1. Objectives: The student will be able to: Draw an accurate free body diagram locating each of the forces acting on an object or a system of objects. Use free body diagrams and Newton's laws of motion to solve word problems. Use the methods of vector algebra to determine the net force acting on an object.

2. The Statue of Liberty has a mass of 225,000 kg. How much does she weigh? Wg W = m * g Write the formula: Substitute known values: W9.80 m/s² W = (225,000 kg) * 9.80 m/s² Present solution with units: W 2205000 N W = 2205000 N

3. 2205000 The Statue of Liberty weighs 2205000 Newtons, which is 495,704 pounds! If she is exerting a 495,704 lb force down, the how much does Liberty Island have to push up on her to maintain static equilibrium? 495,704 lb, according to Newton’s 3rd law Note: 1 N = 0.22 lbs

4. A scalar is simply a number, a magnitude alone. A force is usually shown as a vector, which includes both magnitude and a direction. Force (or free-body) diagrams show the relative magnitude and direction of all forces acting upon an object. The object must be isolated and “free” of its surroundings.

5. This is a free-body diagram of the Statue of Liberty. She is represented by a simple box. The forces acting on her are labeled with a magnitude and the arrow shows direction. Notice the surrounding objects are stripped away and the forces acting on the object are shown. 495,704 lb

6. W W here represents the force of the weight of the statue. N N is the normal force, which represents the force Liberty Island is pushing back up on the statue. The island has a great resistance to compression. The ground is exerting a force upward on the statue perpendicular, or normal, to the surface. 495,704 lb N = W =

7. Think of the diagram on an XY plane. If “up” is assumed to be the positive direction, then N is positive and W is negative. lb 495,704 lb N = N = W = W = (Positive y-direction) +y +x (Positive x-direction)

8. The first line of this calculation reads, The sum of the Forcesin the positive y direction WN “The sum of the Forces in the positive y direction is W + N” (  is the Greek symbol for “sum” ) +   F y WN +   F y = W + N - lb+ lb  F y = (-495704 lb) + (+495704 lb )  F y = 0 lb 495,704 lb N = N = W = W = (Positive y-direction) +y +x (Positive x-direction) The sum of the forces in the y is zero. The forces acting on the object cancel each other out.

9. aWe know F = m * a, where “a” is acceleration. a000If a = 0, then F = m * 0 = 0. When  F = 0, the object is not accelerating. We we can then say that the forces acting on the object cancel each other out and it is in a state of static equilibrium.

10. Sitting Gorilla Free Body Diagram of the Sitting Gorilla (The box represents the gorilla, W = weight of the gorilla, N = Normal force) W N Create a free body diagram (FBD) for each of the following situations. Draw a FBD of the gorilla:

11. This is also an acceptable diagram. N W Sitting Gorilla Create a free body diagram (FBD) for each of the following situations. Draw a FBD of the gorilla:

12. Parrot on wooden swing hung by ropes Draw a FBD of the wooden swing: Free Body Diagram of the wooden swing (The box represents the wooden swing, W = weight of the swing and the parrot, T represents the ropes that are in tension supporting the weight) W T2T1

13. Bungee jumping from crane Draw a FBD of bucket the bungee jumper leaped from:

14. Bungee jumping from crane Draw a FBD of bucket the bungee jumper leaped from: Free Body Diagram of the bucket (T represents the tensile force of the cable the bucket is suspended from, and W is the weight of the diver and the bucket) WT

15. Traffic Light supported by cables Draw a FBD of the ring at point C: A B C D

16. Traffic Light supported by cables Draw a FBD of the ring at point C: A B C D Free Body Diagram of the ring at point C (T represents the force of the cables that are in tension acting on the ring) T CA T CD T CB

17. Draw a FBD of the traffic light: Traffic Light supported by cables A B C D

18. Draw a FBD of the traffic light: Free Body Diagram of the traffic light (T CD represents the force of the cables acting on the light and W is the weight acting on the light) WT CD Traffic Light supported by cables A B C D

19. Pin-Connected Pratt Through Truss Bridge Draw a FBD of the pin at point A: A B E D C

20. Pin-Connected Pratt Through Truss Bridge Draw a FBD of the pin at point A: A B E D C Free Body Diagram of pin A (If you consider the third dimension, then there is an additional force acting on point A into the paper: The force of the beam that connects the front of the bridge to the back of the bridge.) T AE T AC T AB T AD

21. 4-7 Solving Problems with Newton’s Laws – Free-Body Diagrams (work out problem) 1. Draw a sketch. 2. For one object, draw a free-body diagram, showing all the forces acting on the object. Make the magnitudes and directions as accurate as you can. Label each force. If there are multiple objects, draw a separate diagram for each one. 3. Resolve vectors into components. 4. Apply Newton’s second law to each component. 5. Solve.

22. Example 4-11 (Solve)

23. 4-7 Solving Problems with Newton’s Laws – Free-Body Diagrams (Solve Example 4-12) When a cord or rope pulls on an object, it is said to be under tension, and the force it exerts is called a tension force.

24. An Atwood’s machine is two masses connected by a strong light string that are hung over an ideal pulley (light and frictionless). The masses have identical velocity and acceleration magnitudes at every instant. If we define up on the left and down on the right as positive directions, then the masses have identical velocities and accelerations period. This simplifies the analysis a lot.

25. Atwood Machine 100 kg Would this move?

26. Atwood Machine 200 kg 100 kg Would this move? Which way? Forces? Fg = w FTFT FTFT Let’s break this up and analyze each mass…

27. The most natural choice that makes up on the left and down on the right positive directions is to focus on the pulley. Up on the left and down on the right are the same to the pulley – both are clockwise rotations. This the coordinate system to use for this problem. cw

28. Let’s look at the clockwise forces that act on the system of two masses and the affect they have on their acceleration. cw

29. Atwood Device m 1 m 2 m1gm1g T T m2gm2g Assume m 1 < m 2 and that the clockwise direction is +. If the rope & pulley have negligible mass, and if the pulley is frictionless, then T is the same throughout the rope. If the rope doesn’t stretch, a is the same for both masses.

30. Atwood Analysis Remember, clockwise has been defined as +. 2 nd Law on m 1 : T - m 1 g = m 1 a 2 nd Law on m 2 : m 2 g - T = m 2 a Add equations: m 2 g – m 1 g = m 1 a + m 2 a (The T ’s cancel out.) Solve for a : m 2 – m 1 m 1 + m 2 a = g m 1 m 2 m1gm1g T T m2gm2g

31. Atwood as a system Treated as a system (rope & both masses), tension is internal and the T ’s cancel out (one clock-wise, one counterclockwise). F net = (total mass)  a implies (force in + direction) - (force in - direction) = m 2 g - m 1 g = (m 1 + m 2 ) a. Solving for a gives the same result. Then, knowing a, T can be found by substitution. m 1 m 2 m1gm1g T T m2gm2g

32. Atwood: Unit Check m 2 – m 1 m 1 + m 2 a = g kg - kg kg + kg m s 2 m = Whenever you derive a formula you should check to see if it gives the appropriate units. If not, you screwed up. If so, it doesn’t prove you’re right, but it’s a good way to check for errors. Remember, you can multiply or divide scalar quantities with different units, but you can only add or subtract quantities with the same units! units:

33. Besides units, you should also check a formula to see if what happens in extreme & special cases makes sense. m 2 >> m 1 : In this case, m 1 is negligible compared to m 2. If we let m 1 = 0 in the formula, we get a = (m 2 / m 2 )g = g, which makes sense, since with only one mass, we have freefall. m 2 << m 1 : This time m 2 is negligible compared to m 1, and if we let m 2 = 0 in the formula, we get a = (-m 1 / m 1 )g = -g, which is freefall in the negative (counterclockwise) direction. m 2 = m 1 : In this case we find a = 0 / (2m 1 )g = 0, which is what we would expect considering the device is balanced. Note: The masses in the last case can still move but only with constant velocity! Atwood: Checking Extremes m1 m1 m2 m2 m 2 – m 1 m 1 + m 2 a = g

34. Do sample Atwood Machine Problem

35. An Atwood’s machine has m1 = 2 kg, m2 = 3 kg, hung from an ideal pulley. What is the acceleration of the masses? Calculate the tension in the string attached to each mass.

36. Modified Atwood Machine and Lab for 2 nd Law Will need to solve for acceleration, draw FBD, and solve for tension. Do example. Discuss theoretical aspect of the lab.

37. Homework for Chapter 4 Problems # 19, 23, 25, 26, 31