# Chap6 - CHAPTER 6 USING NEWTON’S LAWS ActivPhysics can help with these problems All Activities in Section 2 “Forces and Motion” and Section 4

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Unformatted text preview: CHAPTER 6 USING NEWTON’S LAWS ActivPhysics can help with these problems: All Activities in Section 2 “Forces and Motion” and Section 4 “Circular Motion.” Section 6-1: Using Newton’s Second Law Problem 1. Two forces, both in the x- y plane, act on a 1.5-kg mass, which accelerates at 7 . 3 m / s 2 in a direction 30 ◦ counter-clockwise from the x-axis. One force has magnitude 6.8 N and points in the + x direction. Find the other force. Solution Newton’s second law for this mass says F net =F 1 + F 2 = m a , where we assume no other significant forces are acting. Since the acceleration and the first force are given, one can solve for the second, F 2 = m a − F 1 = (1 . 5 kg)(7 . 3 m / s 2 )( ˆ ı cos30 ◦ +ˆ sin30 ◦ ) − (6 . 8 N) ˆ ı = (2 . 68 ˆ ı+ 5 . 48 ˆ ) N . This has magnitude 6.10 N and direction 63 . 9 ◦ CCW from the x-axis. Problem 2. Two forces act on a 3.1-kg mass, which undergoes acceleration a= . 91 ˆ ı − . 27 ˆ m / s 2 . If one of the forces is F 1 = − 1 . 2 ˆ ı − 2 . 5 ˆ N , what is the other force? Solution As in the previous problem, F 2 = m a − F 1 = (3 . 1 kg)(0 . 91 ˆ ı − . 27 ˆ )(m / s 2 ) − ( − 1 . 2 ˆ ı − 2 . 5 ˆ ) N = (4 . 02 ˆ ı+ 1 . 66 ˆ )N . Problem 3. A 3700-kg barge is being pulled along a canal by two mules, as shown in Fig. 6-59. The tension in each tow rope is 1100 N, and the ropes make 25 ◦ angles with the forward direction. What force does the water exert on the barge (a) if it moves with constant velocity and (b) if it accelerates forward at 0 . 16 m / s 2 ? Solution The horizontal forces on the barge are the two tensions and the resistance of the water, as shown on Figure 6-59. The net force is in the x direction, so 2 T cos25 ◦ − F res = ma x , since T 1 = T 2 = T. (a) If a x = 0 , F res = 2(1100 N)cos25 ◦ = 1 . 99 kN . (b) If a x = 0 . 16 m / s 2 , F res = 1 . 99 kN − (3700 kg) × (0 . 16 m / s 2 ) = 1 . 40 kN . 25 ° 25 ° F res T 1 T 2 x figure 6-59 Problem 3 Solution. Problem 4. At what angle should you tilt an air table to simulate motion on the moon’s surface, where g = 1 . 6 m / s 2 ? Solution The acceleration down an incline is a || = g sin θ (see Example 6-1). To replicate the moon’s surface gravity, the angle of tilt should be θ = sin − 1 (1 . 62 / 9 . 81) = 9 . 51 ◦ (see Appendix E). Problem 5. A block of mass m slides with acceleration a down a frictionless slope that makes an angle θ to the horizontal; the only forces acting on it are the force of gravity F g and the normal force N of the slope. Show that the magnitude of the normal force is given by N = m radicalbig g 2 − a 2 . Solution Choose the x-axis down the slope (parallel to the acceleration) and the y-axis parallel to the normal....
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## This note was uploaded on 10/31/2010 for the course PHYSICS 69054 taught by Professor Huang during the Fall '09 term at University of California, Berkeley.

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Chap6 - CHAPTER 6 USING NEWTON’S LAWS ActivPhysics can help with these problems All Activities in Section 2 “Forces and Motion” and Section 4

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