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Unformatted text preview: 0.414 L or – 2.414 L The negative solution is discarded because the third particle lies on the +x axis between m
and 2m. Thus, D = 0.414 L .
____________________________________________________________________________________________ 38. REASONING In each case the object is in equilibrium. According to Equation 4.9b,
ΣFy = 0, the net force acting in the y (vertical) direction must be zero. The net force is
composed of the weight of the object(s) and the normal force exerted on them.
SOLUTION
a. There are three vertical forces acting on the crate: an upward normal force +FN that the
floor exerts, the weight –m1g of the crate, and the weight –m2g of the person standing on the
crate. Since the weights act downward, they are assigned negative numbers. Setting the sum
of these forces equal to zero gives FN + (−m1g ) + (−m2 g ) = 0
14444244444
4
3
ΣFy
The magnitude of the normal force is
FN = m1g + m2g = (35 kg + 65 kg)(9.80 m/s2) = 980 N
b. There are only two vertical forces acting on the person: an upward normal force +FN that
the crate exerts and the weight –m2g of the person. Setting the sum of these forces equal to
zero gives
FN + (−m2 g ) = 0
144
244
3
ΣFy The magnitude of the normal force is
FN = m2g = (65 kg)(9.80 m/s2) = 640 N
____________________________________________________________________________________________ Chapter 4 Problems 39. 185 SSM REASONING The book is kept from falling as long as the total static frictional
force balances the weight of the book. The forces that act on the book are shown in the
following freebody diagram, where P is the pressing force applied by each hand.
fsMAX fsMAX P P W In this diagram, note that there are two pressing forces, one from each hand. Each hand also
applies a static frictional force, and, therefore, two static frictional forces are shown. The
maximum static frictional force is related in the usual way to a normal force FN, but in this
problem the normal force is provided by the pressing force, so that FN = P.
SOLUTION Since the frictional forces balance the weight, we have
2 f sMAX = 2 ( µ s FN ) = 2 ( µ s P ) = W
Solving for P, we find that P= W
31 N
=
= 39 N
2µ s 2 ( 0.40 ) ____________________________________________________________________________________________ 40. REASONING AND SOLUTION
a. The apparent weight of the person is given by Equation 4.6 as
FN = mg + ma
= (95.0 kg)(9.80 m/s2 + 1.80 m/s2) = 1.10 × 103 N
b. FN = (95.0 kg)(9.80 m/s2) = 931 N c. FN = (95.0 kg)(9.80 m/s2 – 1.30 m/s2) = 808 N ____________________________________________________________________________________________ 186 FORCES AND NEWTON'S LAWS OF MOTION 41. REASONING As the drawing shows, the normal
FN
force FN points perpendicular to the hill, while the weight
W points vertically down. Since the car does not leave the
surface of the hill, there is no acceleration in this
perpendicular direction. Therefore, the magnitude of the
θ
perpendicular component of the weight W cos θ must equal
W cos θ
θ
the magnitude of the normal force, FN = W cos θ . Thus, the
W
magnitude of the normal force is less than the magnitude of
the weight. As the hill becomes steeper, θ increases, and cos θ decreases. Consequently, the
normal force decreases as the hill becomes steeper. The magnitude of the normal force does
not depend on whether the car is traveling up or down the hill.
SOLUTION
a. From the REASONING, we have that FN = W cos θ . The ratio of the magnitude of the
normal force to the magnitude W of the weight is
FN
W = W cos θ
= cos 15° = 0.97
W b. When the angle is 35°, the ratio is
W cos θ
= cos 35° = 0.82
W
W
______________________________________________________________________________
FN = 42. REASONING The reading on the bathroom scale is
Freebody diagrams of the man
proportional to the normal force it exerts on the man.
When he simply stands on the scale, his acceleration
FN1
is zero, so the normal force pushing up on him
balances the downward pull of gravity: FN1 = mg (see
P FN2
the freebody diagram). Thus, the first reading on the
scale is his actual mass m, the ratio of the normal
force the scale exerts on him to the acceleration due to
gravity: First reading = m = FN1/g = 92.6 kg. With the
mg
mg
chinup bar helping to support him, the normal force
exerted on him by the scale decreases, and the second Standing
Pulling down
reading is the ratio of the reduced normal force FN2 to
the acceleration due to gravity: Second reading = FN2/g = 75.1 kg. Lastly, we note that the
magnitude P of the force the chinup bar exerts on the man is exactly equal to the magnitude
P of the force that the man exerts on the chinup bar. This prediction is due to Newton’s
third law. Therefore, it is a value for P that we seek.
SOLUTION When the man is pulling down on the chinup bar, there are two upward forces
acting on him (see the second part of the drawing), and he is still at rest, so the sum of these Chapter 4 Problems 187 two forces balances the downward pull of gravity: FN2 + P = mg,...
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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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