1. Today: (Ch. 3) Tomorrow: (Ch. 4) Forces and Motion in Two and Three Dimensions Equilibrium and Examples Projectile Motion

2. Example A passenger weighing 598 N rides in an elevator. The gravitational field strength is 9.8 N/kg. What is the apparent weight of the passenger in each of the following situations? In each case the magnitude of elevator’s acceleration is 0.5 m/s 2. (a)The passenger is on the 1 st floor and has pushed the button for the 15 th floor i. e. the elevator is beginning to move upward. (b)The elevator is slowing down as it nears the 15 th floor.

3. Statics and Equilibrium Statics  Deals with objects at rest Statics is an area of mechanics dealing with problems in which both the velocity and acceleration are zero The object is also said to be in translational equilibrium  Often the “translational” is dropped

4. According to Newton’s 2 nd law Equilibrium and Newton’s law This vector quantity can be written in terms of two perpendicular components When object in equilibrium the net force acting on it is zero: For object to be in equilibrium

5. Equilibrium on an inclined plane h d φ φ y x φ Identify all the forces in the system Choose the axes, which are convenient to work with…

6. Using Newton’s Second Law Newton’s Second Law Determine all the individual forces acting on the object Construct a free body diagram Add the individual forces as vectors Use Newton’s Second Law to find the acceleration Acceleration can be used to determine velocity and displacement

7. Equilibrium Example Forces acting in A –Gravity and normal force in y- direction –Force exerted by person (push) and static friction in x-direction Free body diagram in B Forces in x and y components and apply condition of equilibrium ΣF x = 0 and ΣF y = 0 For y-direction: N - m g = 0 For x-direction: F push - F friction = 0

8. Equilibrium Example 2 All the forces do not all align with the x- or y-axes Find the x- and y-components of all forces that are not on an axis Applying Newton’s Second Law: –ΣF x = T x – F friction = T cos θ – F friction = 0 –ΣF y = N – mg + T y = N – mg + T sin θ = 0

9. Equilibrium Example 3 Both sections of the rope exert a tension force at the center where the walker is standing The walker and the rope are at rest The forces acting at the center –Tension on the right and on the left & Weight of the walker

10. Problem Solving Strategy for Statics Problems Recognize the principle –For static equilibrium, the sum of the forces must be zero –Use Sketch the problem –Show the given information in the picture –Include a coordinate system Identify the relationships –Use all the forces to construct a free body diagram –Express all the forces on the object in terms of their x- and y-components –Apply ΣF x = 0 and ΣF y = 0 –May also include ΣF z = 0

11. Problem Solving Strategy for Statics Problems, cont. Solve –Solve all the equations –The number of equations must equal the number of unknown quantities Check –Consider what your answer means –Check that your answer makes sense

12. Inclines (Hills) Normal force (N) : perpendicular to the incline (plane) Friction force : up the incline –Opposite to motion The force due to gravity acts straight down Coordinate system –Axes parallel and perpendicular to the incline (If) Acceleration : along the incline Components of the gravitational force The normal force is not equal to mg

13. Angle of Incline To Not Slip The minimum frictional force to keep the object from slipping is F friction = m g sin θ Since this is static friction, F friction ≤ μ static N Assuming it is just in equilibrium (so F friction = μ static N), the angle of the incline at which the object is on the verge of slipping is tan θ = μ s

14. Equilibrium Example, Flag Find out tension and angle (two unknowns so 2D problem) Free body diagram –Horizontal and vertical directions : coordinate system –Tension has x- and y- components Equations for equilibrium in the x- and y-directions

15. Motion in Two Dimensions Two dimensions is just like one dimension, done twice: for the x components, then for the y components. The x and y motions are independent of each other! The x part of the motion occurs exactly as if the y part did not exist. And, the y part of the motion occurs exactly as if the x part did not exist.

16. Projectile Motion The x and y motions are independent of each other! In general, for the x components:  assume that there is no air resistance, so that a x = 0  then v x = constant = v 0x In general, for the y components:  a y = 9.8 m/s 2 down  then v y changes

17. Acceleration due to gravity is always 9.8 m/s 2 downward, throughout the path of the projectile! x v y θ

18. Projectile Motion Consider objects in motion and the forces acting on them Projectile motion is one example of this type of motion We will ignore the force from air drag –For now Components of gravity are F grav, x = 0, F grav, y = - m g

19. Calculate the distance between the window and the ceiling using kinematics A roofing tile falls from rest off the roof of a building. An observer from across the street notices that it takes 0.54 s for the tile to pass between two windowsills that are 2.5 m apart. How far is the sill of the upper window from the roof of the building? HW Question

20. The Three Equations

21. Tomorrow: (ch 4) Projectile Motion Reference Frames & Relative Velocity Example Involving Newton’s Law