Physics Solution Manual for 1100 and 2101

# We will calculate the acceleration of the submarine

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Unformatted text preview: N −1 Chapter 4 Problems 207 Therefore, the kite’s height relative to the person is h = L sin θ = ( 43 m ) sin 48o = 32 m 69. REASONING The weight of the part of the washcloth off the table is moff g. At the instant just before the washcloth begins to slide, this weight is supported by a force that has magnitude equal to fsMAX, which is the static frictional force that the table surface applies to the part of the washcloth on the table. This force is transmitted “around the bend” in the washcloth hanging over the edge by the tension forces between the molecules of the washcloth, in much the same way that a force applied to one end of a rope is transmitted along the rope as it passes around a pulley. SOLUTION Since the static frictional supports the weight of the washcloth off the table, we have fsMAX = moff g. The static frictional force is fsMAX = µsFN . The normal force FN is applied by the table to the part of the washcloth on the table and has a magnitude equal to the weight of that part of the washcloth. This is so, because the table is assumed to be horizontal and the part of the washcloth on it does not accelerate in the vertical direction. Thus, we have f sMAX = µ s FN = µ s mon g = moff g The magnitude g of the acceleration due to gravity can be eliminated algebraically from this result, giving µsmon = moff . Dividing both sides by mon + moff gives µs mon mon + moff moff = mon + moff or µ s fon = foff where we have used fon and foff to denote the fractions of the washcloth on and off the table, respectively. Since fon + foff = 1, we can write the above equation on the left as ( ) µ s 1 – foff = f off or foff = µs 0.40 = = 0.29 1 + µs 1 + 0.40 ____________________________________________________________________________________________ 208 FORCES AND NEWTON'S LAWS OF MOTION 70. REASONING In addition to the upward buoyant force B and the downward resistive force R, a downward gravitational force mg acts on the submarine (see the free-body diagram), where m denotes the mass of the submarine. Because the submarine is not in contact with any rigid surface, no normal force is exerted on it. We will calculate the acceleration of the submarine from Newton’s second law, using the free-body diagram as a guide. +y B R SOLUTION Choosing up as the positive direction, we sum the forces mg acting on the submarine to find the net force ΣF. According to Free-body diagram Newton’s second law, the acceleration a is of the submarine ( ) 2 ΣF B − R − mg 16 140 N − 1030 N − (1450 kg ) 9.80 m/s a= = = = +0.62 m/s 2 m m 1450 kg where the positive value indicates that the direction is upward. 71. REASONING According to Newton’s second law, the acceleration has the same direction as the net force and a magnitude given by a = ΣF/m. SOLUTION Since the two forces are perpendicular, the magnitude of the net force is given by the Pythagorean theorem as ΣF = ( 40.0 N )2 + ( 60.0 N )2 . Thus, according to Newton’s second law, the magnitude of the acceleration is ΣF a= = m ( 40.0 N )2 + ( 60.0 N )2 4.00 kg = 18.0 m/s 2 The direction of the acceleration vector is given by 60.0 N = 56.3° above the +x axis 40.0 N ______________________________________________________________________________ θ = tan –1 72. REASONING a. Since the fish is being pulled up at a constant speed, it has no acceleration. According to Newton’s second law, the net force acting on the fish must be zero. We will use this fact to determine the weight of the heaviest fish that can be pulled up. b. When the fish has an upward acceleration, we can still use Newton’s second law to find the weight of the heaviest fish. However, because the fish has an acceleration, we will see that the maximum weight is less than that in part (a). Chapter 4 Problems 209 SOLUTION a. There are two forces acting on the fish (taking the upward vertical direction to be the +y direction): the maximum force of +45 N due to the line, and the weight –W of the fish (negative, because the weight points down). Newton’s second law ( ΣFy = 0, Equation 4.9b ) gives ΣFy = +45 N − W = 0 b. Since the fish has an upward or W = 45 N acceleration ( ΣFy = ma y , Equation 4.2b ) becomes ay, Newton’s second law ΣFy = +45 N − W ′ = ma y Where W′ is the weight of the heaviest fish that can be pulled up with an acceleration. Solving this equation for W′ gives W ′ = +45 N − ma y The mass m of the fish is the magnitude W′ of its weight divided by the magnitude g of the acceleration due to gravity (see Equation 4.5), or m = W′/g. Substituting this relation for m into the previous equation gives W′ W ′ = +45 N − ay g Solving this equation for W′ yields 45 N 45 N = = 37 N ay 2.0 m/s 2 1+ 1+ g 9.80 m/s 2 ______________________________________________________________________________ W′ = 73. SSM REASONING If we assume that the acceleration is constant, we can use Equation 2.4 ( v = v0 + at ) to find the acceleration of the car. Once the acceleration is known, Newton's second law ( ∑ F = ma ) can be used to...
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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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