, is equal to the fireman’s mass times his acceleration. His mass,
and therefore his weight, is known, but his acceleration is not. We will
turn to one of the equations of kinematics from Chapter 3 to determine
the acceleration.
SOLUTION Newton’s second law ( ΣFy = ma y , Equation 4.2b ) can 219 +y fk W
Freebody diagram for
the fireman be applied to this situation:
ΣFy = + f k − W = ma y The magnitude W of the fireman’s weight can be expressed in terms of his mass as W = mg
(Equation 4.5), where g is magnitude of the acceleration due to gravity. Solving the equation
above for the magnitude of the kinetic frictional force, and using W = mg, gives
f k = ma y + W = ma y + mg (1) Since the initial and final velocities, v0y and vy, and the displacement y are known, we will
use Equation 3.6b from the equations of kinematics to relate these variables to the
2
acceleration: v 2 = v0 y + 2a y y . Solving this equation for ay and substituting the result into
y
Equation 1 gives
2 v 2 − v0 y
y
fk = m 2y + mg We note that the fireman slides down the pole, so his displacement is negative, or
y = −4.0 m. The magnitude of the kinetic frictional force is, then,
2 v 2 − v0 y
y
fk = m 2y + mg (1.4 m/s )2 − ( 0 m/s )2 = ( 86 kg ) + ( 86 kg ) ( 9.80 m/s 2 ) = 820 N
2 ( −4.0 m ) ______________________________________________________________________________ 220 FORCES AND NEWTON'S LAWS OF MOTION 88. REASONING
Consider the forces that act on each block. Only one
+
−
force contributes to the horizontal net force acting on
block 1, as shown in the freebody diagram. This is the
−P
force −P with which block 2 pushes on block 1. The
minus sign in the freebody diagram indicates the
Freebody diagram for
direction of the force is to the left. This force is part of
block 1
the actionreaction pair of forces that is consistent with
Newton’s third law. Block 1 pushes forward and to the
right against block 2, and block 2 pushes backward and to the left against block 1 with an
oppositely directed force of equal magnitude.
Two forces contribute to the horizontal net force acting
+
−
on block 2, as shown in the freebody diagram. One is
the force P with which block 1 pushes on block 2.
P
fk
According to Newton’s third law, this force has the
same magnitude but the opposite direction as the force
Freebody diagram for
with which block 2 pushes on block 1. The other force
block 2
is the kinetic frictional force fk, which points to the left,
in opposition to the relative motion between the block and the surface on which it slides.
Both blocks decelerate, the magnitude of the deceleration being the same for each block.
They have the same deceleration, because they are pressed together. Since the blocks are
moving to the right in the drawing, the acceleration vector points to the left, for it reflects
the slowing down of the motion. In Case A and in Case B we will apply Newton’s second
law separately to each block in order to relate the net force to the acceleration.
SOLUTION Referring to the freebody diagram for block 1, we write Newton’s second law
as follows:
−P
(1)
{ = m1 ( −a )
Net force
on block 1 where a is the magnitude of the acceleration. The minus sign appears on the right side of
this equation because the acceleration, being a deceleration, points to the left, in the negative
direction. Referring to the freebody diagram for block 2, we write Newton’s second law as
follows:
(2)
P − f k = m2 ( − a )
13
2
Net force
on block 2 Solving Equation (1) for a gives a = P / m1 . Substituting this result into Equation (2) gives −P P − f k = m2 m 1 or P= m1 f k m1 + m2 Chapter 4 Problems 221 Substituting this result for P into a = P / m1 gives
a= m1 f k
fk
P
=
=
m1 m1 ( m1 + m2 ) m1 + m2 We can now use these results to calculate P and a in both cases.
a. Case A Case B m1 f k P= b. Case A = ( 3.0 kg ) ( 5.8 N ) = 2.9 N m1 f k P= = ( 6.0 kg )( 5.8 N ) = 3.9 N m1 + m2
m1 + m2 − fk a= m1 + m2 = 3.0 kg + 3.0 kg 6.0 kg + 3.0 kg −5.8 N
= −0.97 m/s 2
3.0 kg + 3.0 kg The magnitude of the acceleration is 0.97 m/s 2 .
Case B a= − fk m1 + m2 = −5.8 N
= −0.64 m/s 2
6.0 kg + 3.0 kg The magnitude of the acceleration is 0.64 m/s 2 . 89. SSM REASONING The tension in each coupling bar is responsible for accelerating the
objects behind it. The masses of the cars are m1, m2, and m3....
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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