Physics Solution Manual for 1100 and 2101

Chapter 4 problems force x component y component f1

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Unformatted text preview: re slipping occurs is F a= 17.0 kg Newton's second law applied to the 5.00 kg block is F – µ sm1g = m1a = (5.00 kg) F 17.0 kg Hence F = 41.6 N ____________________________________________________________________________________________ 95. SSM REASONING The magnitude of the gravitational force that each part exerts on the other is given by Newton’s law of gravitation as F = Gm1m2 / r 2 . To use this expression, we need the masses m1 and m2 of the parts, whereas the problem statement gives the weights W1 and W2. However, the weight is related to the mass by W = mg, so that for each part we know that m = W/g. SOLUTION The gravitational force that each part exerts on the other is F= Gm1m2 r2 = G (W1 / g ) (W2 / g ) r2 ( 6.67 × 10–11 N ⋅ m2 / kg 2 ) (11 000 N )(3400 N ) = = 2 (9.80 m/s2 ) (12 m )2 1.8 × 10 –7 N ____________________________________________________________________________________________ 226 FORCES AND NEWTON'S LAWS OF MOTION 96. REASONING The magnitudes of the initial (v0 = 0 m/s) and final (v = 805 m/s) velocities are known. In addition, data is given for the mass and the thrust, so that Newton’s second law can be used to determine the acceleration of the probe. Therefore, kinematics Equation 2.4 (v = v0 + at) can be used to determine the time t. SOLUTION Solving Equation 2.4 for the time gives t= v − v0 a Newton’s second law gives the acceleration as a = (ΣF)/m. Using this expression in Equation 2.4 gives t= ( v − v0 ) ( ΣF ) / m = m ( v − v0 ) ΣF = ( 474 kg )(805 m/s − 0 m/s ) = 6.8 ×106 s 56 × 10−3 N Since one day contains 8.64 × 104 s, the time is ( t = 6.8 ×106 s 97. 1 day ) 8.64 ×104 s = 79 days SSM REASONING AND SOLUTION According to Equation 3.3b, the acceleration of the astronaut is a y = (v y − v0 y ) / t = v y / t . The apparent weight and the true weight of the astronaut are related according to Equation 4.6. Direct substitution gives vy FN = { + ma y = m ( g + a y ) = m g + mg { t True Apparent weight weight 45 m/s 2 = (57 kg) 9.80 m/s 2 + = 7.3 ×10 N 15 s ____________________________________________________________________________________________ 98. REASONING According to Newton's second law ( ∑ F = ma ), the acceleration of the object is given by a = ∑ F / m , where ∑ F is the net force that acts on the object. We must first find the net force that acts on the object, and then determine the acceleration 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. Chapter 4 Problems Force x-Component y-Component F1 40.0 N 0N F2 (60.0 N) cos 45.0° = 42.4 N (60.0 N) sin 45.0° = 42.4 N ∑ F = F1 + F2 82.4 N 227 42.4 N The magnitude of ∑ F is given by the Pythagorean theorem as ΣF = (82.4 N) 2 + (42.4) 2 = 92.7 N The angle θ that ∑ F makes with the +x axis is 42.4 N θ = tan −1 = 27.2° 82.4 N ΣF θ 42.4 N 82.4 N According to Newton's second law, the magnitude of the acceleration of the object is a= ∑ F 92.7 N = = 30.9 m/s 2 m 3.00 kg Since Newton's second law is a vector equation, we know that the direction of the right hand side must be equal to the direction of the left hand side. In other words, the direction of the acceleration a is the same as the direction of the net force ∑ F . Therefore, the direction of the acceleration of the object is 27.2° above the +x axis . ____________________________________________________________________________________________ 99. SSM REASONING In order to start the crate moving, an external agent must supply a force that is at least as large as the maximum value fsMAX = µs FN , where µs is the coefficient of static friction (see Equation 4.7). Once the crate is moving, the magnitude of the frictional force is very nearly constant at the value f k = µk FN , where µ k is the coefficient of kinetic friction (see Equation 4.8). In both cases described in the problem statement, there are only two vertical forces that act on the crate; they are the upward normal force FN, and the downward pull of gravity (the weight) mg. Furthermore, the crate has no vertical acceleration in either case. Therefore, if we take upward as the positive direction, Newton's second law in the vertical direction gives FN − mg = 0 , and we see that, in both cases, the magnitude of the normal force is FN = mg . 228 FORCES AND NEWTON'S LAWS OF MOTION SOLUTION a. Therefore, the applied force needed to start the crate moving is fsMAX = µs mg = (0.760)(60.0 kg)(9.80 m/s 2 ) = 447 N b. When the crate moves in a straight line at constant speed, its velocity does not change, and it has zero acceleration. Thus, Newton's second law in the horizontal direction becomes P − fk = 0, where P is the required pushing force. Thus, the applied force required to keep the crate sliding across the dock at a constant speed is P = f k = µk mg = (0.410)(60.0 kg)(9.80 m/s 2 ) = 241 N ____________________________________________________________________________________________ 100. REASONING Newton’s secon...
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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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