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Physics 11 HW #4 Solutions Chapter 4: Focus On Concepts: 5, 7, 12, 23 Problems: 4, 49, 54, 64, 74, 98, 110, 115 Focus On Concepts 4-5 (c) Newton’s second law gives the answer directly, provided the net force is calculated by vector addition of the two given forces. The direction of the net force gives the direction of the acceleration. Focus On Concepts 4-7 (e) Answers a and b are false, according to the third law, which states that whenever one body exerts a force on a second body, the second body exerts an oppositely directed force of equal magnitude on the first body. It does not matter whether one of the bodies is stationary or whether it collapses. Answer c is false, because according to the third law, Sam and his sister experience forces of equal magnitudes during the push-off. Since Sam has the greater mass, he flies off with the smaller acceleration, according to the second law. Answer d is false, because in catching and throwing the ball each astronaut applies a force to it, and, according to the third law, the ball applies an oppositely directed force of equal magnitude to each astronaut. These reaction forces accelerate the astronauts away from each other, so that the distance between them increases. Focus On Concepts 4-12 (d) What matters is the direction of the elevator’s acceleration. When the acceleration is upward, the apparent weight is greater than the true weight. When the acceleration is downward, the apparent weight is less than the true weight. In both possibilities the acceleration points upward. Focus On Concepts 4-23 (b) Since the boxes move at a constant velocity, they have no acceleration and are, therefore, in equilibrium. According to Newton’s second law, the net force acting on each box must be zero. Thus, Newton’s second law applied to each box gives two equations in two unknowns, the magnitude of the tension in the rope between the boxes and the kinetic frictional force that acts on each box. Note that the frictional forces acting on the boxes are identical, because the boxes are identical. Solving these two equations shows that the tension is one-half of the applied force. Problem 4-4 REASONING According to Newton’s second law, Equation 4.1, the average net force ΣF is equal to the product of the object’s mass m and the average acceleration a . The average acceleration is equal to the change in velocity divided by the elapsed time (Equation 2.4), where the change in velocity is the final velocity v minus the initial velocity v0. SOLUTION The average net force exerted on the car and riders is ) ( v − v0 45 m/s − 0 m/s ma = m = = 5.5 × 103 kg 3.5 × 104 N ∑F = t − t0 7.0 s Problem 4-49 SSM REASONING AND SOLUTION The free-body diagram is shown at the right. The forces that act on the picture are the pressing force P, the normal force FN exerted on the picture by the wall, the weight mg of the picture, and the force of static friction fsMAX . The maximum magnitude for the frictional force is given by Equation 4.7: fsMAX = µs FN . The picture is in equilibrium, and, if we take the directions to the right and up as positive, we have in the x direction ∑ Fx =P − FN =0 or P =FN fsMAX − mg= 0 ∑ F= y or fsMAX= mg and in the y direction Therefore, MAX fs= m= mg s FN But since FN = P , we have ms P = mg Solving for P, we have mg (1.10 kg)(9.80 m/s 2 ) P = = = 0.660 ms 16.3 N Problem 4-54 REASONING The drawing assumes that upward is the +y direction and shows the I-beam together with the three forces that act on it: its weight W and the tension T in each of the cables. Since the I-beam is moving upward at a constant velocity, its acceleration is zero and it is in equilibrium. Therefore, according to Equation 4.9b, the net force in the vertical (or y) direction must be zero, so that ∑ Fy = 0 . This relation T T 70.0° 70.0° will allow us to find the magnitude of the tension. W = –8.00 × 103 N SOLUTION With up as the +y direction, the vertical component of the tension in each cable is T sin 70.0° , and Equation 4.9b becomes Solving this equation for the tension gives T = 4260 N . Problem 4-64 REASONING The block is in equilibrium in each case, because the block has a constant velocity and is not accelerating. A zero acceleration is the hallmark of equilibrium. At equilibrium, the net force is zero (i.e., the forces balance to zero), and we will obtain the magnitude of the pushing force by utilizing this fact as it pertains to the vertical or y direction. We will use Equation 4.9b 0 ) for this purpose. ( ΣFy = +y +y +x +x Note that the direction of the kinetic frictional force is not the same in each case. The fk frictional force always opposes the relative θ θ fk P motion between the surface of the block and P W W the wall. Therefore, when the block slides upward, the frictional force points downward. Free-body diagram for Free-body diagram for upward motion downward motion When the block slides downward, the frictional force points upward. These directions are shown in the free-body diagrams (not to scale) for the two cases. In these drawings W is the weight of the block and fk is the kinetic frictional force. In each case the magnitude of the frictional force is the same. It is given by Equation 4.8 as fk = μkFN, where μk is the coefficient of kinetic friction and FN is the magnitude of the normal force. The coefficient of kinetic friction does not depend on the direction of the motion. Furthermore, the magnitude of the normal force in each case is the component of the pushing force that is perpendicular to the wall, or FN = P sin θ. SOLUTION Using Equations 4.9b to describe the balance of forces that act on the block in the y direction and referring to the free-body diagrams, we have Upward motion Σ= Fy P cos θ − W −= fk 0 Downward motion Σ= Fy P cos θ − W += fk 0 According to Equation 4.8, the magnitude of the kinetic frictional force is fk = μkFN, where we have pointed out in the REASONING that the magnitude of the normal force is FN = P sin θ. Substituting into the equations for ΣFy in the two cases, we obtain Upward motion Σ= Fy P cos θ − W − µk P sin= θ 0 Downward motion Σ= Fy P cos θ − W + µk P sin= θ 0 Solving each case for P, we find that a. Upward motion = P W 39.0 N = = 52.6 N cos θ − µk sin θ cos 30.0° − ( 0.250 ) sin 30.0° b. Downward motion = P W 39.0 N = = 39.4 N cos θ + µk sin θ cos 30.0° + ( 0.250 ) sin 30.0° Problem 4-74 REASONING In the absence of air resistance, the two forces acting on the sensor are its weight W and the tension T in the towing cable (see the free-body diagram). We see that Tx is the only horizontal force acting on the sensor, and therefore Newton’s second law ΣFx = max (Equation 4.2a) gives Tx = max. Because the vertical component of the sensor’s acceleration is zero, the vertical component of the cable’s tension T must balance the sensor’s weight: Ty = W = mg. We thus have sufficient information to calculate the horizontal and vertical components of the tension force T, and therefore to calculate its 2 magnitude T from the Pythagorean theorem: T= Tx2 + Ty2 . SOLUTION Given that Tx = max and that Ty = mg, the Pythagorean theorem yields the magnitude T of the tension in the cable: T= Tx2 + Ty2 = = (129 kg) ( ma )2 + ( mg )2 = m2 a 2 + m2 g 2 = m a 2 + g 2 m/s 2 ) ( 2.84 m/s2 ) + (9.80= 2 2 1320 N Problem 4-98 REASONING 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; ax = (ΣFx)/m. For completeness, however, the free-body diagram will include the vertical forces also (the normal force FN and the weight W). SOLUTION The free-body diagram is shown at the right, where +y F1 = 59.0 N F2 = 33.0 N FN θ = 70.0° When F1 is replaced by its x and y components, we obtain the free body diagram in the following drawing. F2 +x θ W F1 Choosing right to be the positive direction, we have ΣFx F1 cos θ − F2 ax = = m m ax = ( 59.0 N ) cos 70.0° − ( 33.0 N ) = 7.00 kg −1.83 m/s +y FN 2 F1cos θ F2 Thus, the horizontal acceleration has a magnitude of 1.83 m / s2 , and the minus sign indicates that it points to +x F1sin θ the left . W Problem 4-110 REASONING Since the mountain climber is at rest, she is in equilibrium and the net force acting on her must be zero. Three forces comprise the net force, her weight, and the tension forces from the left and right sides of the rope. 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, the tension TL in the left side of the rope and the tension TR in the right side. These two equations will be solved simultaneously to give values for the two unknowns. SOLUTION Using W to denote the weight of the mountain climber and choosing right and upward to be the positive directions, we have the following free-body diagram for the climber: For the x components of the forces we have = ΣFx TR sin 80.0° − TL = sin 65.0° 0 +y TL 65.0º 80.0º For the y components of the forces we have = ΣFy TR cos80.0° + TL cos 65.0 = ° −W 0 TR +x W Solving the first of these equations for TR, we find that TR = TL sin 65.0° sin 80.0° Substituting this result into the second equation gives TL sin 65.0° cos80.0° + TL cos= 65.0° − W 0= or TL 1.717 W sin 80.0° Using this result in the expression for TR reveals that sin 65.0° = TR T= L sin 80.0° sin 65.0° 1.580 W sin 80.0° (1.717W= ) Since the weight of the climber is W = 535 N, we find that = = TL 1.717 W 1.717 (= 535 N ) 919 N = TR 1.580 = W 1.580 (= 535 N ) 845 N Problem 4-115 SSM REASONING Let us assume that the skater is moving horizontally along the +x axis. The time t it takes for the skater to reduce her velocity to vx = +2.8 m/s from v0x = +6.3 m/s can be obtained from one of the equations of kinematics: v= v0 x + ax t x (3.3a) The initial and final velocities are known, but the acceleration is not. We can obtain the acceleration from Newton’s second law ( ΣFx = max , Equation 4.2a ) in the following manner. The kinetic frictional force is the only horizontal force that acts on the skater, and, since it is a resistive force, it acts opposite to the direction of the motion. Thus, the net force in the x direction is ΣFx = − f k , where fk is the magnitude of the kinetic frictional force. Therefore, the acceleration of the skater is ax = ΣFx /m = − fk / m . The magnitude of the frictional force is f k = µk FN (Equation 4.8), where µk is the coefficient of kinetic friction between the ice and the skate blades and FN is the magnitude of the normal force. There are two vertical forces acting on the skater: the upward-acting normal force FN and the downward pull of gravity (her weight) mg. Since the skater has no vertical acceleration, Newton's second law in the vertical direction gives (taking upward as the positive direction) ΣFy = FN − mg = 0 . Therefore, the magnitude of the normal force is FN = mg and the magnitude of the acceleration is = ax − f k − mk FN − mk m g = = = − mk g m m m SOLUTION Solving the equation v= v0 x + ax t for the time and substituting the expression above for x the acceleration yields = t vx − v0 x vx − v0 x 2.8 m/s − 6.3 m/s = = = 4.4 s ax − mk g − ( 0.081) ( 9.80 m/s 2 )