Download Ch.3 PROBLEMS Sections 3.1 and 3.2 3.1 (I) What net force must be

Selected solutions to Ch.3 Problems
Ch.3 PROBLEMS
Sections 3.1 and 3.2
3.1 (I) What net force must be exerted on a 7.0-kg sack of potatoes to give it an
acceleration of 3.5 m/sec2?
Solution. a =
f
m ⎞
kg • m
⎛
∴ f = ma = 7kg ⎜ 3.5 2 ⎟ = 24.5
= 24.5 Newtons
⎝
⎠
m
sec
sec 2
3.2. (I) A net force of 30 N is applied to an object, which is then observed to
accelerate at 0.25 m/sec2. Calculate the mass of the object.
kg • m
f
f
sec 2 = 120kg
Solution. a= ∴ m= =
m
a .25m/sec 2
30
3.3. (I) Calculate the acceleration of a rocket of mass 1.2 x 106 kg if the net force
on it is 2.0 x 106 N. Express in (a) meters per second-squared and (b) as a multiple
of the acceleration of gravity, g.
f
2 x10 6 N
m
a=
=
=1.67 2
Solution 3.3a.
6
m 1.2 x10 kg
sec
⎛
⎞
m ⎜ g ⎟
= 0.170g
Solution 3.3b. 1.67 2 i⎜
s 9.8 m ⎟
⎜⎝
⎟
s2 ⎠
3.4. (I) (a) What is the weight in newtons of a 50-kg person on earth? (b) The
acceleration due to gravity on the moon is 1.67 m/s2. What would this person's
weight be on the moon?
Solution 3.4a. Weight = mg = 50kg(
Solution 3.4b. Weight on the moon.
9.8m
kg m
) = 490
= 490 newtons
2
sec
sec 2
Weight = mg = mgmoon = 50kg(
1.67m
) = 83.5 newtons
sec 2
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3.5 (I) Calculate the mass in kilograms of a flea that has a weight of 5 x 10-6 N.
wt 5 x10 −6 kg m/sec 2
=
= 5.1x10 −7 kg
Solution 3.5. Wt = mg therefore, m =
2
g
9.8 m/sec
TABLE 3.1 I COEFFICIENTS OF KINETIC AND STATIC FRICTIONa
µk Kinetic Friction Rubber on dry concrete
0.7
Rubber on wet concrete
0.5
Wood on wood 0.3
Waxed wood on wet snow 0.1
Metal on wood 0.3
Steel on steel (dry) 0.3
Steel on steel (oiled) 0.03 Teflon on steel 0.04 Bone lubricated with synovial fluid 0.015 Shoes on wood 0.7
Shoes on ice 0.05 Ice on ice 0.03 Steel on ice 0.02 µS
Static Friction
1.0
0.7
0.5
0.14
0.5
0.6
0.05
0.04
0.016
0.9
0.1
0.1
0.4
"Values are approximate.
3.6 (I) How much kinetic friction will there be in a knee joint if the weight
supported by the joint is 500 newtons. (Use table 3.1 for coefficient of friction).
Solution 3.6. From table 3.1, the coefficient of friction, µ, of synovial fluid
is 0.015.
f= µFN = 0.015(500 N)= 7.5 newtons.
3.7. (I) If a steel spatula experiences a frictional force of 0.20 N when scraping
against a Teflon frying pan, what is the normal force between the spatula and the
pan? (Use table 3.1 for coefficient of friction).
f 0.2 N
= 5.0 newtons
Solution 3.7. friction = FN ⋅ µ Therefore, FN = =
µ 0.04
3.8. (I) A frictional force of 300 N is observed between two moving pieces of steel
when a normal force of 1000 N exists between them. Please refer to table 3.1 in
order to determine if the steel oiled or dry?
f
300 n
friction
=
F
µ
∴
µ
=
=
= 0.30
N
Solution 3.8
FN 1000 n
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Ch.3 PROBLEMS continued
Sections 3.1 and 3.2
3.9. (I) What is the tension in a strand of spider thread when a spider of mass 1.0 x
10-4 kg hangs motionlessly from it? Hint: The tension will have the same
magnitude as the spider’s weight.
Tension
Weight
Solution. 3.9. Tension = weight = mg= 1 x 10- 4 kg(9.8 m/sec2)
= 9.8 x 10-
4
newtons
3.11. (II) An orderly exerts a horizontal force of 50 N on a gurney with a patient on
it. The gurney and patient have a total mass of 90 kg. If the gurney and patient
accelerate at 0.35 m/sec2, what is the magnitude of the frictional force opposing the
motion?
friction
90 kg
50 newtons
Solution. 3.11.
f 50 n + f
m
m
a= =
= 0.35 2 ∴ f = 90 kg(.35 2 ) − 50n = −18.5 newtons
m
90kg
sec
sec
The negative sign indicates that the frictional force is backward. You should write
your answer as friction = 18.5 newtons backward.
Selected solutions to Ch.3 Problems
Ch.3 PROBLEMS continued Section 3.3
3.20. (II) Two movers push horizontally on a refrigerator. One pushes due north
with a force of 150 N and the other pushes due east with a force of 200 N. Using
trigonometric functions, determine the magnitude and direction of the resultant
force on the refrigerator.
Resultant force
150 N
200 N
Solution 3.20a.
We will use the Pythagorean
theorem to find the magnitude of the resultant force.
Resultant magnitude = 150 2 + 200 2 = 250 newtons
Solution 3.20b. The direction of the resultant can be found by using the trig.
rule for the tangent of an angle.
opposite 150
Tan θ =
=
= 0.75 ∴ θ = 36.9
adjacent 200
The direction of the force is 36.9˚ North of East. That is equivalent
to a bearing of 53.1 degrees.
3.22. (I) Using trigonometric functions, determine the
magnitudes of the two forces, F I and F 2, which
add up to the total force shown in Figure 3.25.
Fig. 3.25
Solution 3.22, F1
opposite
Sin35˚=
50N
opposite = Sin35˚i50N = 28.7Newtons
F1 = 28.7 Newtons
Solution 3.22, F2 Cos35˚=
adjacent
; Adjacent = Cos35˚i50N = 41.0N
50N
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3.23. (II) Figure 3.26 represents a sled viewed from above being pulled by two
children and experiencing a frictional force. Find the acceleration of the 9.0-kg
sled.
Solution 3.23
Baby step 1. Determine the forward and left components of F1
Forward component of F1: Cos30˚i10N = 8.66N
Left component of F1:
Sin30˚i10N = 5.0N
Baby step 2. Determine the forward and right components of F2.
Forward component of F2: Cos45˚i14.1N = 9.97N
Right component of F2: Sin45˚i14.1N = 9.97N
Baby step 3. Total up the forward components and total up the left-right
components. Assign forward as positive and backward as negative. Let
right be positive and left be negative. Don’t forget the frictional
force.
Newtons
Forward-----Backward
Right-----Left
8.66
-5
9.97
9.97
friction
-10
0
Total
8.63
4.97
Total forward components = 8.63 Newtons forward.
Total left-right components = 4.97 newtons to the right.
Selected solutions to Ch.3 Problems
Baby step 4. Draw a diagram and determine the magnitude and direction
of the resultant force due to the boys and friction.
Magnitude =
8.63 N
ø
8.632 + 4.97 2 = 9.96Newtons
4.97 N
Direction = Tan −1 ⎛⎜
4.97 ⎞
= 29.9˚
⎝ 8.63 ⎟⎠
Baby step 5. Use Newton’s second law to determine the acceleration of the
sled.
kg ⋅ m
9.96
( 27˚ to the right of forward )
2
F
∑
s
a=
=
m
9kg
m
a
=
1.11
Ans.
s 2 , 29.9˚ to the right of forward.
Section 3.4
Selected solutions to Ch.3 Problems
3.38. (II) Even when the head is held
erect, its center of gravity is not
directly over its major point of
support, the atlanto-occipital joint.
The splenius muscles in the back of
the neck must therefore exert a force
to keep the head erect. Calculate the
force they must exert, using the
information in Figure 3.36, if the /
mass of the head is 5.0 kg.
center of gravity
5 cm
FM
2.5 cm
weight = mg =
5kgi9.8
m
= 49Newtons
s2
The weight of an object can be viewed as acting in a straight line from the center of
gravity of the object, toward the center of the earth.
FM is the unknown force produced by the splenius muscle. The mass of the head
‘m’= 5 kg. We can solve this problem by using the concept that when a system is
in rotational equilibrium, the clockwise and counterclockwise torques are equal.
Solution 3.38.
Clockwise torque = Counterclockwise torque
49Ni2.5cm = FM i5cm
Ans. FM =
49Ni2.5cm
= 24.5N No wonder my neck aches.
5cm
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3.39. (II) A man stands on his toes by
exerting an upward force through the
Achilles tendon, as in Figure 3.37.
Calculate the force in the Achilles
tendon if he stands on one foot and
has a mass of 80 kg. (Yes, this is
another torque problem).
Solution 3.39.
Note: that 80 kg is his mass, not his
weight. First determine his weight.
weight = mg = 80kgi9.8
m
= 784Newtons
s2
Clockwise torque = Counterclockwise torque
FA i4.0cm = 784Ni12.0cm
Ans. FA =
784Ni12.0cm
= 2, 350 newtons
4.0cm
3.41. (II) What force must the woman in Figure 3.38 exert on the floor with her
hands in order to do a pushup?
m
= 490newtons
s2
Clockwise torque = Counterclockwise torque
490Ni0.90m = Freaction i1.50m
Ans. Freaction =
Solution 3.41
wt = mg = 50kgi9.8
490Ni0.90m
= 294newtons
1.50m
Selected solutions to Ch.3 Problems
Section 3.5
3.45. (II) What acceleration is experienced by materials 10 cm from the center of
rotation of a centrifuge that spins at 4000 revolutions per minute? Express your
answer in m/s2.
Solution 3.45.
Baby step 1.Convert 10 cm to meters.
1m
10cm
= 0.10m
100cm
Baby step 2. Objects moving in a circular path, undergo centripetal
acceleration. Write down the equation for centripetal acceleration.
v2
ac =
r
Note that we do not know the tangential velocity, v. We have been given the
angular velocity in revolutions per minute.
Baby step 3. Determine the tangential velocity in meters per second. Use
the fact that distance traveled in one revolution is 2πr, (the circumference of a
circle).
r e v 2π i0.10m 1min
m
v = 4000
i
i
= 41.9
min
rev
60 sec =
s
Baby step 4.
2
m⎞
⎛
41.9 ⎟
2
⎜
⎝
v
m
m
s⎠
a
=
=
=
17,
546
=
17,
500
Ans. c
r
0.10m
s2
s2
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3.46. (I) (a) How much sideways force must the wheels of a 950-kg car exert to
cause it to round a corner of radius 200 m at a speed of 35 m/sec? (b) What
acceleration does the driver experience? Express your answer in meters per
second-squared and as a multiple of g.
Solution 3.46b.
Baby step 1. Let’s determine the centripetal acceleration first.
⎛ m⎞
⎜⎝ 35 ⎟⎠
v
m
s
ac =
=
= 6.13 2
r
200m
s
2
Baby step 2. Convert from m/s2 to number of ‘g’s.
m 1g
6.13
i
= 0.625g
s 2 9.8 m
s2
Solution 3.46a.
Newton’s 2nd law tells us that ∑ F = ma .
∑ Fcentripetal = miac = 950kgi6.13
m
= 5, 820Newtons
s2
3.47. (I) (a) Calculate the acceleration of a 0.55-m-diameter car tire while the car
is traveling at a constant speed of 25 m/sec. (b) Compare this with the acceleration
experienced by a jet car tire of diameter 1.0 m when setting a land speed record of
310 m/sec.
2
⎛ m⎞
25 ⎟
v 2 ⎜⎝
m
s⎠
= 2, 270 2
Solution 3.47a. ac = =
r
0.275m
s
Solution 3.47b.
m 2
v2 ( 310 sec )
m
ac = =
=192, 000 2
r
.5 m
sec
m
acceleration of jet tire
s 2 = 84.6
=
m
acceleration of car tire
2270 2
s
192, 000
Selected solutions to Ch.3 Problems
3.48 (II) What centripetal acceleration (in meters per second squared) is
experienced by the passengers in a jet airplane making a level turn of radius 1. 0
km at a speed of 400 km/hr?
Solution 3.48.
Baby step 1.Convert the speed to m/s.
km 1000m 1hr
m
400
i
i
= 111
hr 1km 3600s
s
Baby step 2. Convert 1.0 km to meters.
1kmi
1000m
= 1000m
1km
Baby step 3. Write down the centripetal acceleration equation and do the
math.
m 2
v2 (111 sec )
m
=
=12.3 2
Ans. a c =
r
1000 m
sec
3.49. (II) What centripetal force is exerted by the rope on a 1.2-kg tether ball
swung in a 2.0-m-diameter circle at 45 revolutions per minute?
Solution 3.49.
v2
Baby step 1. ac =
r We don’t have the tangential velocity. Remember
that every revolution is once around a circle. The circumference of the circle
is 2π ir . Let’s calculate v. Note that the radius is 1.0 meters.
Baby step 2. We are ready to use the centripetal acceleration equation.
45
rev 2π i1.0m 1min
m
i
= 4.71
1min 1rev 60s
s
2
m⎞
⎛
4.71 ⎟
2
⎜
⎝
v
m
s⎠
ac =
=
= 22.2 2
r
1.0m
s
Baby step 3. Determine the centripetal force.
Ans.
∑F
centripetal
= miac = 1.2kgi22.2
m
= 26.6newtons
s2
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3.50. (II) At how many revolutions per minute must an astronaut-training
centrifuge rotate to produce an acceleration of 10g if the radius of rotation is 25 m?
Solution 3.50.
Baby step 1. Convert 10g to m/s/s.
m
s 2 = 98 m
g
s2
9.8
10g
Baby step 2.Determine the tangential velocity.
v2
ac =
therefore v= ac ir
r
v = 98
Baby step 3.Convert tangential velocity to angular velocity in revolutions
per minute.
m
m
i25m
=
49.5
s2
s
Ans. ω = 49.5
m 1Re v 60s
rev
i
i
= 18.9
s 2π i25m 1min
min