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Unformatted text preview: e net force and of the acceleration are ΣF = a= ( ΣFx ) 2 () + ΣFy 2 2 () + ΣFy 2 ( 4.30 ×106 N ) + (1.73 ×106 N )
2 = m ( ΣFx ) 2 4.50 ×10 kg
5 = 10.3 m/s 2 The direction of the acceleration is the same as the direction of the net force. Thus, it is
directed above the horizontal at an angle of
6 ΣFy −1 1.73 × 10 N = tan = 21.9°
6 4.30 × 10 N ΣFx θ = tan −1 14. REASONING The acceleration of the sky diver can be obtained directly from Newton’s
second law as the net force divided by the sky diver’s mass. The net force is the vector sum
of the sky diver’s weight and the drag force.
SOLUTION From Newton’s second law, ΣF = ma (Equation
4.1), the sky diver’s acceleration is
a= f ΣF
m The freebody diagram shows the two forces acting on the sky
diver, his weight W and the drag force f. The net force is
ΣF = f − W . Thus, the acceleration can be written as
f −W
a=
m +
− W
Freebody diagram Chapter 4 Problems 169 The acceleration of the sky diver is
a= f − W 1027 N − 915 N
=
= +1.20 m/s 2
m
93.4 kg Note that the acceleration is positive, indicating that it points upward .
____________________________________________________________________________________________ 15. REASONING Newton’s second law, ΣF = ma, states that a net force of ΣF must act on an
object of mass m in order to impart an acceleration a to the object. In the impact shock test
the box is subjected to a large deceleration and, hence, a correspondingly large net force.
To determine the net force we will determine the deceleration in a kinematics calculation
and use it in Newton’s second law. SOLUTION According to Newton’s second law, the net force is ΣF = ma, where the
acceleration can be determined with the aid of Equation 2.4 (v = v0 + at). According to this
equation
v − v0
a=
t
Substituting this result for the acceleration into the second law gives v − v0 ΣF = ma = m t
Since the initial velocity (v0 = +220 m/s), final velocity (v = 0 m/s), and the duration of the
collision (t = 6.5 × 10−3 s) are known, we find v − v0 0 m/s − 220 m/s 6
ΣF = m = ( 41 kg ) = −1.39 × 10 N
−3 6.5 ×10 s t
The minus sign indicates that the net force points opposite to the direction in which the box
is thrown, which has been assumed to be the positive direction. The magnitude of the net
force is 1.39 × 106 N , which is over three hundred thousand pounds. 16. REASONING Since there is only one force acting on the man in the horizontal direction, it
is the net force. According to Newton’s second law, Equation 4.1, the man must accelerate
under the action of this force. The factors that determine this acceleration are (1) the
magnitude and (2) the direction of the force exerted on the man, and (3) the mass of the
man. 170 FORCES AND NEWTON'S LAWS OF MOTION When the woman exerts a force on the man, the man exerts a force of equal magnitude, but
opposite direction, on the woman (Newton’s third law). It is the only force acting on the
woman in the horizontal direction, so, as is the case with the man, she must accelerate. The
factors that determine her acceleration are (1) the magnitude and (2) the direction of the
force exerted on her, and (3) the her mass.
SOLUTION
a. The acceleration of the man is, according to Equation 4.1, equal to the net force acting on
him divided by his mass.
aman = ΣF 45 N
=
= 0.55 m / s 2 (due east)
m 82 kg b. The acceleration of the woman is equal to the net force acting on her divided by her mass. awoman = ΣF 45 N
=
= 0.94 m / s 2 (due west)
m 48 kg ______________________________________________________________________________________ 17. REASONING Equations 3.5a ( x = v0 xt + 12 axt 2 ) and 3.5b ( y = v0 yt + 12 ayt 2 ) give the displacements of an object under the influence of constant accelerations ax and ay. We can
add these displacements as vectors to find the magnitude and direction of the resultant
displacement. To use Equations 3.5a and 3.5b, however, we must have values for ax and ay.
We can obtain these values from Newton’s second law, provided that we combine the given
forces to calculate the x and y components of the net force acting on the duck, and it is here
that our solution begins.
SOLUTION Let the directions due east and due north, respectively, be the +x and +y
directions. Then, the components of the net force are
ΣFx = 0.10 N + ( 0.20 N ) cos 52° = 0.2231 N
ΣFy = – ( 0.20 N ) sin 52° = –0.1576 N According to Newton’s second law, the components of the acceleration are
ax =
ay = ΣFx
m
ΣFy
m = 0.2231 N
= 0.08924 m/s 2
2.5 kg = –0.1576 N
= –0.06304 m/s 2
2.5 kg Chapter 4 Problems 171 From Equations 3.5a and 3.5b, we now obtain the displacements in the x and y directions: ( ) x = v0 xt + 1 axt 2 = ( 0.11 m/s )( 3.0 s ) + 1 0.08924 m/s 2 ( 3.0 s ) = 0.7316 m
2
2 ( 2 ) y = v0 y t + 1 a y t 2 = ( 0 m/s )( 3.0 s ) + 1 –0.06304 m/s2 ( 3.0 s ) = –0.2837 m
2
2
2 The magnitude of the resultant displacement is
r = x2 + y 2 = ( 0.7316 m )2 + ( –0.2837 m )2 = 0.78 m The directio...
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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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