Physics Solution Manual for 1100 and 2101

Ssm reasoning and solution using equation 101 we

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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...
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