PHYS 101: Solutions to Chapter 4 Home Work

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Kelley Bridges
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1 PHYS 101: Solutions to Chapter 4 Home ork 3. EASONING In each case, we will appl Newton s second law. emember that it is the net force that appears in the second law. he net force is the vector sum of both forces. SOUION a. e will use Newton s second law, F = ma, to find the force F. aking the positive direction to be to the right, we have F 1 F F = ma so F = ma F 1 F = (3.0 kg)(+5.0 m/s ) (+9.0 N) = 6N b. Appling Newton s second law again gives F = ma F 1 = (3.0 kg)(5.0 m/s ) (+9.0 N) = 4 N c. An application of Newton s second law gives F = ma F 1 = (3.0 kg)(0 m/s ) (+9.0 N) = 9.0 N 7. SSM EASONING AND SOUION he acceleration required is v v m/s a.5 m/s 50.0 m Newton's second law then gives the magnitude of the net force as F = ma = (1580 kg)(.5 m/s ) = 3560 N 11. EASONING Newton s second law gives the acceleration as a = (ΣF)/m. Since we seek onl the horizontal acceleration, it is the component of this equation that we will use; a = (ΣF )/m. For completeness, however, the free-bod diagram will include the vertical forces also (the normal force F N and the weight ).

2 SOUION he free-bod diagram is shown at the right, where + F 1 = 59.0 N F = 33.0 N F N θ = 70.0 hen F 1 is replaced b its and components, we obtain the free bod diagram in the following drawing. F θ F 1 + Choosing right to be the positive direction, we have F F1cos F a m m a 59.0 N cos N 7.00 kg 1.83 m/s + F N he minus sign indicates that the horizontal acceleration points to the left. F F 1 cos θ F 1 sin θ SSM EASONING o determine the acceleration we will use Newton s second law F = ma. wo forces act on the rocket, the thrust and the rocket s weight, which is mg = ( kg)(9.80 m/s ) = N. Both of these forces must be considered when determining the net force F. he direction of the acceleration is the same as the direction of the net force. SOUION In constructing the free-bod diagram for the rocket we choose upward and to the right as the positive directions. he free-bod diagram is as follows: he component of the net force is + F cos N cos N 55.0º = sin 55.0º + he component of the net force is F sin N sin N N he magnitudes of the net force and of the acceleration are = cos 55.0º PHYS 101: Solutions to Chapter 4 Home ork pg. /9

3 F F F 6 6 F F N N a m kg 10.3 m/s he direction of the acceleration is the same as the direction of the net force. hus, it is directed above the horizontal at an angle of F tan tan N F N 1. EASONING AND SOUION a. According to Equation 4.5, the weight of the space traveler of mass m = 115 kg on earth is 3 mg (115 kg) (9.80 m/s ) N b. In interplanetar space where there are no nearb planetar objects, the gravitational force eerted on the space traveler is zero and g = 0 m/s. herefore, the weight is = 0 N. Since the mass of an object is an intrinsic propert of the object and is independent of its location in the universe, the mass of the space traveler is still m =115 kg. 4. EASONING Newton s law of universal gravitation indicates that the gravitational force that each uniform sphere eerts on the other has a magnitude that is inversel proportional to the square of the distance between the centers of the spheres. herefore, the maimum gravitational force between two uniform spheres occurs when the centers of the spheres are as close together as possible, and this occurs when the surfaces of the spheres are touching. hen, the distance between the centers of the spheres is the sum of the two radii. SOUION hen the bowling ball and the billiard ball are touching, the distance between their centers is r = r Bowling + r Billiard. Using this epression in Newton s law of universal gravitation gives Gm m Gm m F r r r Bowling Billiard Bowling Billiard Bowling Billiard N m / kg 7. kg0.38 kg 0.11 m 0.08 m N 9 PHYS 101: Solutions to Chapter 4 Home ork pg. 3/9

4 35. EASONING he gravitational force that the sun eerts on a person standing on the earth is given b Equation 4.3 as Fsun GMsunm / rsun-earth, where M sun is the mass of the sun, m is the mass of the person, and r sun-earth is the distance from the sun to the earth. ikewise, the gravitational force that the moon eerts on a person standing on the earth is given b Fmoon GM moonm / rmoon-earth, where M moon is the mass of the moon and r moon-earth is the distance from the moon to the earth. hese relations will allow us to determine whether the sun or the moon eerts the greater gravitational force on the person. SOUION aking the ratio of F sun to F moon, and using the mass and distance data from the inside of the tet s front cover, we find GM m sun sun sun-earth sun moon-earth F r M r F GM m M r moon moon moon sun-earth rmoon-earth kg m kg m 178 herefore, the sun eerts the greater gravitational force. 38. EASONING In each case the object is in equilibrium. According to Equation 4.9b, F = 0, the net force acting in the (vertical) direction must be zero. he net force is composed of the weight of the object(s) and the normal force eerted on them. SOUION a. here are three vertical forces acting on the crate: an upward normal force +F N that the floor eerts, the weight m 1 g of the crate, and the weight m g of the person standing on the crate. Since the weights act downward, the are assigned negative numbers. Setting the sum of these forces equal to zero gives he magnitude of the normal force is F ( m g) ( m g) 0 N 1 F F N = m 1 g + m g = (35 kg + 65 kg)(9.80 m/s ) = 980 N b. here are onl two vertical forces acting on the person: an upward normal force +F N that the crate eerts and the weight m g of the person. Setting the sum of these forces equal to zero gives he magnitude of the normal force is F ( m g) 0 N F F N = m g = (65 kg)(9.80 m/s ) = 640 N PHYS 101: Solutions to Chapter 4 Home ork pg. 4/9

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.

5 39. SSM EASONING he book is kept from falling as long as the total static frictional force balances the weight of the book. he forces that act on the book are shown in the following free-bod diagram, where P is the pressing force applied b each hand. 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. he maimum static frictional force is related in the usual wa to a normal force F N, but in this problem the normal force is provided b the pressing force, so that F N = P. SOUION Since the frictional forces balance the weight, we have Solving for P, we find that MAX s s N s f F P 31 N P 0.40 s 39 N 47. EASONING he magnitude of the kinetic frictional force is given b Equation 4.8 as the coefficient of kinetic friction times the magnitude of the normal force. Since the slide into second base is horizontal, the normal force is vertical. It can be evaluated b noting that there is no acceleration in the vertical direction and, therefore, the normal force must balance the weight. o find the plaer s initial velocit v 0, we will use kinematics. he time interval for the slide into second base is given as t = 1.6 s. Since the plaer comes to rest at the end of the slide, his final velocit is v = 0 m/s. he plaer s acceleration a can be obtained from Newton s second law, since the net force is the kinetic frictional force, which is known from part (a), and the mass is given. Since t, v, and a are known and we seek v 0, the appropriate kinematics equation is Equation.4 (v = v 0 + at). SOUION a. Since the normal force F N balances the weight mg, we know that F N = mg. Using this fact and Equation 4.8, we find that the magnitude of the kinetic frictional force is fk kfn kmg kg 9.8 m/s 390 N b. Solving Equation.4 (v = v 0 + at) for v 0 gives v 0 = v at. aking the direction of the plaer s slide to be the positive direction, we use Newton s second law and Equation 4.8 for the kinetic frictional force to write the acceleration a as follows: PHYS 101: Solutions to Chapter 4 Home ork pg. 5/9

6 F kmg a kg m m he acceleration is negative, because it points opposite to the plaer s velocit, since the plaer slows down during the slide. hus, we find for the initial velocit that v0 v kg t 0 m/s m/s 1.6 s 7.7 m/s 6. EASONING Since the mountain climber is at rest, she is in equilibrium and the net force acting on her must be zero. hree forces comprise the net force, her weight, and the tension forces from the left and right sides of the rope. e will resolve the forces into components and set the sum of the components and the sum of the components separatel equal to zero. In so doing we will obtain two equations containing the unknown quantities, the tension in the left side of the rope and the tension in the right side. hese two equations will be solved simultaneousl to give values for the two unknowns. SOUION Using to denote the weight of the mountain climber and choosing right and upward to be the positive directions, we have the following free-bod diagram for the climber: For the components of the forces we have + F sin 80.0 sin For the components of the forces we have 65.0º 80.0º + F cos80.0 cos Solving the first of these equations for, we find that sin 65.0 sin 80.0 Substituting this result into the second equation gives sin 65.0 cos80.0 cos or sin 80.0 Using this result in the epression for reveals that sin 65.0 sin sin 80.0 sin 80.0 Since the weight of the climber is = 535 N, we find that N 919 N N 845 N PHYS 101: Solutions to Chapter 4 Home ork pg. 6/9

67. SSM EASONING hen the biccle is coasting straight down the hill, the forces that act on it are the normal force F N eerted b the surface of the hill, the force of gravit mg, and the force of air

7 67. SSM EASONING hen the biccle is coasting straight down the hill, the forces that act on it are the normal force F N eerted b the surface of the hill, the force of gravit mg, and the force of air resistance. hen the biccle climbs the hill, there is one additional force; it is the applied force that is required for the bicclist to climb the hill at constant speed. e can use our knowledge of the motion of the biccle down the hill to find. Once is known, we can analze the motion of the biccle as it climbs the hill. SOUION he figure to the left below shows the free-bod diagram for the forces during the downhill motion. he hill is inclined at an angle above the horizontal. he figure to the right shows these forces resolved into components parallel to and perpendicular to the line of motion. F N + + F N mg mg sin mg cos Since the bicclist is traveling at a constant velocit, his acceleration is zero. herefore, according to Newton's second law, we have F 0 and F 0. aking the direction up the hill as positive, we have F mg sin 0, or hen the bicclist climbs the same hill at constan speed, an applied force P must push the sstem up the hill. Since the speed is the same, the magnitude o the force of air resistance will remain 03 N. However the air resistance will oppose the motion b pointing down the hill. he figure at the right shows the resolved forces that act on the sstem during the uphi motion. mg sin (80.0 kg)(9.80 m/ s ) sin N Using the same sign convention as above, we have F P mg sin 0, or P = +mg sin 03 N 03 N 406 N PHYS 101: Solutions to Chapter 4 Home ork pg. 7/9

8 76. EASONING AND SOUION Newton's second law applied to object 1 (4 N) gives = m 1 a 1 Similarl, for object (185 N) m g = m a If the string is not to break or go slack, both objects must have accelerations of the same magnitude. hen a 1 = a and a = a. he above equations become = m 1 a (1) m g = m a () a. Substituting Equation (1) into Equation () and solving for a ields he masses of objects 1 and are mg a m m m / g 4 N / 9.80 m/s 43.1 kg m / g 185 N / 9.80 m/s 18.9 kg he acceleration is kg9.80 m/s mg a m m 43.1 kg 18.9 kg.99 m/s b. Using this value in Equation (1) gives m1 a 43.1 kg.99 m/s 19 N PHYS 101: Solutions to Chapter 4 Home ork pg. 8/9

9 83. SSM EASONING he free-bod diagrams for obin (mass = m) and for the chandelier (mass = M) are given at the right. he tension in the rope applies an upward force to both. obin accelerates upward, while the chandelier accelerates downward, each acceleration having the same magnitude. Our solution is based on separate applications of Newton s second law to obin and the chandelier. obin mg Chandelier Mg SOUION Appling Newton s second law, we find mg ma and Mg Ma obin Hood Chandelier In these applications we have taken upward as the positive direction, so that obin s acceleration is a, while the chandelier s acceleration is a. Solving the obin-hood equation for gives mg ma Substituting this epression for into the Chandelier equation gives a. obin s acceleration is M m mg ma Mg Ma or a g M m 195 kg 77.0 kg M m a g 9.80 m/s 4.5 m/s M m 195 kg 77.0 kg b. Substituting the value of a into the epression for gives mg ma 77.0 kg 9.80 m/s 4.5 m/s 1080 N PHYS 101: Solutions to Chapter 4 Home ork pg. 9/9

5. REASONING AND SOLUTION An object will not necessarily accelerate when two or more forces are applied to the object simultaneously.

5. REASONING AND SOLUTION An object will not necessarily accelerate when two or more forces are applied to the object simultaneously. The applied forces may cancel so the net force is zero; in such a case,