218
measure because the earth’s mass is so great.
20. The person’s normal force on the box is paired
with the box’s normal force on the person. The dirt’s
frictional force on the box pairs with the box’s fric-
tional force on the dirt. The earth’s gravitational force
o n the box matches the box’s gravitational force on
the earth.
Chapter 6
5. (a) The easiest strategy is to find the time spent
aloft, and then find the range. The vertical motion and
the horizontal motion are independent. The vertical
motion has acceleration —g, and the cannonball
spends enough time in the air to reverse its vertical
velocity component completely, so we have
.
v
y
= v
yf
—v
yi
= —2v sin
.
.
The time spent aloft is therefore
.
t=
.
v
y
/ a
y
= 2v sin
.
/ g .
During this time, the horizontal distance traveled is
R= v
x
.
t
= 2 v
2
sin
.
cos
.
/ g .
(b) The range becomes zero at both
.
=0 and at
.
=90
°
. The
.
=0 case gives zero range because the
ball hits the ground as soon as it leaves the mouth of
the cannon. A 90 degree angle gives zero range
because the cannonball has no horizontal motion.
Chapter 8
8. We want to find out about the velocity vector v
BG
of
the bullet relative to the ground, so we need to add
Annie’s velocity relative to the ground v
AG
to the
bullet’s velocity vector v
BA
relative to her. Letting the
positive x axis be east and y north, we have
v
BA,x
= (140 mi/hr) cos 45
°
= 100 mi/hr
v
BA,y
= (140 mi/hr) sin 45
°
= 100 mi/hr
and
v
AG,x
= 0
v
AG,y
= 30 mi/hr .
The bullet’s velocity relative to the ground therefore
has components
v
BG,x
= 100 mi/hr and
v
BG,y
= 130 mi/hr .
Its speed on impact with the animal is the magnitude
of this vector
|v
BG
|=
(100mi/hr)
2
+(130mi/hr)
2
= 160 mi/hr
(rounded off to 2 significant figures).
9. Since its velocity vector is constant, it has zero
acceleration, and the sum of the force vectors acting
on it must be zero. There are three forces acting on
the plane: thrust, lift, and gravity. We are given the
first two, and if we can find the third we can infer its
mass. The sum of the y components of the forces is
zero, so
0= F
thrust,y
+F
lift,y
+F
W,y
= |F
thrust
| sin
.
+ |F
lift
| cos
.
— mg .
The mass is
m= (|F
thrust
| sin
.
+ |F
lift
| cos
.
) / g
= 6.9x10
4
kg
10. (a) Since the wagon has no acceleration, the total
forces in both the x and y directions must be zero.
There are three forces acting on the wagon: F
T
, F
W
,
and the normal force from the ground, F
N
. If we pick a
coordinate system with x being horizontal and y
vertical, then the angles of these forces measured
counterclockwise from the x axis are 90
°
-
.
, 270
°
, and
90
°
+
.
, respectively. We have
F
x,total
= F
T
cos(90
°
-
.
) + F
W
cos(270
°
) + F
N
cos(90
°
+
.
)
F
y,total
= F
T
sin(90
°
-
.
) + F
W
sin(270
°
) + F
N
sin(90
°
+
.
) ,
which simplifies to
0 = F
T
sin
.
– F
N
sin
.
0 = F
T
cos
.
– F
W
+ F
N
cos
.
.
The normal force is a quantity that we are not given
and do not with to find, so we should choose it to
eliminate. Solving the first equation for F
N
=(sin
./
sin
.
)F
T
, we eliminate F
N
from the second equation,
0 = F
T
cos
.
– F
W
+ F
T
sin
.
cos
./
sin
.
and solve for F
T
, finding
F
T
=F
W
cos
.
+sin
.
cos
.
/sin
.
.
Multiplying both the top and the bottom of the fraction
by sin
.
, and using the trig identity for sin(
.
+
.
) gives
the desired result,
F
T
=sin
.
sin
.
+
.
F
W
(b) The case of
.
=0, i.e. pulling straight up on the