Physics Solution Manual for 1100 and 2101

76 m 12 ms 2 150 kg 1 130 kg 180 kg 180 kg980

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Unformatted text preview: tude is 98 N, because it supports the 98-N weight hanging from the pulley system. Other forces also act on the arm at the shoulder joint, but we can ignore them. The reason is that their lines of action pass directly through the axis at the shoulder joint, which is the axis that we will use to determine torques. Thus, these forces have zero lever arms and contribute no torque. SOLUTION The magnitude of each individual torque is the magnitude of the force times the corresponding lever arm. The forces and their lever arms are as follows: Force Lever Arm 98 N 0.61 m 47 N 0.28 M (0.069 m) sin 29° Each torque is positive if it causes a counterclockwise rotation and negative if it causes a clockwise rotation about the axis. Thus, since the net torque must be zero, we see that ( 98 N ) ( 0.61 m ) − ( 47 N ) ( 0.28 m ) − M ( 0.069 m ) sin 29° = 0 Solving for M gives M= ( 98 N ) ( 0.61 m ) − ( 47 N ) ( 0.28 m ) = 1400 N ( 0.069 m ) sin 29° Chapter 9 Problems 489 75. SSM WWW REASONING AND SOLUTION The figure below shows the massless board and the forces that act on the board due to the person and the scales. 2.00 m person's feet person's head center of gravity of person x 425 N 315 N W a. Applying Newton's second law to the vertical direction gives 315 N + 425 N – W = 0 or W = 7.40 ×102 N, downward b. Let x be the position of the center of gravity relative to the scale at the person's head. Taking torques about an axis through the contact point between the scale at the person's head and the board gives (315 N)(2.00 m) – (7.40 × 102 N)x = 0 or x = 0.851 m 76. REASONING AND SOLUTION a. The net torque about an axis through the point of contact between the floor and her shoes is 2 Στ = − (5.00 × 10 N)(1.10 m)sin 30.0° + FN(cos 30.0°)(1.50 m) = 0 FN = 212 N b. Newton's second law applied in the horizontal direction gives Fh − FN = 0, so Fh = 212 N . c. Newton's second law in the vertical direction gives Fv − W = 0, so Fv = 5.00 × 102 N . 77. SSM REASONING When the modules pull together, they do so by means of forces that are internal. These pulling forces, therefore, do not create a net external torque, and the angular momentum of the system is conserved. In other words, it remains constant. We will use the conservation of angular momentum to obtain a relationship between the initial and 490 ROTATIONAL DYNAMICS final angular speeds. Then, we will use Equation 8.9 (v = rω) to relate the angular speeds ω0 and ωf to the tangential speeds v0 and vf. SOLUTION Let L be the initial length of the cable between the modules and ω0 be the initial angular speed. Relative to the center-of-mass axis, the initial momentum of inertia of the two-module system is I0 = 2M(L/2)2, according to Equation 9.6. After the modules pull together, the length of the cable is L/2, the final angular speed is ωf, and the momentum of inertia is If = 2M(L/4)2. The conservation of angular momentum indicates that I f ω f = I 0ω 0 123 123 Final angular momentum Initial angular momentum LM F I O = LM F I O L L 2 ω 2 ω M GJ P M GJ P M4PM2P N HKQ N HKQ 2 2 f 0 ω f = 4ω 0 According to Equation 8.9, ωf = vf/(L/4) and ω0 = v0/(L/2). With these substitutions, the result that ωf = 4ω0 becomes vf L/4 =4 Fv I or G/ 2 J L HK 0 b g v f = 2 v 0 = 2 17 m / s = 34 m / s 78. REASONING The drawing shows the rod in its initial (dashed lines) and final (solid lines) orientations. Since both friction and air resistance are absent, the total mechanical energy is conserved. In this case, the total mechanical energy is E= 1 Iω 2 224 13 4 Rotational kinetic energy + Center of mass mgh 1 24 43 Pivot Gravitational potential energy In this expression, m is the rod’s mass, I is its moment of inertia, ω is its angular speed, and h is the height of its center of mass above a reference level that we take to be the ground. Since the rod is uniform, its center of mass is located at the center of the rod. The energy-conservation principle will guide our solution. hf Center of mass h0 v0 Chapter 9 Problems 491 SOLUTION Conservation of the total mechanical energy dictates that 1 I ωf2 + mghf 2 144 244 3 = 1 2 I ω0 + mgh0 24 1442444 3 Final total mechanical energy (1) Initial total mechanical energy Since the rod is momentarily at rest in its final orientation, we know that ωf = 0 rad/s, so that Equation (1) becomes 2 mghf = Iω0 + mgh0 1 2 1 2 or 2 I ω0 = mg ( hf − h0 ) (2) We can relate the initial angular speed ω0 to the initial linear speed v0 by using v Equation 8.9: ω0 = 0 . With this substitution, the fact that hf − h0 = L (see the drawing), L 1 3 and the fact that I = mL2 (see Table 9.1 in the text), Equation (2) indicates that 1 2 2 I ω0 = mg ( hf − h0 ) Solving for v0 gives ( 11 mL2 23 or ( ) 2 v0 = mgL L or 12 v 60 = gL ) v0 = 6 gL = 6 9.80 m/s2 ( 0.80 m ) = 6.9 m/s 79. REASONING The drawing shows the forces acting on the board, which has a length L. Wall 2 exerts a normal force P2 on the lower end of the board. The maximum force of static friction that wall 2 can apply to the lower end of the board is µsP2 and is directed upward in the drawing. The weight W acts downw...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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