Unformatted text preview: es. In these drawings W is the weight of the block and fk is the kinetic
frictional force. In each case the magnitude of the frictional force is the same. It is given by Equation 4.8 as
fk = µkFN, where µk is the coefficient of kinetic friction and FN is the magnitude of the
normal force. The coefficient of kinetic friction does not depend on the direction of the
motion. Furthermore, the magnitude of the normal force in each case is the component of
the pushing force that is perpendicular to the wall, or FN = P sin θ.
SOLUTION Using Equations 4.9b to describe the balance of forces that act on the block in
the y direction and referring to the freebody diagrams, we have
Upward motion ΣFy = P cos θ − W − f k = 0 Downward motion ΣFy = P cos θ − W + f k = 0 According to Equation 4.8, the magnitude of the kinetic frictional force is fk = µkFN, where
we have pointed out in the REASONING that the magnitude of the normal force is
FN = P sin θ. Substituting into the equations for ΣFy in the two cases, we obtain
Upward motion ΣFy = P cos θ − W − µk P sin θ = 0 Downward motion ΣFy = P cos θ − W + µ k P sin θ = 0 Solving each case for P, we find that
a. Upward motion P= W
39.0 N
=
= 52.6 N
cos θ − µk sin θ cos 30.0° − ( 0.250 ) sin 30.0° Chapter 4 Problems 205 b. Downward motion P= 67. W
39.0 N
=
= 39.4 N
cos θ + µk sin θ cos 30.0° + ( 0.250 ) sin 30.0° SSM REASONING When the bicycle is coasting straight down the hill, the forces that
act on it are the normal force FN exerted by the surface of the hill, the force of gravity mg,
and the force of air resistance R. When the bicycle climbs the hill, there is one additional
force; it is the applied force that is required for the bicyclist to climb the hill at constant
speed. We can use our knowledge of the motion of the bicycle down the hill to find R.
Once R is known, we can analyze the motion of the bicycle as it climbs the hill. SOLUTION The figure to the left below shows the freebody diagram for the forces
during the downhill motion. The hill is inclined at an angle θ above the horizontal. The
figure to the right shows these forces resolved into components parallel to and perpendicular
to the line of motion.
+y
+x
FN
F
N
R
R
mg sin θ
mg mg cos θ θ Since the bicyclist is traveling at a constant velocity, his acceleration is zero. Therefore,
according to Newton's second law, we have ∑ Fx = 0 and ∑ Fy = 0 . Taking the direction up
the hill as positive, we have ∑ Fx = R − mg sin θ = 0 , or
R = mg sin θ = (80.0 kg)(9.80 m / s 2 ) sin 15.0° = 203 N
When the bicyclist climbs the same hill at
constant speed, an applied force P must push the
system up the hill. Since the speed is the same,
the magnitude of the force of air resistance will
remain 203 N. However, the air resistance will
oppose the motion by pointing down the hill.
The figure at the right shows the resolved forces
that act on the system during the uphill motion. FN P mg sinθ
mg cosθ
R Using the same sign convention as above, we have ∑ Fx = P − mg sin θ − R = 0 , or P = R + mg sin θ = 203 N + 203 N = 406 N ____________________________________________________________________________________________ 206 FORCES AND NEWTON'S LAWS OF MOTION 68. REASONING Because the kite line is straight, the distance between the kite and the person
holding the line is L = 43 m, as shown in the right part of the drawing. In order to find the
height h of the kite, we need to know the angle θ that the kite line makes with the horizontal
(see the drawing). Once that is known, h = L sin θ will give the kite’s height relative to the
person. The key to finding the angle θ is the realization that the tension force T exerted on
the kite by the line is parallel to the line itself. Therefore, the tension T is directed at an
angle θ below the horizontal (see the freebody diagram below). To find the angle θ, it is
sufficient, then, to find one of the components of the tension force, either Tx or Ty, because
the magnitude T of the tension is known. With values for Tx or Ty and T, we can use either
the sine or cosine function to determine θ.
Both the wind’s force f and the tension T have horizontal and vertical components, while
the weight force W is purely vertical. Therefore, there are only two horizontal forces acting
on the kite (Tx, fx), but three vertical forces (Ty, fy, W), so it will be easier to calculate Tx, the
horizontal component of the tension. Because the kite is stationary, the horizontal
component of the tension must balance the horizontal component of the force exerted on the
kite by the wind:
Tx = fx (1) f +y
fy L
Tx 56°
fx +x h θ
W
T Ty
θ Freebody diagram
of the kite Height of the kite SOLUTION From the freebody diagram of the kite, we can see that the x components of
the tension and air resistance forces are Tx = T cos θ and fx = f cos 56°. Substituting these
expressions into Equation (1), we find that Tx = f x or T cos θ = f cos 56o or cos θ = f cos 56o
T o f cos 56o −1 (19 N ) cos 56
o
θ = cos = 48 = cos T
16...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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