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Physics 103: Lecture 8, Pg 1 Physics 103: Lecture 8 Application of Newton's Laws l Survey results l Leftovers from Chapter 4 l How to solve problems.

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Presentation on theme: "Physics 103: Lecture 8, Pg 1 Physics 103: Lecture 8 Application of Newton's Laws l Survey results l Leftovers from Chapter 4 l How to solve problems."— Presentation transcript:

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2 Physics 103: Lecture 8, Pg 1 Physics 103: Lecture 8 Application of Newton's Laws l Survey results l Leftovers from Chapter 4 l How to solve problems

3 Physics 103: Lecture 8, Pg 2 Survey Results Are the preflights helpful? Is the interactive learning helpful? Are the homeworks on the web helpful? YES NO YES NO YES NO Has the course web site been problematic? YES NO With 80% approval rating the web format is here to stay for this course.

4 Physics 103: Lecture 8, Pg 3 But, all is not well! l Most requests çPlease work out problems in class fully »Will address this to some extent in class »Better forum for that is the discussion »Please ask questions l Some requests çSlow down and explain more l Most complaints çThere is too much work for the course çHomework is too hard and takes too much time l Some complaints çWe need more problems assigned for practice! l Some comments çDon’t go over pre-flights in lecture çClarify the concepts (from pre-flights) in lecture l Thanks for taking time to answer çI will try to balance the requests and address complaints

5 Physics 103: Lecture 8, Pg 4 From Lecture 1: Course Philosophy l Basic Course Philosophy çread about it (text) çuntangle it (lectures) çplay with it (labs) çchallenge yourself (homework) çclose the loop (discussion) Do make use of discussion, office hours, etc. Try the homework during the weekend so that you can ask questions during the week Homework is posted Friday and is due the following Friday. Exam worries: Will post practice exams by Monday You can bring a single 8.5”x11” sheet with equations to exam

6 Physics 103: Lecture 8, Pg 5 Preflight 8: Question 1 Consider a person standing in an elevator that is accelerating upward. The upward normal force N exerted by the elevator floor on the person is a) larger than b) identical to c) less than the downward weight W of the person. Example 4.11 in the book! Person is accelerating upwards - net upwards force is non zero mg N

7 Physics 103: Lecture 8, Pg 6 Preflight 8: Question 3 You are pushing a wooden crate across the floor at constant speed. You decide to turn the crate on end, reducing by half the surface area in contact with the floor. In the new orientation, to push the same crate across the same floor with the same speed, the force that you apply must be about a) four times as great b) twice as great c) equally as great d) half as great e) one-fourth as great as the force required before you changed the crate orientation. Frictional force does not depend on the area of contact. It depends only on the normal force and the coefficient of friction for the contact.

8 Physics 103: Lecture 8, Pg 7Tension Tension is a force along the length of a medium Tension can be transmitted around corners If there is no friction in the pulleys, T remains the same

9 Physics 103: Lecture 8, Pg 8 Example 8: Pulley Problem I What is the tension in the string? A) T

10 Physics 103: Lecture 8, Pg 9 Example 9: Pulley Problem II What is the tension in the string? A) T

11 Physics 103: Lecture 8, Pg 10 Preflight 8: Question 5 In the 17th century, Otto von Guricke, a physicist in Magdeburg, fitted two hollow bronze hemispheres together and removed the air from the resulting sphere with a pump. Two eight-horse teams could not pull the halves apart even though the hemispheres fell apart when air was readmitted! Suppose von Guricke had tied both teams of horses to one side and bolted the other side to a heavy tree trunk. In this case, the tension on the hemisphere would be a) twice what it was b) exactly what it was c) half what it was FhFh T TFhFh 2F h 2T

12 Physics 103: Lecture 8, Pg 11 Reminder: Procedure for Solving Problems l Identify force using Free Body Diagram l Set up axes l Write F net =ma for each axis l Calculate acceleration components l Use kinematic equations l Solve! the most crucial step!

13 Physics 103: Lecture 8, Pg 12 Problem 4.27 l What force does a trampoline have to apply to a 45 kg gymnast to accelerate her up at 7.5 m/s 2 ? mg FtFt

14 Physics 103: Lecture 8, Pg 13 Problem 4.21 l A flea jumps by exerting a force of 1.2 x N straight down on the ground. A breeze blowing on the flea parallel to ground exerts a force of 0.5 x N on the flea. Find the direction and magnitude of the acceleration of the flea if its mass is 6 x kg. mg F f

15 Physics 103: Lecture 8, Pg 14 Problem 4.33 Suppose you have a 120 kg wooden crate resting on a wood floor. (a) What maximum force can you exert horizontally on the crate without moving it? mg N What kind of friction is impeding movement? static kinetic (b) If you continue to exert this force once the crate starts to slip what will its acceleration be? What is the net force acting on the crate? Force exerted Force exerted - Static friction Force exerted - Kinetic friction F fkfk

16 Physics 103: Lecture 8, Pg 15 Example 3 T=50 N M = 5 kg  k = 0.2 Find acceleration of block  =50 0 Hints: Draw FBD Resolve T in x and y components Normal force is smaller than when pulling along horizontal Tension component along horizontal is also smaller Answer: 6.0 m/s 2 Similar to example 4.8 in book

17 Physics 103: Lecture 8, Pg 16 Example 6 Find acceleration and normal force Choose axis and calculate components Answers: a= g sin(  ) F N =mg cos (  )   mg cos (  ) mg sin (  )

18 Physics 103: Lecture 8, Pg 17 Problems 4.59 and A contestant in a winter games event pushes a 45 kg block of ice across a frozen lake. Calculate the minimum force, F, he must exert to get the bock moving. What is the force, F’, he needs to exert if he were to pull the bock with a rope over his shoulder at the same angle above the horizontal? Is F = F’? 1) Yes 2) No 25 o F cos25 o F’ cos25 o -F sin25 o F’ sin25 o F’F  s (ice-on-ice)=0.1

19 Physics 103: Lecture 8, Pg 18 Problem 4.59 and 4.60 Continued F’ 25 o F cos25 o F’ cos25 o -F sin25 o F’ sin25 o F  s (ice-on-ice)=0.1 Case A Case B mg N N’


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