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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

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.

Page1 / 23

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online