Week 4 Homework/Recitation: 9/21/2017 Chapter4: Problems 3, 5, 11, 16, 24, 38, 52, 77, 78, 98. is shown in the drawing. F 2

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1 Week 4 Homework/Recitation: 9/1/017 Chapter4: Problems 3, 5, 11, 16, 4, 38, 5, 77, 78, Two horizontal forces, F 1 and F, are acting on a box, but only F 1 is shown in the drawing. F can point either to the right or to the left. The box moves only along the x axis. There is no friction between the box and the surface. Suppose that F 1 = N and the mass of the box is 3.0 kg. Find the magnitude and direction of F when the acceleration of the box is (a) +5.0 m/s, (b) 5.0 m/s, and (c) 0 m/s. RESONING In each case, we will apply Newton s second law. Remember that it is the net force that appears in the second law. The net force is the vector sum of both forces. SOLUTION a. We will use Newton s second law, ΣF x, to find the force F. Taking the positive x direction to be to the right, we have so F F = (3.0 kg)(+5.0 m/s ) (+9.0 N) = + 6N b. pplying Newton s second law again gives F = (3.0 kg)( 5.0 m/s ) (+9.0 N) = 4 N c. n application of Newton s second law gives F = (3.0 kg)(0 m/s ) (+9.0 N) = 9.0 N

2 5. ssm person in a kayak starts paddling, and it accelerates from 0 to 0.60 m/s in a distance of 0.41 m. If the combined mass of the person and the kayak is 73 kg, what is the magnitude of the net force acting on the kayak? RESONING The magnitude ΣF of the net force acting on the kayak is given by Newton s second law as Σ F = ma (Equation 4.1), where m is the combined mass of the person and kayak, and a is their acceleration. Since the initial and final velocities, v 0 and v, and the displacement x are known, we can employ one of the equations of kinematics from Chapter to find the acceleration. v = v + ax from the equations of kinematics for SOLUTION Solving Equation.9 ( ) the acceleration, we have 0 v0 v a =, Hence x ( ) ( ) ( ) v v m/s 0 m/s Σ F = ma = m = 73 kg = 3 N x ( 0.41 m) 11. Only two forces act on an object (mass = 3.00 kg), as in the drawing. Find the magnitude and direction (relative to the x axis) of the acceleration of the object. RESONING ccording to Newton's second law ( F= ma). We must first find the net force that acts on the object, and then determine a using Newton's second law. SOLUTION The following table gives the x and y components of the two forces that act on the object. The third row of that table gives the components of the net force. Force x-component y-component 40.0 N 0 N F (60.0 N) cos 45.0 = 4.4 N (60.0 N) sin 45.0 = 4.4 N F= F1+ F 8.4 N 4.4 N The magnitude of F is given by the Pythagorean theorem as Σ F = (8.4 N) + (4.4) = 9.7 N The angle θ that F makes with the +x axis is θ N = tan = N ccording to Newton's second law, the magnitude of the acceleration of the object is F 9.7 N a = = = 30.9 m/s m 3.00 kg The direction of the acceleration of the object is 7. above the + x axis.

3 16. Two skaters, a man and a woman, are standing on ice. Neglect any friction between the skate blades and the ice. The mass of the man is 8 kg, and the mass of the woman is 48 kg. The woman pushes on the man with a force of 45 N due east. Determine the acceleration (magnitude and direction) of (a) the man and (b) the woman. RESONING Since there is only one force acting on the man in the horizontal direction, it is the net force. ccording to Newton s second law, Equation 4.1, the man must accelerate under the action of this force. The factors that determine this acceleration are (1) the magnitude and () the direction of the force exerted on the man, and (3) the mass of the man. When the woman exerts a force on the man, the man exerts a force of equal magnitude, but opposite direction, on the woman (Newton s third law). It is the only force acting on the woman in the horizontal direction, so, as is the case with the man, she must accelerate. The factors that determine her acceleration are (1) the magnitude and () the direction of the force exerted on her, and (3) the her mass. SOLUTION a. The acceleration of the man is, according to Equation 4.1, equal to the net force acting on him divided by his mass. a man ΣF 45 N = = = m 8 kg 0.55 m / s (due east) b. The acceleration of the woman is equal to the net force acting on her divided by her mass. a woman ΣF 45 N = = = m 48 kg 0.94 m / s (due west)

4 4. The weight of an object is the same on two different planets. The mass of planet is only sixty percent that of planet B. Find the ratio r /r B of the radii of the planets. RESONING Newton s law of gravitation shows how the weight W of an object of mass m is related to the mass M and radius r of the planet on which the object is located: W = GMm/r. In this expression G is the universal gravitational constant. Using the law of gravitation, we can express the weight of the object on each planet, set the two weights equal, and obtain the desired ratio. SOLUTION ccording to Newton s law of gravitation, we have The mass m of the object, being an intrinsic property, is the same on both planets and can be eliminated algebraically from this equation. The universal gravitational constant can likewise be eliminated algebraically. s a result, we find that = B or = M B B B M M M r r r r r r B M = = 0.60 = 0.77 M B

5 kg crate rests on a horizontal floor, and a 65-kg person is standing on the crate. Determine the magnitude of the normal force that (a) the floor exerts on the crate and (b) the crate exerts on the person. RESONING In each case the object is in equilibrium. ccording to Equation 4.9b, ΣF y = 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 +F N that the floor exerts, 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, they are assigned negative numbers. Setting the sum of these forces equal to zero gives The magnitude of the normal force is F N = m 1 g + m g = (35 kg + 65 kg)(9.80 m/s ) = 980 N b. There are only two vertical forces acting on the person: an upward normal force +F N that the crate exerts and the weight m g of the person. Setting the sum of these forces equal to zero gives The magnitude of the normal force is F N = m g = (65 kg)(9.80 m/s ) = 640 N

6 5. mmh The helicopter in the drawing is moving horizontally to the right at a constant velocity v The weight of the helicopter is W = N. The lift force L generated by the rotating blade makes an angle of 1.0 with respect to the vertical. (a) What is the magnitude of the lift force? (b) Determine the magnitude of the air resistance R that opposes the motion. RESONING The free-body diagram for the helicopter is shown in the drawing. Since the velocity is constant, the acceleration is zero and the helicopter is at equilibrium. Therefore, according to Newton s second law, the net force acting on the helicopter is zero. SOLUTION Since the net force is zero, the components of the net force in the vertical and horizontal directions are separately zero. Referring to the free-body diagram, we can see, then, that L cos 1.0 W = 0 (1) L sin 1.0 R = 0 () W N a. Equation (1) gives L = = = cos 1.0 cos N b. Equation () gives R= Lsin 1.0 = ( N) sin 1.0 = N

7 77. car is towing a boat on a trailer. The driver starts from rest and accelerates to a velocity of +11 m/s in a time of 8 s. The combined mass of the boat and trailer is 410 kg. The frictional force acting on the trailer can be ignored. What is the tension in the hitch that connects the trailer to the car? RESONING The only horizontal force acting on the boat and trailer is the tension in the hitch; therefore, it is the net force. ccording to Newton s second law, the tension (or the net force) equals the mass times the acceleration. The mass is known, and the acceleration can be found by applying an appropriate equation of kinematics from Chapter 3. SOLUTION ssume that the boat and trailer are moving in the +x direction. Newton s second law is Σ Fx = max (see Equation 4.a), where the net force is just the tension +T in the hitch, so Σ Fx = T. Thus, T (1) Since the initial and final velocities, v 0x and v x, and the time t are known, we may use Equation 3.3a from the equations of kinematics to relate these variables to the acceleration: v = v + a t (3.3a) x 0x x Solving Equation (3.3a) for a x and substituting the result into Equation (1), we find that vx v0x 11 m/s 0 m/s T = max = m = ( 410 kg) = 160 N t 8 s

8 78. 9-kg motorcycle is accelerating up along a ramp that is inclined 30.0 above the horizontal. The propulsion force pushing the motorcycle up the ramp is 3150 N, and air resistance produces a force of 50 N that opposes the motion. Find the magnitude of the motorcycle s acceleration. RESONING The forces acting on the motorcycle are the normal force F N, the 3150-N propulsion force, the 50-N force of air resistance, and the weight, which is mg = (9 kg)(9.80 m/s = 860 N. ll of these forces must be considered when determining the net force for use with Newton s second law to determine the acceleration. In particular, we note that the motion occurs along the ramp and that both the propulsion force and air resistance are directed parallel to the ramp surface. In contrast, the normal force and the weight do not act parallel to the ramp. The normal force is perpendicular to the ramp surface, while the weight acts vertically downward. However, the weight does have a component along the ramp. SOLUTION In drawing the free-body diagram for the motorcycle we choose the +x axis to be parallel to the ramp surface and upward, the +y direction being perpendicular to the ramp surface. The diagram is shown at the right. Since the motorcycle accelerates along the ramp and we seek only that acceleration, we can ignore the forces that point along the y axis (the normal force F N and the y component of the weight). The x component of Newton s second law is ( ) Σ F = 3150 N 860 N sin N = ma x Solving for the acceleration a x gives x 30.0º +y F N 50 N +x 3150 N 30.0º 860 N (860 N) sin 30.0º a x ( ) 3150 N 860 N sin N = = 5.03 m/s 9 kg _

9 98. mmh Two forces, F 1 and F, act on the 7.00-kg block shown in the drawing. The magnitudes of the forces are = 59.0 N and F = 33.0 N. What is the horizontal acceleration (magnitude and direction) of the block? RESONING Newton s second law gives the acceleration as a = (ΣF)/m. Since we seek only the horizontal acceleration, it is the x component of this equation that we will use; a x = (ΣF x )/m. For completeness, however, the free-body diagram will include the vertical forces also (the normal force F N and the weight W). SOLUTION The free-body diagram is shown at the right, where = 59.0 N F = 33.0 N θ = y F N When is replaced by its x and y components, we obtain the free body diagram in the following drawing. F W θ +x Choosing right to be the positive direction, we have ΣFx F1cosθ F a x = = m m a x ( 59.0 N) cos 70.0 ( 33.0 N) = = 1.83 m/s 7.00 kg Thus, the horizontal acceleration has a magnitude of 1.83 m / s, and the minus sign indicates that it points to the left. F + F cos sin W +