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Homework #3 12. Two forces are applied to a car in an effort to move it, as shown in Figure P4.12. (a) What is the resultant of these two forces? (b) If the car has a mass of 3 000 kg, what acceleration does it have? Ignore friction. 14. The force exerted by the wind on the sails of a sailboat is 390 N north. The water exerts a force of 180 N east. If the boat (including its crew) has a mass of 270 kg, what are the magnitude and direction of its acceleration? Figure P4.12 15. Find the tension in each cable supporting the 600-N cat burglar in Figure P4.15. Figure P4.15 Figure P4.16 16. Find the tension in the two wires that support the 100-N light fixture in Figure P4.16. 17. A 150-N bird feeder is supported by three cables as shown in Figure P4.17. Find the tension in each cable. 18. The leg and cast in Figure P4.18 weigh 220 N (w1). Determine the weight w2 and the angle α needed so that no force is exerted on the hip joint by the leg plus the cast. Figure P4.17 Figure P4.18 21. The distance between two telephone poles is 50.0 m. When a 1.00-kg bird lands on the telephone wire midway between the poles, the wire sags 0.200 m. Draw a free-body diagram of the bird. How much tension does the bird produce in the wire? Ignore the weight of the wire. Homework #3 Solutions 4.12 (a) Choose the positive y-axis in the forward direction. We resolve the forces into their components as Force 400 N 450 N Resultant x-component 200 N –78.1 N y-component 346 N 443 N Fx 122 N Fy 790 N The magnitude and direction of the resultant force is FR Fx 2 Fy 2 F 799 N , tan1 x 8.77 to right of y-axis. Fy Thus, FR 799 N at8.77 to the rightofthe forw ard direction (b) The acceleration is in the same direction as FR and has magnitude a 4.14 FR 799 N 0.266 m s2 m 3000 kg Since the two forces are perpendicular to each other, their resultant is: FR 180 N 2 390 N 2 430 N , at tan1 390 N 65.2 N ofE 180 N Thus, a FR 430 N 1.59 m s2 m 270 kg a 1.59 m s2 at65.2 N ofE or 4.15 Since the burglar is held in equilibrium, the tension in the vertical cable equals the burglar’s weight of 600 N Now, consider the junction in the three cables: Fy 0 , giving T2 sin 37.0 600 N 0 or T2 997 N in the inclined cable Also, Fx 0 which yields T2 cos37.0 T1 0 or 4.16 T1 997 N cos37.0 796 N in the horizontalcable From Fx 0 , T1 cos40.0 T2 cos40.0 0 or T1 T2 Then, Fy 0 gives 2 T1 sin 40.0 100 N 0 yielding T1 T2 77.8 N 4.17 From Fx 0 , T1 cos30.0 T2 cos60.0 0 or T2 1.73 T1 (1) Then Fy 0 becomes T1 sin30.0 1.73 T1 sin60.0 150 N 0 which gives T1 75.0 N in the rightside cable Finally, Equation (1) above gives T2 130 N in the leftside cable 4.18 If the hip exerts no force on the leg, the system must be in equilibrium with the three forces shown in the freebody diagram. Thus Fx 0 becomes w 2 cos 110 N cos40 (1) From Fy 0 , we find w 2 sin 220 N 110 N sin40 (2) Dividing Equation (2) by Equation (1) yields 220 N 110 N sin 40 61 110 N cos40 tan1 Then, from either Equation (1) or (2), w 2 1.7 102 N 4.21 m 1.00 kg and m g 9.80 N 0.200 m 0.458 25.0 m tan1 Require that Fy 0 , 2T sin m g T 9.80 N 613 N 2sin