Unformatted text preview: ard
at the board's center. Wall 1 applies a normal force P1
to the upper end of the board. We take upward and to
the right as our positive directions. Wall 1 P1 Wall 2 1 µ s P2
W θ P2 SOLUTION Then, since the horizontal forces balance to zero, we have P1 – P2 = 0 (1) µsP2 – W = 0 (2) The vertical forces also balance to zero: 492 ROTATIONAL DYNAMICS Using an axis through the lower end of the board, we now balance the torques to zero: L
W ( cos θ ) − P L ( sin θ ) = 0
1
2 (3) Rearranging Equation (3) gives
tan θ = W
2P
1 (4) But W = µsP2 according to Equation (2), and P2 = P1 according to Equation (1). Therefore,
W = µsP1, which can be substituted in Equation (4) to show that
tan θ =
or µs P
1
2P
1 = µs
2 = 0.98
2 θ = tan–1(0.49) = 26° From the drawing at the right,
cos θ = d
L L Therefore, the longest board that can be propped between
the two walls is
L= d
1.5 m
=
=
cos θ cos 26° 1.7 m θ d Chapter 9 Problems 493 80. REASONING The drawing shows the drum, pulley, and the crate, as well as the tensions
in the cord
r2
T1 Pulley
m2 = 130 kg T2 T2 T1
r1 Crate
m3 = 180 kg Drum
m1 = 150 kg R r1 = 0.76 m Let T1 represent the magnitude of the tension in the cord between the drum and the pulley.
Then, the net torque exerted on the drum must be, according to Equation 9.7, Στ = I1α1, where I1 is the moment of inertia of the drum, and α1 is its angular acceleration. If we
assume that the cable does not slip, then Equation 9.7 can be written as c hF I
GJ
HK a
− T1 r1 + τ = m1 r12
14 4
2 3 123 r1
2
∑τ
I1 1 3 (1) α1 where τ is the counterclockwise torque provided by the motor, and a is the acceleration of
the cord (a = 1.2 m/s2). This equation cannot be solved for τ directly, because the tension T1
is not known.
We next apply Newton’s second law for rotational motion to the pulley in the drawing: c hF I
GJ
HK a
1
+ T1 r − T2 r2 = 2 m2 r22
14 2244
4
3 1 24 r2
43
13
2
I2
∑τ (2) α2 where T2 is the magnitude of the tension in the cord between the pulley and the crate, and I2
is the moment of inertial of the pulley.
Finally, Newton’s second law for translational motion (ΣFy = m a) is applied to the crate,
yielding 494 ROTATIONAL DYNAMICS + T2 − m3 g = m3 a
14 24 3
4
4
∑ Fy (3) SOLUTION Solving Equation (1) for T1 and substituting the result into Equation (2), then
solving Equation (2) for T2 and substituting the result into Equation (3), results in the
following value for the torque τ = r1 a ( m1 + 1 m2 + m3 ) + m3 g 2 ( ) = (0.76 m) (1.2 m/s 2 ) 150 kg + 1 130 kg + 180 kg + (180 kg)(9.80 m/s 2 ) = 1700 N ⋅ m
2 CHAPTER 10 SIMPLE HARMONIC MOTION AND ELASTICITY
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. 0.12 m
2. (c) The restoring force is given by Equation 10.2 as F = −kx, where k is the spring constant
(positive). The graph of this equation is a straight line and indicates that the restoring force
has a direction that is always opposite to the direction of the displacement. Thus, when x is
positive, F is negative, and vice versa. 1m
. Greater
2π k
values for the mass m and smaller values for the spring constant k lead to greater values for
the period. 3. (b) According to Equations 10.4 and 10.11, the period T is given by T = 4. (d) The maximum speed in simple harmonic motion is given by Equation 10.8 ( vmax = Aω ) .
Thus, increases in both the amplitude A and the angular frequencyω lead to an increase in the
maximum speed.
5. (e) The maximum acceleration in simple harmonic motion is given by Equation 10.10
( amax = Aω 2 ) . A decrease in the amplitude A decreases the maximum acceleration, but this
decrease is more than offset by the increase in the angular frequency ω, which is squared in
Equation 10.10.
6. 1.38 m/s
7. (b) The velocity has a maximum magnitude at point A, where the object passes through the
position where the spring is unstrained. The acceleration at point A is zero, because the
spring is unstrained there and is not applying a force to the object. The velocity is zero at
point B, where the object comes to a momentary halt and reverses the direction of its travel.
The magnitude of the acceleration at point B is a maximum, because the spring is maximally
stretched there and, therefore, applies a force of maximum magnitude to the object.
8. 0.061 m
9. +7.61 m/s2
10. 0.050 m 496 SIMPLE HARMONIC MOTION AND ELASTICITY 11. (c) The principle of conservation of mechanical energy applies in the absence of
nonconservative forces, so that KE + PE = constant. Thus, the total energy is the same at all
points of the oscillation cycle. At the equilibrium position, where the spring is unstrained,
the potential energy is zero, and the kinetic energy is KEmax; thus, the total energy is KEmax.
At the extreme ends of the cycle, where the object comes to a momentary halt, the kinetic
energy is zero, and the potential energy is PEmax; thus, the total energy is also PEmax. Since
both KEmax and PEmax equal the total energy, it must be true that KEmax = PEmax.
12. (e) In simple harmonic motion the speed and, henc...
View
Full Document
 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

Click to edit the document details