Physics Solution Manual for 1100 and 2101

53 890 n 310 n 1 s 1 053

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Unformatted text preview: re as shown in the free-body diagram. The third force acting on the bottle is W, its weight or the gravitational force exerted on it by Free-body diagram of the wine bottle the earth. We will apply Newton’s second law and analyze the vertical forces in this free-body diagram to determine the magnitude F of the forces F1 and F2, using the fact that the bottle is in equilibrium. SOLUTION The vertical components of the forces exerted by the surfaces are F1y = F1 sin 45.0° and F2y = F2 sin 45.0°. But the forces F1 and F2 have the same magnitude F, so the two vertical components become F1y = F2y = F sin 45.0°. Because the bottle is in equilibrium, the upward forces must balance the downward force: Chapter 4 Problems F1 y + F2 y = W or F sin 45.0o + F sin 45.0o = W 199 2 F sin 45.0o = W or Therefore, ( ) (1.40 kg ) 9.80 m/s2 W mg F= = = = 9.70 N 2 sin 45.0o 2sin 45.0o 2 sin 45.0o 60. REASONING The free-body diagram in the drawing at the right shows the forces that act on the clown (weight = W). In this drawing, note that P denotes the pulling force. Since the rope passes around three pulleys, forces of magnitude P are applied both to the clown’s hands and his feet. The normal force due to the floor is FN , and the maximum static frictional force is fsMAX. At the instant just before the clown’s feet move, the net vertical and net horizontal forces are zero, according to Newton’s second law, since there is no acceleration at this instant. FN P fsMAX P W SOLUTION According to Newton’s second law, with upward and to the right chosen as the positive directions, we have FN + P – W = 0 144 244 3 Vertical forces and f sMAX – P = 0 14243 Horizontal forces From the horizontal-force equation we find P = fsMAX. But fsMAX = µsFN . From the vertical-force equation, the normal force is FN = W – P . With these substitutions, it follows that P = f sMAX = µs FN = µ s (W – P ) Solving for P gives P= µ sW ( 0.53) (890 N ) = = 310 N 1 + µs 1 + 0.53 ____________________________________________________________________________________________ 61. REASONING Since the boxes are at rest, they are in equilibrium. According to Equation 4.9b, the net force in the vertical, or y, direction is zero, ΣFy = 0. There are two unknowns in this problem, the normal force that the table exerts on box 1 and the tension in the rope that connects boxes 2 and 3. To determine these unknowns we will apply the relation ΣFy = 0 twice, once to the boxes on the left of the pulley and once to the box on the right. 200 FORCES AND NEWTON'S LAWS OF MOTION SOLUTION There are four forces acting on the two boxes on the left. The boxes are in equilibrium, so that the net force must be zero. Choosing the +y direction as being the upward direction, we have that (1) − W1 − W + FN + T = 0 14442 2444 3 ΣFy where W1 and W2 are the magnitudes of the weights of the boxes, FN is the magnitude of the normal force that the table exerts on box 1, and T is the magnitude of the tension in the rope. We know the weights. To find the unknown tension, note that the box 3 is also in equilibrium, so that the net force acting on it must be zero. T 3 T FN W3 2 1 W2 − W3 + T = 0 so that T = W3 1 24 43 ΣFy Substituting this expression for T into Equation (1) and solving for the normal force gives W1 FN = W1 + W2 − W3 = 55 N + 35 N − 28 N = 62 N ____________________________________________________________________________________________ 62. REASONING Since the mountain climber is at rest, she is in equilibrium and the net force acting on her must be zero. Three forces comprise the net force, her weight, and the tension forces from the left and right sides of the rope. We will resolve the forces into components and set the sum of the x components and the sum of the y components separately equal to zero. In so doing we will obtain two equations containing the unknown quantities, the tension TL in the left side of the rope and the tension TR in the right side. These two equations will be solved simultaneously to give values for the two unknowns. SOLUTION Using W to denote the weight of the mountain climber and choosing right and upward to be the positive directions, we have the following free-body diagram for the climber: For the x components of the forces we have ΣFx = TR sin 80.0° − TL sin 65.0° = 0 +y TL For the y components of the forces we have ΣFy = TR cos 80.0° + TL cos 65.0° − W = 0 Solving the first of these equations for TR, we find that 65.0º 80.0º W TR +x Chapter 4 Problems TR = TL 201 sin 65.0° sin 80.0° Substituting this result into the second equation gives TL sin 65.0° cos 80.0° + TL cos 65.0° − W = 0 sin 80.0° or TL = 1.717 W Using this result in the expression for TR reveals that TR = TL sin 65.0° sin 65.0° = (1.717W ) = 1.580 W sin 80.0° sin 80.0° Since the weight of the climber is W = 535 N, we find that TL = 1.717 W = 1.717 ( 535 N ) = 919 N TR = 1.580 W = 1.580 ( 535 N ) = 845 N 63. SSM REASONING There are four forces that act on the chandelier; they are the forces of tension T in each of the three wires, and the...
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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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