This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full 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 61: Using Newton’s Second Law Problem 1. Two forces, both in the x y plane, act on a 1.5kg mass, which accelerates at 7 . 3 m / s 2 in a direction 30 ◦ counterclockwise from the xaxis. 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 xaxis. Problem 2. Two forces act on a 3.1kg 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 3700kg barge is being pulled along a canal by two mules, as shown in Fig. 659. 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 659. 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 659 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 61). 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 xaxis down the slope (parallel to the acceleration) and the yaxis 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.
 Fall '09
 Huang

Click to edit the document details