Unformatted text preview: be found from analyzing the forces that are
perpendicular to the incline. Taking up to be positive, we have
∑ Fy = P sin θ + FN – mg cos θ = 0 or FN = mg cos θ – P sin θ Equation (1) then becomes
P cos θ + mg sin θ − µ s (mg cos θ – P sin θ ) = 0
Solving for the coefficient of static friction, we find that µs = P cos θ + mg sin θ (535 N) cos 20.0° + (225 kg)(9.80 m/s 2 ) sin 20.0°
=
= 0.665
mg cos θ – P sin θ (225 kg)(9.80 m/s 2 ) cos 20.0° – (535 N) sin 20.0° ____________________________________________________________________________________________ 242 FORCES AND NEWTON'S LAWS OF MOTION 119. REASONING The freebody diagram for the box is shown in the following drawing on the
left. On the right the same drawing is repeated, except that the pushing force P is resolved
into its horizontal and vertical components. P P sin θ FN θ FN P cos θ fk fk
mg mg Since the block is moving at a constant velocity, it has no acceleration, and Newton’s
second law indicates that the net vertical and net horizontal forces must separately be zero.
SOLUTION Taking upward and to the right as the positive directions, we write the zero net
vertical and horizontal forces as follows:
FN – mg – P sin θ = 0
144424443 P cos θ – f = 0
144 k 3
244 Vertical Horizontal From the equation for the horizontal forces, we have P cos θ = fk . But the kinetic frictional force is fk = µkFN . Furthermore, from the equation for the vertical forces, we have
FN = mg + P sin θ . With these substitutions, we obtain P cos θ = f k = µk FN = µk ( mg + P sin θ )
Solving for P gives P= µk mg
cos θ − µk sin θ The necessary pushing force becomes infinitely large when the denominator in this
expression is zero. Hence, we find that cos θ – µk sin θ = 0 , which can be rearranged to
show that
sin θ
1
1
= tanθ =
or θ = tan –1 = 68°
µk
cos θ 0.41 ____________________________________________________________________________________________ CHAPTER 5 DYNAMICS OF UNIFORM CIRCULAR MOTION
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
____________________________________________________________________________________________ 1. (c) The velocity of car A has a constant magnitude (speed) and direction. Since its velocity
is constant, car A does not have an acceleration. The velocity of car B is continually
changing direction during the turn. Therefore, even though car B has a constant speed, it has
an acceleration (known as a centripetal acceleration). 2. (d) The centripetal (or “centerseeking”) acceleration of the car is perpendicular to its
velocity and points toward the center of the circle that the road follows. 3. (b) The magnitude of the centripetal acceleration is equal to v2/r, where v is the speed of the
object and r is the radius of the circular path. Since the radius of the track is smaller at A
compared to B, the centripetal acceleration of the car at A has a greater magnitude. 4. (a) The magnitude ac of the centripetal acceleration is given by ac = v2/r. 5. (d) The acceleration (known as the centripetal acceleration) and the net force (known as the
centripetal force) have the same direction and point toward the center of the circular path. 6. (a) According to the discussion in Example 7 in Section 5.3, the maximum speed that the
cylinder can have is given by vmax = µs gr , where µs is the coefficient of static friction, g
is the acceleration due to gravity, and r is the radius of the path. 7. (d) The radius of path 1 is twice that of path 2. The tension in the cord is the centripetal
force. Since the centripetal force is inversely proportional to the radius r of the path, T1 must
be onehalf of T2. 8. (a) The centripetal force is given by Fc = mv /r. The centripetal forces for particles 1, 2 and 2 2 2 2 3 are, respectively, 4m0v0 /r0, 3m0v0 /r0, and 2m0v0 /r0.
9. (d) The centripetal force is directed along the radius and toward the center of the circular
path. The component FN sin θ of the normal force is directed along the radius and points
toward the center of the path.
2 10. (a) The magnitude of the centripetal force is given by Fc = mv /r. The two cars have the
same speed v and the radius r of the turn is the same. The cars also have the same mass m,
even though they have different weights due to the different accelerations due to gravity.
Therefore, the centripetal accelerations are the same. 244 DYNAMICS OF UNIFORM CIRCULAR MOTION 11. (e) The centripetal force acting on a satellite is provided by the gravitational force. The
2
magnitude of the gravitational force is inversely proportional to the radius squared (1/r ), so
if the radius is doubled, the gravitational force is one fourth as great; 1/22 = 1/4.
12. The orbital speed is v = 1.02 × 103 m/s.
13. (b) The magnitude of the centripetal force acting on the astronaut is equal to her apparent
2
weight. The centripetal force is given by Equation 5.3 as Fc = mv /r, which depends on the
square (v2) of the astronaut’s speed and inversely (1/r) on the radius of the ring. According
to Equation 5.1, r...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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