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as shown in the freebody diagram. The third force acting on the
bottle is W, its weight or the gravitational force exerted on it by Freebody diagram of
the wine bottle
the earth. We will apply Newton’s second law and analyze the
vertical forces in this freebody diagram to determine the
magnitude F of the forces F1 and F2, using the fact that the bottle is in equilibrium.
SOLUTION The vertical components of the forces exerted by the surfaces are
F1y = F1 sin 45.0° and F2y = F2 sin 45.0°. But the forces F1 and F2 have the same magnitude
F, so the two vertical components become F1y = F2y = F sin 45.0°. Because the bottle is in
equilibrium, the upward forces must balance the downward force: Chapter 4 Problems F1 y + F2 y = W or F sin 45.0o + F sin 45.0o = W 199 2 F sin 45.0o = W or Therefore, ( ) (1.40 kg ) 9.80 m/s2
W
mg
F=
=
=
= 9.70 N
2 sin 45.0o 2sin 45.0o
2 sin 45.0o 60. REASONING The freebody diagram in the drawing at the right
shows the forces that act on the clown (weight = W). In this
drawing, note that P denotes the pulling force. Since the rope
passes around three pulleys, forces of magnitude P are applied
both to the clown’s hands and his feet. The normal force due to
the floor is FN , and the maximum static frictional force is fsMAX.
At the instant just before the clown’s feet move, the net vertical
and net horizontal forces are zero, according to Newton’s second
law, since there is no acceleration at this instant. FN P
fsMAX P W SOLUTION According to Newton’s second law, with upward and to the right chosen as
the positive directions, we have
FN + P – W = 0
144
244
3
Vertical forces and f sMAX – P = 0
14243
Horizontal forces From the horizontalforce equation we find P = fsMAX. But fsMAX = µsFN . From the
verticalforce equation, the normal force is FN = W – P . With these substitutions, it follows
that P = f sMAX = µs FN = µ s (W – P )
Solving for P gives
P= µ sW ( 0.53) (890 N )
=
= 310 N
1 + µs
1 + 0.53 ____________________________________________________________________________________________ 61. REASONING Since the boxes are at rest, they are in equilibrium. According to Equation
4.9b, the net force in the vertical, or y, direction is zero, ΣFy = 0. There are two unknowns
in this problem, the normal force that the table exerts on box 1 and the tension in the rope
that connects boxes 2 and 3. To determine these unknowns we will apply the relation
ΣFy = 0 twice, once to the boxes on the left of the pulley and once to the box on the right. 200 FORCES AND NEWTON'S LAWS OF MOTION SOLUTION There are four forces acting on the two boxes on
the left. The boxes are in equilibrium, so that the net force must
be zero. Choosing the +y direction as being the upward direction,
we have that
(1)
− W1 − W + FN + T = 0
14442
2444
3
ΣFy
where W1 and W2 are the magnitudes of the weights of the boxes,
FN is the magnitude of the normal force that the table exerts on
box 1, and T is the magnitude of the tension in the rope. We
know the weights. To find the unknown tension, note that the box
3 is also in equilibrium, so that the net force acting on it must be
zero. T
3
T
FN W3 2
1 W2 − W3 + T = 0
so that
T = W3
1 24
43
ΣFy
Substituting this expression for T into Equation (1) and solving
for the normal force gives W1 FN = W1 + W2 − W3 = 55 N + 35 N − 28 N = 62 N
____________________________________________________________________________________________ 62. REASONING Since the mountain climber is at rest, she is in equilibrium and the net force
acting on her must be zero. Three forces comprise the net force, her weight, and the tension
forces from the left and right sides of the rope. We will resolve the forces into components
and set the sum of the x components and the sum of the y components separately equal to
zero. In so doing we will obtain two equations containing the unknown quantities, the
tension TL in the left side of the rope and the tension TR in the right side. These two
equations will be solved simultaneously to give values for the two unknowns.
SOLUTION Using W to denote the weight of the mountain climber and choosing right and
upward to be the positive directions, we have the following freebody diagram for the
climber:
For the x components of the forces we have
ΣFx = TR sin 80.0° − TL sin 65.0° = 0 +y TL For the y components of the forces we have
ΣFy = TR cos 80.0° + TL cos 65.0° − W = 0 Solving the first of these equations for TR, we find that 65.0º 80.0º W TR
+x Chapter 4 Problems TR = TL 201 sin 65.0°
sin 80.0° Substituting this result into the second equation gives
TL sin 65.0°
cos 80.0° + TL cos 65.0° − W = 0
sin 80.0° or TL = 1.717 W Using this result in the expression for TR reveals that
TR = TL sin 65.0°
sin 65.0°
= (1.717W )
= 1.580 W
sin 80.0°
sin 80.0° Since the weight of the climber is W = 535 N, we find that
TL = 1.717 W = 1.717 ( 535 N ) = 919 N
TR = 1.580 W = 1.580 ( 535 N ) = 845 N 63. SSM REASONING There are four forces that act on the chandelier; they are the forces
of tension T in each of the three wires, and the...
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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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