Physics Solution Manual for 1100 and 2101

# The forces that act on the book are shown in the

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Unformatted text preview: 0.414 L or – 2.414 L The negative solution is discarded because the third particle lies on the +x axis between m and 2m. Thus, D = 0.414 L . ____________________________________________________________________________________________ 38. REASONING In each case the object is in equilibrium. According to Equation 4.9b, ΣFy = 0, the net force acting in the y (vertical) direction must be zero. The net force is composed of the weight of the object(s) and the normal force exerted on them. SOLUTION a. There are three vertical forces acting on the crate: an upward normal force +FN that the floor exerts, the weight –m1g of the crate, and the weight –m2g of the person standing on the crate. Since the weights act downward, they are assigned negative numbers. Setting the sum of these forces equal to zero gives FN + (−m1g ) + (−m2 g ) = 0 14444244444 4 3 ΣFy The magnitude of the normal force is FN = m1g + m2g = (35 kg + 65 kg)(9.80 m/s2) = 980 N b. There are only two vertical forces acting on the person: an upward normal force +FN that the crate exerts and the weight –m2g of the person. Setting the sum of these forces equal to zero gives FN + (−m2 g ) = 0 144 244 3 ΣFy The magnitude of the normal force is FN = m2g = (65 kg)(9.80 m/s2) = 640 N ____________________________________________________________________________________________ Chapter 4 Problems 39. 185 SSM REASONING The book is kept from falling as long as the total static frictional force balances the weight of the book. The forces that act on the book are shown in the following free-body diagram, where P is the pressing force applied by each hand. fsMAX fsMAX P P W In this diagram, note that there are two pressing forces, one from each hand. Each hand also applies a static frictional force, and, therefore, two static frictional forces are shown. The maximum static frictional force is related in the usual way to a normal force FN, but in this problem the normal force is provided by the pressing force, so that FN = P. SOLUTION Since the frictional forces balance the weight, we have 2 f sMAX = 2 ( µ s FN ) = 2 ( µ s P ) = W Solving for P, we find that P= W 31 N = = 39 N 2µ s 2 ( 0.40 ) ____________________________________________________________________________________________ 40. REASONING AND SOLUTION a. The apparent weight of the person is given by Equation 4.6 as FN = mg + ma = (95.0 kg)(9.80 m/s2 + 1.80 m/s2) = 1.10 × 103 N b. FN = (95.0 kg)(9.80 m/s2) = 931 N c. FN = (95.0 kg)(9.80 m/s2 – 1.30 m/s2) = 808 N ____________________________________________________________________________________________ 186 FORCES AND NEWTON'S LAWS OF MOTION 41. REASONING As the drawing shows, the normal FN force FN points perpendicular to the hill, while the weight W points vertically down. Since the car does not leave the surface of the hill, there is no acceleration in this perpendicular direction. Therefore, the magnitude of the θ perpendicular component of the weight W cos θ must equal W cos θ θ the magnitude of the normal force, FN = W cos θ . Thus, the W magnitude of the normal force is less than the magnitude of the weight. As the hill becomes steeper, θ increases, and cos θ decreases. Consequently, the normal force decreases as the hill becomes steeper. The magnitude of the normal force does not depend on whether the car is traveling up or down the hill. SOLUTION a. From the REASONING, we have that FN = W cos θ . The ratio of the magnitude of the normal force to the magnitude W of the weight is FN W = W cos θ = cos 15° = 0.97 W b. When the angle is 35°, the ratio is W cos θ = cos 35° = 0.82 W W ______________________________________________________________________________ FN = 42. REASONING The reading on the bathroom scale is Free-body diagrams of the man proportional to the normal force it exerts on the man. When he simply stands on the scale, his acceleration FN1 is zero, so the normal force pushing up on him balances the downward pull of gravity: FN1 = mg (see P FN2 the free-body diagram). Thus, the first reading on the scale is his actual mass m, the ratio of the normal force the scale exerts on him to the acceleration due to gravity: First reading = m = FN1/g = 92.6 kg. With the mg mg chin-up bar helping to support him, the normal force exerted on him by the scale decreases, and the second Standing Pulling down reading is the ratio of the reduced normal force FN2 to the acceleration due to gravity: Second reading = FN2/g = 75.1 kg. Lastly, we note that the magnitude P of the force the chin-up bar exerts on the man is exactly equal to the magnitude P of the force that the man exerts on the chin-up bar. This prediction is due to Newton’s third law. Therefore, it is a value for P that we seek. SOLUTION When the man is pulling down on the chin-up bar, there are two upward forces acting on him (see the second part of the drawing), and he is still at rest, so the sum of these Chapter 4 Problems 187 two forces balances the downward pull of gravity: FN2 + P = mg,...
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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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