Physics Solution Manual for 1100 and 2101

1 according to equation 48 the magnitude fk of the

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Unformatted text preview: from Equation 4.7 fsMAX = µ FN . Because the car’s acceleration has no vertical component, the net vertical force acting on the car must be zero, so the upward normal force FN must balance the two downward forces, the car’s weight W and the downforce D: FN = W + D = mg + D SOLUTION According to Equation 4.7 (2) ( fsMAX = µs FN ) and Equation (2), the maximum static frictional force the track can exert on the car is fsMAX = µ FN = µ ( mg + D ) (3) Now solving Equation (1) for the car’s acceleration ax and then substituting Equation (3) for the static frictional force fsMAX , we obtain ax = = fsMAX − f A m = µs ( mg + D ) − f A m ( 0.87 ) ( 690 kg ) ( 9.80 m/s2 ) + 4060 N − 1190 N 690 kg = 12 m/s 2 Chapter 4 Problems 193 51. REASONING The diagram shows the two applied forces that act on the crate. These two forces, plus the kinetic frictional force fk constitute the net force that acts on the crate. Once the net force has been determined, Newtons’ second law, ΣF = ma (Equation 4.1) can be used to find the acceleration of the crate. SOLUTION The sum of the applied forces is F = F1 + F2. The x-component of this sum is Fx = F1 cos 55.0° + F2 = (88.0 N) cos 55.0° + 54.0 N = 104 N. The y-component of F is Fy = F1 sin 55.0° = (88.0 N) sin 55.0° = 72.1 N. The magnitude of F is F = Fx2 + Fy2 = (104 N )2 + ( 72.1 N )2 +y F1 55.0° +x F2 = 127 N Since the crate starts from rest, it moves along the direction of F. The kinetic frictional force fk opposes the motion, so it points opposite to F. The net force acting on the crate is the sum of F and fk. The magnitude a of the crate’s acceleration is equal to the magnitude ΣF of the net force divided by the mass m of the crate a= ΣF − f k + F = m m (4.1) According to Equation 4.8, the magnitude fk of the kinetic frictional force is given by f k = µk FN , where FN is the magnitude of the normal force. In this situation, FN is equal to the magnitude of the crate’s weight, so FN = mg. Thus, the x-component of the acceleration is ( ) 2 − µk mg + F − ( 0.350 )( 25.0 kg ) 9.80 m/s + 127 N a= = = 1.65 m/s 2 m 25.0 kg The crate moves along the direction of F, whose x and y components have been determined previously. Therefore, the acceleration is also along F. The angle φ that F makes with the xaxis can be found using the inverse tangent function: Fy F x φ = tan −1 F1 sin 55.0° −1 = tan F cos 55.0° + F 1 2 ( 88.0 N ) sin 55.0° = tan −1 = 34.6° above the x axis ( 88.0 N ) cos 55.0° + 54.0 N ____________________________________________________________________________________________ 194 FORCES AND NEWTON'S LAWS OF MOTION 52. REASONING AND SOLUTION horizontal directions gives Newton’s second law applied in the vertical and L cos 21.0° – W = 0 L sin 21.0° – R = 0 (1) (2) a. Equation (1) gives 21.0° L= W 53 800 N = = 57 600 N cos 21.0° cos 21.0° L b. Equation (2) gives R R = L sin 21.0° = ( 57 600 N ) sin 21.0° = 20 600 N W ____________________________________________________________________________________________ 53. SSM REASONING In order for the object to move with constant velocity, the net force on the object must be zero. Therefore, the north/south component of the third force must be equal in magnitude and opposite in direction to the 80.0 N force, while the east/west component of the third force must be equal in magnitude and opposite in direction to the 60.0 N force. Therefore, the third force has components: 80.0 N due south and 60.0 N due east. We can use the Pythagorean theorem and trigonometry to find the magnitude and direction of this third force. SOLUTION The magnitude of the third force is N F3 = (80.0 N) 2 + (60.0 N) 2 = 1.00 × 10 2 N The direction of F3 is specified by the angle θ where 80.0 N θ = tan –1 = 53.1°, south of east 60.0 N E θ 80.0 N θ F3 60.0 N ____________________________________________________________________________________________ Chapter 4 Problems 195 54. REASONING AND SOLUTION a. In the horizontal direction the thrust F is balanced by the resistive force fr of the water. That is, ΣFx = 0 or fr = F = 7.40 × 105 N b. In the vertical direction, the weight, mg, is balanced by the buoyant force, Fb. So ΣFy = 0 gives Fb = mg = (1.70 × 108 kg)(9.80 m/s2) = 1.67 × 109 N ____________________________________________________________________________________________ 55. REASONING The drawing shows the two forces, T and T′ , that act on the tooth. To obtain the net force, we will add the two forces using the method of components (see T′ Section 1.8). +y +x 16.0° 16.0° T SOLUTION The table lists the two vectors and their x and y components: Vector x component y component T +T cos 16.0° −T sin 16.0° T′ −T′ cos 16.0° −T′ sin 16.0° T + T′ +T cos 16.0° − T′ cos 16.0° −T sin 16.0° − T′ sin 16.0° Since we are given that T = T′ = 21.0 N, the sum of the x components of the forces is ΣFx = +T cos 16.0° − T′ cos 16.0° = +(21 N) cos 16.0° − (21 N) cos 16.0° = 0 N The sum of the y component...
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