1 / 14

University Physics: Mechanics

University Physics: Mechanics. Ch 5 . Newton’s Law of Motion. Lecture 7. Dr.-Ing. Erwin Sitompul. http://zitompul.wordpress.com. Homework 5: Two Boxes and A Pulley.

theresam
Download Presentation

University Physics: Mechanics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

  2. Homework 5: Two Boxes and A Pulley A block of mass m1 = 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 m2 = 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?

  3. Solution of Homework 5 FN T T a a m1gsinθ m1gcosθ m2g m1g Forces on m1 along the x axis: Forces on m2 along the y axis: • What is the meaning of negative sign?

  4. Solution of Homework 5 (a) The magnitude of the acceleration of each block (b) The direction of the acceleration of the hanging block Assumption : The acceleration points upward Result : Negative value Conclusion : The true acceleration points downward Down (c) The tension in the cord

  5. 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/s2 and downward at 3.2 m/s2?

  6. 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 FN, 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

  7. Applying Newton’s Law: Problem 3 (c) What does the scale read if the cab acceleration upward 3.2 m/s2 and downward at 3.2 m/s2? • If the cab accelerates upward, the magnitude of acceleration is positive • It the cab accelerates downward, the magnitude of acceleration is negative

  8. Applying Newton’s Law: Problem 3 What does the scale read if, in case accident happens, the cab falls vertically downward?

  9. 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 m1 = 2.8 kg; the other has mass m2 = 1.3 kg. Determine: (a) the magnitude of the blocks’ acceleration. (b) the tension in the cord. • Atwood Machine

  10. Applying Newton’s Law: Problem 4 Mass m1 Mass m2 • The acceleration of m1 and m2 have the same magnitude a, oppose in direction. • We take the acceleration of m1 as negative (downward) and of m2 as positive (upward). T a T a m1g m2g

  11. Applying Newton’s Law: Problem 4 (a) The magnitude of the blocks’ acceleration • What happen if m2 > m1? (b) The tension in the cord

  12. 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 + ct2 – (2 m/s3)t3, 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?

  13. 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?

  14. Homework 6 1. What is the net force acting on the ring in the next figure? New 2. Joe’s Advertising wishes to hang a sign weighing 750.0 N so that cable A, attached to the store makes a 30.0° angle, as shown in the figure. Cable B is horizontal and attached to and adjoining building. What is the tension in cable B?

More Related