1. University Physics: Mechanics Ch5. Newton’s Law of Motion Lecture 7 Dr.-Ing. Erwin Sitompul http://zitompul.wordpress.com

2. 7/2 September–December 2009 University Physics: Mechanics Announcement Industrial Engineering (IE) 18.11.0917.30–18.30: Lecture 7 19.00–20.15: Quiz 2 Material: Lecture 5 & 6 25.11.0917.30–18.30: Lecture 8 & Make-up Quiz 2 19.00–20.15: Discussion Quiz 2 02.12.0917.30–18.30: Lecture 9 19.00–20.15: Quiz 3 Material: Lecture 7 & 8 09.12.0917.30–18.30: Discussion & Make-up Quiz 3: 16.12.0918.30–20.30: Final Examination

3. 7/3 September–December 2009 University Physics: Mechanics Announcement Information Technology (IT) 20.11.0917.30–18.30: Lecture 7 19.00–20.15: Quiz 2 Material: Lecture 5 & 6 27.11.09Eid al-Adha (No class) 04.12.0917.30–18.30: Lecture 8 & Make-up Quiz 2 19.00–20.15: Discussion Quiz 2 20.15–20.45: Quiz 3 Material: Lecture 6 & 7 11.12.0917.30–18.30: Lecture 9 & Make-up Quiz 3 19.00–20.15: Discussion Quiz 3 18.12.0918.30–20.30: Final Examination

4. 7/4 September–December 2009 University Physics: Mechanics Homework 5: Two Boxes and A Pulley A block of mass m 1 = 3.7 kg on a frictionless plane inclined at angle θ = 30° is connected by a cord over a massless, frictionless pulley to a second block of mass m 2 = 2.3 kg. What are: (a) the magnitude of the acceleration of each block, (b) the direction of the acceleration of the hanging block, and (c) the tension in the cord?

5. 7/5 September–December 2009 University Physics: Mechanics a Solution of Homework 5 T T m1gm1g m2gm2g a m 1 gcosθ m 1 gsinθ FNFN Forces on m 1 along the x axis: Forces on m 2 along the y axis: What is the meaning of negative sign?

6. 7/6 September–December 2009 University Physics: Mechanics Solution of Homework 5 (a) The magnitude of the acceleration of each block (b) The direction of the acceleration of the hanging block (c) The tension in the cord Down Assumption: The acceleration points upward Result: Negative value Conclusion: The true acceleration points downward

7. 7/7 September–December 2009 University Physics: Mechanics Applying Newton’s Law: Problem 3 A passenger of mass 71.43 kg stands on a platform scale in an elevator cab. We are concerned with the scale reading when the cab is stationary and when it is moving up or down. (a)Find a general solution for the scale reading, whatever the vertical motion of the cab. (b) What does the scale read if the cab is stationary or moving upward at a constant 0.5 m/s? (c)What does the scale read if the cab acceleration upward 3.2 m/s 2 and downward at 3.2 m/s 2 ?

8. 7/8 September–December 2009 University Physics: Mechanics Applying Newton’s Law: Problem 3 (a)Find a general solution for the scale reading, whatever the vertical motion of the cab. The scale reading is equal to F N, which is the force exerted by the surface of the scale towards the passenger → (b)What does the scale read if the cab is stationary or moving upward at a constant 0.5 m/s? In stationary condition or when moving upward with a constant velocity, the acceleration of passenger is zero

9. 7/9 September–December 2009 University Physics: Mechanics Applying Newton’s Law: Problem 3 (c)What does the scale read if the cab acceleration upward 3.2 m/s 2 and downward at 3.2 m/s 2 ? If the cab accelerates upward, the magnitude of acceleration is positive It the cab accelerates downward, the magnitude of acceleration is negative

10. 7/10 September–December 2009 University Physics: Mechanics Applying Newton’s Law: Problem 3 What does the scale read if, in case accident happens, the cab falls vertically downward?

11. 7/11 September–December 2009 University Physics: Mechanics Applying Newton’s Law: Problem 4 The figure below shows two blocks connected by a cord (of negligible mass) that passes over a frictionless pulley (also of negligible mass). One block has mass m 1 = 2.8 kg; the other has mass m 2 = 1.3 kg. Determine: (a) the magnitude of the blocks’ acceleration. (b) the tension in the cord.

12. 7/12 September–December 2009 University Physics: Mechanics Applying Newton’s Law: Problem 4 T T m1gm1g m2gm2g a a Mass m 1  The acceleration of m 1 and m 2 have the same magnitude a, oppose in direction.  We take the acceleration of m 1 as negative (downward) and of m 2 as positive (upward).

13. 7/13 September–December 2009 University Physics: Mechanics (a) The magnitude of the blocks’ acceleration Applying Newton’s Law: Problem 4 (b) The tension in the cord What happen if m 2 > m 1 ?

14. 7/14 September–December 2009 University Physics: Mechanics Example: Particle Movement A 2 kg particle moves along an x axis, being propelled by a variable force directed along that axis. Its position is given by x = 3 m + (4 m/s)t + ct 2 – (2 m/s 3 )t 3, with x in meters and t in seconds. The factor c is a constant. At t = 3 s, the force on the particle has a magnitude of 36 N and is in the negative direction of the axis. What is c?

15. 7/15 September–December 2009 University Physics: Mechanics Homework 6: The Traffic Light A traffic light weighing 122 N hangs from a cable tied to two other cables fastened to a support, as in the figure below. The upper cables make angles of 37° and 53° with the horizontal. These upper cables are not as strong as the vertical cable and will break if the tension in them exceeds 100N. Will the traffic light remain hanging in this situation, or will one of the cables break?