Physics Solution Manual for 1100 and 2101

# It remains now to determine the magnitude of the

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Unformatted text preview: ng FB = FBC – FAB = 8.497 × 10 –5 N – 5.007 × 10 –5 N = 3.49 × 10 –5 N, right c. Both particles A and B attract particle C to the left, the net force being FC = FAC + FBC = 6.629 × 10 –6 N + 8.497 × 10 –5 N = 9.16 × 10 –5 N, left ____________________________________________________________________________________________ y 107. SSM REASONING AND SOLUTION The system is shown in the drawing. We will let m1 = 21.0 kg , and m2 = 45.0 kg . Then, m1 will move upward, and m2 will move downward. There are two forces that act on each object; they are the tension T in the cord and the weight mg of the object. The forces are shown in the free-body diagrams at the far right. y T x m1 m2 m1g We will take up as the positive direction. If the acceleration of m1 is a, then the acceleration of m2 must be –a. From Newton's second law, we have for m1 T m2g 232 FORCES AND NEWTON'S LAWS OF MOTION ∑ Fy = T − m1g = m1a (1) ∑ Fy = T − m2 g = – m2 a (2) and for m2 a. Eliminating T between these two equations, we obtain a= m2 – m1 45.0 kg – 21.0 kg 2 2 g = (9.80 m/s ) = 3.56 m/s m2 + m1 45.0 kg + 21.0 kg b. Eliminating a between Equations (1) and (2), we find T= 2m1m2 2(21.0 kg)(45.0 kg) 2 g= (9.80 m/s ) = 281 N m1 + m2 21.0 kg + 45.0 kg ____________________________________________________________________________________________ 108. REASONING Static friction determines the magnitude of the applied force at which either the upper or lower block begins to slide. For the upper block the static frictional force is applied only by the lower block. For the lower block, however, separate static frictional forces are applied by the upper block and by the horizontal surface. The maximum magnitude of any of the individual frictional forces is given by Equation 4.7 as the coefficient of static friction times the magnitude of the normal force. SOLUTION We begin by drawing the free-body diagram for the lower block. MAX fs, from A FApplied A MAX B fs, from surface Free-body diagram for lower block B This diagram shows that three horizontal forces act on the lower block, the applied force, and the two maximum static frictional forces, one from the upper block and one from the horizontal surface. At the instant that the lower block just begins to slide, the blocks are in equilibrium and the applied force is balanced by the two frictional forces, with the result that MAX MAX FApplied = fs, from A + fs, from surface (1) According to Equation 4.7, the magnitude of the maximum frictional force from the surface is MAX (2) fs, from surface = µs FN = µs 2mg Chapter 4 Problems 233 Here, we have recognized that the normal force FN from the horizontal surface must balance the weight 2mg of both blocks. It remains now to determine the magnitude of the maximum frictional MAX force fs, from A from the upper block. 47.0 N MAX fs, from B A To this end, we draw the free-body B diagram for the upper block at the Free-body diagram for instant that it just begins to slip due to upper block A the 47.0-N applied force. At this instant the block is in equilibrium, so that the frictional force from the lower block B balances the 47.0-N force. Thus, MAX fs, from B = 47.0 N , and according to Equation 4.7, we have MAX fs, from B = µs FN = µs mg = 47.0 N Here, we have recognized that the normal force FN from the lower block must balance the weight mg of only the upper block. This result tells us that µsmg = 47.0 N. To determine MAX fs, from A we invoke Newtons third law to conclude that the magnitudes of the frictional forces at the A-B interface are equal, since they are action-reaction forces. MAX fs, from A = µs mg . Substituting this result and Equation (2) into Equation (1) gives Thus, MAX MAX FApplied = fs, from A + fs, from surface = µs mg + µs 2mg = 3 ( 47.0 N ) = 141 N ______________________________________________________________________________ 109. REASONING AND SOLUTION If the +x axis is taken in the direction of motion, ΣFx = 0 gives F F – fk – mg sin θ = 0 N where fk fk = µ kFN Then F F – µ kFN – mg sin θ = 0 Also, ΣFy = 0 gives so (1) FN – mg cos θ = 0 FN = mg cos θ (2) Substituting Equation (2) into Equation (1) and solving for F yields W 25.0° 234 FORCES AND NEWTON'S LAWS OF MOTION F = mg( sin θ + µ k cos θ ) F = (55.0 kg)(9.80 m/s2)[sin 25.0° + (0.120)cos 25.0°] = 286 N ____________________________________________________________________________________________ 110. REASONING Since the wire beneath the limb is at rest, it is in equilibrium and the net force acting on it must be zero. Three forces comprise the net force, the 151-N force from the limb, the 447-N tension force from the left section of the wire, and the tension force T from the right section of the wire. 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, which are the horizontal and vertical components of the tension force T. These two equations will be solved simultaneously to give values for the two unknowns. Knowing the components of the tension force, we can determine its magnitude and direction. SOLUTION Let Tx and Ty be the horizontal and vertical components of the tension force. The free-body diagram fo...
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