Physics Solution Manual for 1100 and 2101

0 and the weight is mg where we have chosen up as the

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = 0.187 2 gT2 9.80 m/s2 (1.80 s ) ( ) ______________________________________________________________________________ 21. REASONING AND SOLUTION The centripetal acceleration of the block is ac = v2/r = (28 m/s)2/(150 m) = 5.2 m/s2 The angle θ can be obtained from θ = tan –1 a F I = tan F5.2 m / s I = G J G m/s J g K 9.80 HK H 2 C –1 2 28 ° 22. REASONING The coefficient µs of static friction is related to the magnitude f sM AX of the maximum static frictional force and the magnitude FN of the normal force acting on the car by fsMAX = µs FN (Equation 4.7), so that: fsMAX µs = FN The car is going around an unbanked curve, so the centripetal force Fc = (1) mv 2 r (Equation 5.3) must be horizontal. The static frictional force is the only horizontal force, so 254 DYNAMICS OF UNIFORM CIRCULAR MOTION it serves as the centripetal force. The maximum centripetal force occurs when Fc = f sM AX . Therefore, the maximum speed v the car can have without slipping is related to f sM AX by Fc = fsMAX = mv 2 r (2) Substituting Equation (2) into Equation (1) yields mv 2 µs = r FN (3) In part a the car is subject to two downward-pointing forces, its weight W and the downforce D. The vertical acceleration of the car is zero, so the upward normal force must balance the two downward forces: FN = W + D. Combining this relation with Equation (3), we obtain an expression for the coefficient of static friction: mv 2 mv 2 mv 2 µs = r = r = FN W + D r ( mg + D ) (4) SOLUTION a. Since the downforce is D = 11 000 N, Equation (4) gives the coefficient of static friction as ( 830 kg )( 58 m/s ) mv 2 = = 0.91 r ( mg + D ) (160 m ) ( 830 kg ) 9.80 m/s 2 + 11 000 N 2 µs = ( ) b. The downforce is now absent (D = 0 N). Solving Equation (4) for the speed of the car, we find that v= µs r ( mg + D ) m = µs rg = = µs r ( mg + 0 N ) m = ( 0.91)(160 m ) ( 9.80 m/s2 ) = µs r ( mg ) m 38 m/s Chapter 5 Problems 255 23. REASONING a. The free body diagram shows the swing ride and the two forces that act on a chair: the tension T in the cable, and the weight mg of the chair and its occupant. We note that the chair does not accelerate vertically, so the net force ∑ Fy in the vertical direction must be zero, ∑ Fy = 0 . The net force consists of the upward vertical component of the tension and the downward weight of the chair. The fact that the net force is zero will allow us to determine the magnitude of the tension. 60.0° T +y 15.0 m +x r mg b. According to Newton’s second law, the net force ∑ Fx in the horizontal direction is equal to the mass m of the chair and its occupant times the centripetal acceleration ( ac = v 2 / r ) , so that ∑ Fx = mac = mv2 / r . There is only one force in the horizontal direction, the horizontal component of the tension, so it is the net force. We will use Newton’s second law to find the speed v of the chair. SOLUTION a. The vertical component of the tension is +T cos 60.0°, and the weight is −mg, where we have chosen “up” as the + direction. Since the chair and its occupant have no vertical acceleration, we have that ∑ Fy = 0 , so +T cos 60.0° − mg = 0 144 2444 4 3 ∑ Fy (1) Solving for the magnitude T of the tension gives (179 kg ) ( 9.80 m/s 2 ) mg T= = = 3510 N cos 60.0° cos 60.0° b. The horizontal component of the tension is +T sin 60.0°, where we have chosen the direction to the left in the diagram as the + direction. Since the chair and its occupant have a centripetal acceleration in this direction, we have 256 DYNAMICS OF UNIFORM CIRCULAR MOTION v2 T sin 60.0° = mac = m 14 3 24 r ∑ Fx (2) From the drawing we see that the radius r of the circular path is r = (15.0 m) sin 60.0° = 13.0 m. Solving Equation (2) for the speed v gives v= r T sin 60.0° = m (13.0 m )( 3510 N ) sin 60.0° = 14.9 m/s 179 kg 24. REASONING We will treat this situation as Seawall a circular turn on a banked surface, with the angle θ that the rider leans serving as the banking angle. The banking angle θ is r related to the speed v of the watercraft, the Beginning of turn radius r of the curve and the magnitude g of 2 v the acceleration due to gravity by tan θ = r rg (Equation 5.4). If the rider is closer to the seawall than r, she will hit the wall while making the turn Therefore, the minimum distance at which she must begin the turn is r, the minimum turn radius (see the drawing). SOLUTION Solving Equation 5.4 for r, we obtain the minimum distance: ( 26 m/s ) v2 r= = = 170 m g tan θ 9.80 m/s 2 tan 22o 2 ( ) 25. REASONING The angle θ at which a friction-free curve is banked depends on the radius r of the curve and the speed v with which the curve is to be negotiated, according to 2 Equation 5.4: tan θ = v /( rg). For known values of θ and r, the safe speed is v = rg tan θ Before we can use this result, we must determine tan θ for the banking of the track. SOLUTION The drawing at the right shows a cross-section of the track. From the drawing we have 18 m tan θ = = 0. 34 53 m θ 165 m – 112 m = 53 m 18 m Chapter 5 Problems 257 a. Therefore, the smallest speed at which cars can move on this track without relying on friction is vmin = (...
View Full Document

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.

Ask a homework question - tutors are online