d. c. b. a. e. ct1. x x x x d. c. b. a. ct2 x x x
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D.
C.
B.
A.
E.
CT1
X
X
X
X
D.
C.
B.
A.
CT2
X
X
X
D.
C.
B.
A.
CT3
X The engine or battery exerts a force on the object.
X If an object is moving there is a “force of motion.”.
X An object can’t exert a force on itself.
X
X
X
D.
C.
B.
A.
CT4
X The more active or energetic object exerts more force.
X The bigger or heavier object exerts more force.
X The student uses the effects of a force as an indication of the relative magnitudes of the forces in an interaction.
X
X
X
Newton’s Third Law!
D.
C.
B.
A.
CT5
X If an object moves, the third law pair forces must be unbalanced.
X The student identifies equal force pairs, but indicates that both forces act on the same object. (For the example of a book at rest on a table, the gravitational force down on the book and the normal force up by the table on the book are identified as an action-reaction pair.)
X The bigger or heavier object exerts more force.
Newton’s Third Law!
X
X
X
Ch6: Applications of Newton’s Laws
Equilibrium is defined as F = 0.
Remember Galileo!
Ropes, strings, cords, etc. are assumed massless unless otherwise stated.
Thus tensions are the same throughout the rope, string, cord, etc.
Pulleys are assumed massless and mounted on frictionless bearings unless otherwise stated.
Thus pulleys only change the direction of the force.
Applications of Newton’s Laws - Method
Draw picture of the problem.
Choose body (bodies) to isolate.
Draw Free Body Diagrams (FBDs) for isolated bodies.
Choose and label coordinate axes.
Apply Newton’s 2nd Law: Fx = max and Fy = may
Solve for F, m or a.
Work out kinematics.
Check solution is reasonable.
6-1 Frictional Forces
Force of Friction - Model Static Friction
fs,max = sn
0 fs fs,max
fs sn
(as needed to
maintain
equilibrium)
Kinetic Friction
fk = kn
(opposing motion)
A.
B.
C.
D.
E.
CT6
P6.8 (P.171)
You have P6.7
6-2 Strings and SpringsHooke’s Law for springs:
Fx = -kx
Fx is the force of the spring x is the extension (>0) or compression (<0) of the springk is the spring constant
P6.19 (p.171)
You have P6.20
Concept Question 7P6.19b:If the mass of the backpack is doubled, the answer to part a will
A. double.
B. halve.
C. stay the same.
6-3 Translational Equilibrium
P6.32 (p.173)Like your 6.37
except your a 0.
6-3 Translational Equilibrium
P6.35 (p.173)
6-4 Connected Objects
P6.39 (p.174)
F.
E.
D.
C.
B.
A.
6-5 Circular MotionCT8
The red and black triangles are similar triangles:
r/r = v/v
r/t (1/r) = v/t (1/v)
Take the limit of both sides as t 0.
v/r = acp/v
acp = v2/r (inward towards center of circle)
acp and v are always changing.
constant speed in a circle
ri = rf = r
vi = vf = v
CT9: When I whirl a nurf ball in a vertical circle attached to a rubber band, which statement is true? A. The rubber band will contract to provide an outward force on the nurf ball. B. The rubber band will contract because of the inward force on the nurf ball. C. The rubber band will not change in length. D. The rubber band will stretch because of the outward force on the nurf ball. E. The rubber band will stretch to provide an inward force on the nurf ball.
A.
B.
C.
D.
CT10
P6.51 (p.175)
Like your P6.50
except n = the
apparent weight
in P6.50
CT11: At the top of the path when I whirl a bucket of water over my head, the water in the bucket will
A. stay in the bucket because it is forced outward and stopped by the bottom of the bucket. B. stay in the bucket because gravity is temporarily suspended. C. stay in the bucket only if I whirl the bucket with enough speed so that the bottom of the bucket must supply an inward force. D. stay in the bucket at any speed that I whirl the bucket. E. not stay in the bucket.
P6.61 (p.175)
CT12: If the mass of the bucket and water are doubled, at the top of the path
A. both vmin and the tension in the string will double.B. both vmin and the tension in the string will remain the same. C. vmin will remain the same and the tension in the string will double. D. vmin will double and the tension in the string will remain the same