Download Chapter 2 Review Solutions SPH4C

CHAPTER 2 SUMMARY
MAKE A SUMMARY
(Page 109)
Concept maps will vary. Students should list appropriate concepts under each heading and also include additional
branches. For example, branching off of “simple machines” can be a new heading entitled “compound machines.”
Students should also attempt to make connections between the lists under each heading. For example, under “functions”
students should list the function of each simple and compound machine they have listed. They can then connect each
function to other lists, including domestic applications, industrial applications, mechanical advantage, and efficiency.
Students should write equations under appropriate headings, such as torques on levers, mechanical advantage, and
percent efficiency. The heading torques on levers should include the equations involved and the three classes of levers.
The heading mechanical advantage can also include the different equations and how they are used. Percent efficiency
should include the equation and connections to the applications as well as reasons for low efficiency (friction).
There are many applications students might come up with for the last three headings including the human arm or leg,
household tools, appliances, and any other industrial machine. It is important that they are linked to the previous
concepts and to the simple machines involved.
CHAPTER 2 SELF-QUIZ
(Page 110)
True/False
1.
F; You are using a third-class lever.
2.
F; If there is static equilibrium the effort force exceeds the load force.
3.
T
4.
F; Torque is measured in newton-metres but IMA has no units.
5.
T
6.
F; Scissors are an example of a first-class lever.
7.
T
8.
T
Multiple Choice
9.
(c)
10. (b)
11. (a)
12. (a)
13. (d)
14. (c)
15. (d)
CHAPTER 2 REVIEW
(Pages 111–113)
Understanding Concepts
1. (a) The main function is to change the magnitude of the force. It is a first-class lever.
(b) The main function is to change the magnitude of the force. It is a first-class lever.
(c) The main function is to change the direction of the force. It is a first-class lever.
(d) The main function is to change the distance. It is a first-class lever.
2. (a) Two examples of a third-class lever in the human body are the forearm and the jaw.
(b) One example of a second-class lever in the human body is the foot.
3. (a) The lab jack is a first-class lever.
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(b) The locking pliers are first and second-class levers.
(c) The Hamster dragster is made up of a wheel and axle and a pulley.
4. The angle that yields torque is 90º.
5. Large buses and trucks experience greater friction and require a higher IMA to turn the larger and heavier steering
wheel. Cars experience less friction and there is a smaller turning distance for the driver, which means the steering
wheel is smaller.
6. (a)
(b) FL = 6.8 × 102 N
dL = 1.2 m
dE = 1.6 m
FE = ?
FE =
FLdL
dE
(6.8 × 10 N) (1.2 m)
=
2
1.6 m
FE = 5.1× 102 N
The magnitude of the effort force required is 5.1 × 102 N.
7. (a) m = 95 kg
g = 9.8 N/kg
FL = ?
FL = Fg
= mg
= (95 kg)(9.8 N/kg)
FL = 9.3 × 102 N
The magnitude of the load force is 9.3 × 102 N.
(b) FL = 9.3 × 102 N
dL = 0.60 m
FE = 5.2 × 102 N
dE =?
dE =
FLdL
FE
(9.3 × 10 N) (0.60 m)
=
2
5.2 × 102 N
dE = 1.1 m
The distance from the effort force to the load is 1.1 m.
8. (a) The arm is a third-class lever.
(b) FE = 1.5 × 102 N
dL = 31.5 cm = 0.315 m
dE = 4.0 cm = 0.040 m
FL = ?
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FL =
FEdE
dL
(1.5 × 10 N) (0.040 m)
=
2
0.315 m
FL = 19 N
The magnitude of the load force is 19 N.
(c) FL = Fg = 19 N
g = 9.8 N/kg
m=?
m=
Fg
g
19 N
=
9.8 N/kg
m = 1.9 kg
The mass of the load is 1.9 kg.
9. (a)
The pull tab system is a first-class lever.
(b) The IMA will depend on the measurements of the diagram.
dE = 2.0 cm
dL = 0.5 cm
IMA = ?
IMA =
dE
dL
2.0 cm
0.5 cm
IMA = 4.0
=
The IMA is 4.0.
10. (a) This is a first-class lever.
(b) First, the load force must be calculated.
m = 4.0 kg
g = 9.8 N/kg
FL(load) = ?
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FL (load) = Fg
= mg
= (4.0 kg)(9.8 N/kg)
FL (load) = 39.2 N
Use this value to determine the load force of the forearm.
m = 1.8 kg
g = 9.8 N/kg
FL(forearm) = ?
FL (forearm) = Fg
= mg
= (1.8 kg)(9.8 N/kg)
FL (forearm) = 17.6 N
We may now calculate the effort force.
FL(load) = 39.2 N
dL(load) = 0.28 m
FL(forearm) = 17.6 N
dL(forearm) = 0.11 m
dE = 0.024 m
FE = ?
FE =
=
(FLdL )forearm + (FLdL )load
dE
(17.6 N)(0.11 m) + (39.2 N)(0.28 m)
0.024 m
FE = 5.4 × 10 N
2
The effort force required to hold the shot is 5.4 × 102 N.
(c) FL = 17.6 N + 39.2 N = 56.8 N
FE = 5.4 × 102 N
AMA = ?
AMA =
FL
FE
56.8 N
5.4 × 102 N
AMA = 0.10
=
The AMA of the arm is 0.10.
11. (a) FL = 5200 N
FE = 650 N
AMA = ?
AMA =
FL
FE
5200 N
650 N
AMA = 8.0
=
The AMA is 8.0.
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(b) dE = 3.6 m
dL = 0.40 m
IMA = ?
IMA =
dE
dL
3.6 m
0.40 m
IMA = 9.0
=
The IMA is 9.0.
AMA
× 100%
IMA
8.0
=
× 100%
9.0
% eff = 89%
(c) % eff =
The percent efficiency is 89%.
12. (a) Since the diameter is twice the radius the factor of 2 will cancel out in the ratio for IMA. Therefore the diameters
are used.
dE = 105 mm
dL = 35 mm
IMA = ?
IMA =
dE
dL
105 mm
35 mm
IMA = 3.0
=
The IMA is 3.0.
dE = 50 mm
dL = 65 mm
IMA = ?
IMA =
dE
dL
50 mm
65 mm
IMA = 0.77
=
The IMA is 0.77.
For the lowest IMA,
dE = 35 mm
dL = 105 mm
IMA = ?
IMA =
dE
dL
35 mm
105 mm
IMA = 0.33
=
The lowest IMA is 0.33.
(b) For every situation, a first-class lever is involved.
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(c) dE = 14 cm
IMA = 3.0
dL = ?
dE
IMA
14 cm
=
3.0
dL = 4.7 cm
dL =
The load force moved 4.7 cm.
dE = 14 cm
IMA = 0.77
dL = ?
dE
IMA
14 cm
=
0.77
dL = 18 cm
dL =
The load force moved 18 cm.
dE = 14 cm
IMA = 0.33
dL = ?
dE
IMA
14 cm
=
0.33
dL = 42 cm
dL =
The load force moved 42 cm.
Applying Inquiry Skills
13. Answers will vary. Torque is at a maximum when the crank is vertical and is at a minimum when the crank is
horizontal.
14. (a) The sticks will be pulled together no matter how much force is exerted by the two students holding the sticks
apart.
(b) The number of support strands on each stick is 9 or 10 (as shown), which provides an IMA to the student pulling
on the rope.
15. (a) The simple machines that make up this model include levers, pulleys, wheel and axles, and gears.
(b) You could make measurements of effort and load forces and distances, radius of pulleys, and the number of
teeth in the gears.
Making Connections
16. (a) The zipper was invented by Elias Howe in 1851 but marketed for the first time 44 years later by Whitcomb
Judson. A Canadian Gideon Sundback designed the modern zipper by 1913.
(b) The wedges are located inside the slide. The slide exerts a torque on the track by bending the track, opening up
the spaces between the teeth so that they can fit and lock together the hooks and hollows. This provides a
mechanical advantage because the teeth are rotated by a torque to fit the teeth together.
17. Answers will vary but should show proper use of safety equipment, follow proper workplace rules, use machines
safely and properly, know where safety devices are located, and be aware of fire escape routes.
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