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Unformatted text preview: or P = mg − FN 2 . Since
the second reading on the scale is equal to FN2/g, the normal force the scale exerts on him is
FN2 = (Second reading)g. Thus we obtain the magnitude P of the force the man exerts on
the chinup bar:
P = mg − FN2 = mg − (Second reading) g = ( m − Second reading ) g ( ) = ( 92.6 kg − 75.1 kg ) 9.80 m/s 2 = 172 N 43. REASONING As shown in the freebody diagram below, three forces act on the car: the
static frictional force fs (directed up the hill), the normal force FN (directed perpendicular to
the road), and its weight mg. As it sits on the hill, the car has an acceleration of zero
2
(ax = ay = 0 m/s ). Therefore, the net force acting on the car in the x direction must be zero ( ΣFx = 0 ) and the net force in the y direction must be zero ( ΣFy = 0 ) . These two relations will allow us to find the normal force and the static frictional force.
SOLUTION
a. Applying Newton’s second
( ΣFy = 0) yields +y law to the y direction FN fs ΣFy = + FN − mg cos15° = 0 (4.2b) where the term −mg cos 15° is the y component of the
car’s weight (negative, because this component points
along the negative y axis). Solving for the magnitude FN
of the normal force, we obtain 15º
+x
mg
15º FN = mg cos15° = (1700 kg ) ( 9.80 m/s 2 ) cos15° = 1.6 × 104 N
b. Applying Newton’s second law to the x direction ( ΣFx = 0 ) gives
ΣFx = + mg sin15° − fs = 0 (4.2a) where the term mg sin 15° is the x component of the car’s weight. Solving this expression
for the static frictional force gives fs = mg sin15° = (1700 kg ) ( 9.80 m/s2 ) sin15° = 4.3 ×103 N 188 FORCES AND NEWTON'S LAWS OF MOTION 44. REASONING
a. Since the refrigerator does not move, the static frictional force must be equal in
magnitude, but opposite in direction, to the horizontal pushing force that the person exerts
on the refrigerator.
b. The magnitude of the maximum static frictional force is given by Equation 4.7 as
fsMAX = µs FN . This is also the largest possible force that the person can exert on the
refrigerator before it begins to move. Thus, the factors that determine this force magnitude
are the coefficient of static friction µs and the magnitude FN of the normal force (which is
equal to the weight of the refrigerator in this case).
SOLUTION
a. Since the refrigerator does not move, it is in equilibrium, and the magnitude of the static
frictional force must be equal to the magnitude of the horizontal pushing force. Thus, the
magnitude of the static frictional force is 267 N . The direction of this force must be
opposite to that of the pushing force, so the static frictional force is in the +x direction .
b. The magnitude of the largest pushing force is given by Equation 4.7 as
fsMAX = µs FN = µs mg = (0.65)(57 kg)(9.80 m/s 2 ) = 360 N
____________________________________________________________________________________________ 45. SSM REASONING AND SOLUTION Four forces act on the sled. They are the pulling
force P, the force of kinetic friction f k , the weight mg of the sled, and the normal force FN
exerted on the sled by the surface on which it slides. The following figures show freebody
diagrams for the sled. In the diagram on the right, the forces have been resolved into their x
and y components.
y P sin θ y FN FN P
fk P cos θ fk θ
x mg x mg Chapter 4 Problems 189 Since the sled is pulled at constant velocity, its acceleration is zero, and Newton's second
law in the direction of motion is (with right chosen as the positive direction)
∑ Fx = P cos θ − f k = ma x = 0
From Equation 4.8, we know that f k = µ k FN , so that the above expression becomes
P cos θ − µk FN = 0 (1) In the vertical direction,
∑ Fy = P sin θ + FN − mg = ma y = 0 (2) Solving Equation (2) for the normal force, and substituting into Equation (1), we obtain P cos θ − µk ( mg − P sin θ ) = 0
Solving for µ k , the coefficient of kinetic friction, we find µk = P cos θ
(80.0 N) cos 30.0°
=
= 0.444
mg − P sin θ (20.0 kg) (9.80 m/s 2 ) − (80.0 N) sin 30.0° ____________________________________________________________________________________________ 46. REASONING In each of the three cases under consideration the kinetic frictional force is
given by fk = µkFN. However, the normal force FN varies from case to case. To determine
the normal force, we use Equation 4.6 (FN = mg + ma) and thereby take into account the
acceleration of the elevator. The normal force is greatest when the elevator accelerates
upward (a positive) and smallest when the elevator accelerates downward (a negative). SOLUTION
a. When the elevator is stationary, its acceleration is a = 0 m/s2. Using Equation 4.6, we
can express the kinetic frictional force as f k = µ k FN = µ k ( mg + ma ) = µ k m ( g + a ) ( )( ) = ( 0.360 )( 6.00 kg ) 9.80 m/s 2 + 0 m/s 2 = 21.2 N b. When the elevator accelerates upward, a = +1.20 m/s2. Then,
f k = µ k FN = µ k ( mg + ma ) = µk m ( g + a ) ( )( ) = ( 0.360 )( 6.00 kg ) 9.80 m/s 2 + 1.20 m/s 2 = 23.8 N 190 FORCES AND NEW...
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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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