Forces

1. Forces Homework: Physics: Handout #1 – 9, 12, 14, 15, but all are highly recommended.

2. Forces: push pull A force is a _____________ or a ____________ on an object by another object.

3. Forces can be applied either through direct contact, as shown in the picture below (a), (b), and (c). Forces shown on the right are called “action at a distance”, where no physical contact is required between the objects to cause a force.

4. inertia A force is tied to the motion of an object through ________________ . resist Inertia is the ability of an object to ________________ a change in its motion. mass Inertia is measured as the__________ of an object. The standard unit is the kilogram, kg. The more massive an object is, the more it is able to resist a change in its motion. It is easier to push a 10 kg object into motion than it is a 3000 kg car. dynamics All of this begins the topic of _______________, or why objects move.

5. Newton’s Laws of Motion: The relationship between motion and force was stated clearly by Isaac Newton in his 3 laws. Each law relates a different aspect of motion. Newton’s First Law: If the total, or net, force on an object is zero, then an object will not accelerate. First variation: If an object is at rest, it will continue to remain at rest until acted upon by some external agent. Example: A book placed on a desk will remain on the desk until someone removes it. There are forces acting on the book: gravity pulls down on it and the table pushed upwards on it. These forces balance out, and the object does not move.

6. Newton’s First Law: continued Second variation: If an object is moving, it will continue to move with a constant velocity. Since velocity is a vector, this means the speed (how fast) and the direction remain unchanged. Example: Take a look at a hockey game. When the puck (small black projectile) is hit, the puck travels at nearly the same speed and in a straight line across the ice. All real objects are subject to friction for anything sliding on a surface, and an icy surface is as close to frictionless as we can get.

8. Newton’s Second Law: If the net force on an object is not zero, then an object will accelerate. The acceleration of the object is directly proportional to the total force acting upon it, and the acceleration is inversely proportional to the mass of the object. As an equation: Here, m is the mass of the object, measured in kg. a is the acceleration of the object, measured in m/s2. This is a vector quantity F refers to the amount of force. Fnet or SF is the sum of the forces acting on an object. Force is a vector, so it will have magnitude and direction. The net force and the acceleration point in the same direction.

9. Units: Force is measured in units called newtons, and this unit is represented by the letter N. Newtons can be written in term of fundamental units through the statement of Newton’s Second Law: Example #1: What is the acceleration of a 2.00 kg mass if a force of 4.00 N acts upon it?

10. Example #2: An object with a mass of 3.00 kg moves from rest to a speed of 20.0 m/s in a time of 5.00 s. a. What is the acceleration of the object? and solve for a: Start with: b. What is the net force acting on the object? Both force and acceleration are vectors. The net force points in the same direction as the acceleration of the object.

11. Example #2: c. Let the object’s acceleration point to the right. If there is a drag force of 30.0 N opposing the motion, what other applied force must there be on the object to give the desired acceleration? A picture helps!

12. For one dimension, the vector signs are dropped. It is understood all vectors point along the same line. Just use a (+) sign for the vectors pointing right and a (–) sign for the vectors pointing left. The variables are assumed to be only the magnitude of the vectors.

13. Newton’s Third Law: If one object (object A) exerts a force on another object (object B), then object B exerts the same force back on object A. The force is equal in magnitude, but opposite in direction. Alternate interpretation: For every action there is an equal but opposite reaction. Example: If you drive your fist against a wooden wall as hard as you can, you will possibly leave a mark on the wall with your fist. The mark is made by the force you exerted against the wall. The wall also exerts an equal force on your fist. You perceive this as the severe pain felt in your knuckles. object A object B Drawing: FB on A FA on B

14. Example #3: A student pushes with a 30.0 lb force towards the South on a heavy box. There is a friction (drag) force between the floor and the box that prevents the movement of the box. What is the magnitude and direction of the friction force? Since the box does not move, the net force must be zero.

15. Example #4: A baseball is thrown at a speed of 40.0 m/s towards the right. The ball is hit by a baseball bat, and then travels at 50.0 m/s towards the left. The ball is in contact with the bat for 2.00 ms and the ball has a mass of 200 grams. What is the net force acting on the ball? Initial information: Let right be positive and left be negative. vo = +40.0 m/s, v = – 50.0 m/s, t = 2.00 x 10-3 s, m = 0.200 kg

16. Example #5 A horse is hitched to a wagon. Which statement is correct? A. The force that the horse exerts on the wagon is greater than the force that the wagon exerts on the horse. B. The force that the horse exerts on the wagon is less than the force that the wagon exerts on the horse. C. The force that the horse exerts on the wagon is just as strong as the force that the wagon exerts on the horse. D. The answer depends on the velocity of horse and wagon. E. The answer depends on the acceleration of horse and wagon.

17. Example #5 A horse is hitched to a wagon. Which statement is correct? A. The force that the horse exerts on the wagon is greater than the force that the wagon exerts on the horse . B. The force that the horse exerts on the wagon is less than the force that the wagon exerts on the horse. C. The force that the horse exerts on the wagon is just as strong as the force that the wagon exerts on the horse. D. The answer depends on the velocity of horse and wagon. E. The answer depends on the acceleration of horse and wagon.

18. Example #6 You are standing at rest and begin to walk forward. What force pushes you forward? A. the force of your feet on the ground B. the force of your acceleration C. the force of your velocity D. the force of your momentum E. the force of the ground on your feet

19. Example #6 You are standing at rest and begin to walk forward. What force pushes you forward? A. the force of your feet on the ground B. the force of your acceleration C. the force of your velocity D. the force of your momentum E. the force of the ground on your feet

20. Table Cloth Pull: An example of the Law of Inertia.

21. Tension, Gravity, and Normal Force Homework: Physics Day 1: #1 – 7, 10 Physics Day 2: #8, 9, 11 – 14, 15 a,b,c.

22. Definitions: Forces are any kind of push or pull on a system or on an object. There are some special types of forces that we will define today. The first is the weight of an object. gravity The weight of an object is the force of ___________ acting on the object by Earth. mass The amount of force is directly proportional to the ____________ of the object. The equation for weight is as follows: w = weight of the object. This is a force measured in N (newtons). m = mass of the object, in kg (kilograms). g = acceleration of gravity, 9.80 m/s2. The direction of the force is downwards, towards the center of Earth.

23. Example #1: What is the weight of a 55.0 kg person? Example #2: In the British system, mass is measured in slugs and force is measured in pounds. Weight is still measured as mg, with one slug mass weighing 32.2 pounds. Convert the mass of the person above into slugs. Conversion Factor: 1.0000 kg mass weighs 2.2046 lbs

24. normal Definition: The __________ force on an object is defined as the force exerted on an object by a surface. The term “normal” is another term used in mathematics for perpendicular. The normal force exerted by a surface is always perpendicular to that surface. The normal force is represented by the letter _________ . Do not confuse this with newtons of force! n Problem Solving: Always draw a picture, and label all the forces acting on the object. This is referred to as drawing a “Free Body Diagram”. Only draw the forces acting on the object. By Newton’s third law, the forces on an object result also in forces exerted by the object on the surroundings. We are only interested in the forces ontheobject by the environment.

25. Example #3: A 1.4 kg book is placed on a flat surface at rest. What is the normal force acting on the book? Start with a labeled picture: normal = n Next apply Newton’s Second Law: m = 1.4 kg For one dimension, define up as positive and down as negative. The forces may be written as follows: weight = mg The variables show the magnitude, or size of the vector. The ± sign gives the direction. and

26. The net force becomes: Since the object is at rest, the acceleration of the object is zero. Combine the above equations and solve for n.

27. Example #4: {Classic Physics Problem!!!} A person stands in an elevator on a scale. (Don’t ask why…..) When the elevator is at rest, the scale reads 588 N. What will the scale read if the elevator accelerates upwards at 2.20 m/s2? Solution: The scale is designed to measure a force. The scale does not measure the weight of the person, rather the force exerted upwards on the person to support them. In other words, the scale measures the normal force acting on the person. When the person is at rest, the normal force equals the weight of the person. When the person is accelerated in the elevator, the net force on the person is no longer zero. For an upward acceleration, the normal force must be larger than the weight of the person. Let’s find out why.

28. Before anything, determine the mass of the person: At rest, the person weighs 588 N. This equals mg, so the mass is:

29. up = positive Start by drawing a complete picture and then apply Newton’s Second Law. normal = n Draw in a coordinate axis and choose a positive direction. elevator Apply Newton’s 2nd Law: person, m weight = mg down = negative

30. Substitute numbers: Solve for the new normal force for the accelerated elevator. The person appears to weigh more when the elevator accelerates upwards.

31. Example #5: What is the motion of the elevator if the person’s weight appears to be 360 N? Just follow the same reasoning as above and show again: Solve this expression for a. Since up was chosen as the positive direction, a negative acceleration means that the elevator is accelerating downwards.

32. Example #6: What would the scale read if the elevator was in freefall? Just follow the same reasoning as above and show again: Solve this expression for n. Note that a = – g. For freefall, the scale would read zero. The person would feel weightless.

33. pulling tension Definition: A __________ force is a ____________ force, usually applied through a string or rope. Ropes are to be treated as ideal, meaning that the mass of the ropes can be ignored and the ropes do not stretch or sag when used. The force of tension is then the same throughout the rope.

34. Example #7: Two masses rest on a frictionless horizontal surface. The mass on the left is 60.0 kg and the mass on the right is 40.0 kg. The masses are tied together with a light cord and a 25.0 N force is applied towards the right on the 40.0 kg mass. Find the acceleration of the system and the tension force in the cord. applied force Fa = 25.0N m1 = 60.0 kg m2 = 40.0 kg cord

35. Solution: First treat the two masses as a single object. They move the same: same velocity and acceleration. applied force Fa = 25.0N mtotal = m1 +m2 Apply Newton’s 2nd Law:

36. Now isolate one mass to solve for the force of tension in the cord. If the mass on the left (m1) is chosen, the solution is as follows: T = tension force m1 Apply Newton’s 2nd Law and solve for the tension force.

37. An alternate solution would be to choose the mass on the right (m2). applied force Fa = 25.0N T = tension force m2 Apply Newton’s 2nd Law and solve for the tension force. Let the positive direction be towards the right.

38. Example #8: A mass (m1 = 10.0 kg) hangs from the ceiling of an elevator from a light cord. A second mass (m2 = 20.0 kg) is suspended from the first mass by another cord. If the elevator accelerates upwards at a rate of 4.70 m/s2, determine the tension in each cord. up = positive cord with tension T1 m1 elevator cord with tension T2 m2 down = negative

39. Solution: First treat the two masses as a single object and solve for the tension in the top cord. Start with a simple picture. Apply Newton’s 2nd Law and solve for the tension force. Let the positive direction be upwards. T1 mtot mtotg

40. Now isolate one mass to solve for the force of tension in the lower cord. If the mass on the bottom (m2) is chosen, the solution is as follows: Apply Newton’s 2nd Law and solve for the tension force. Let the positive direction be upwards. T2 m2 m2g

41. If the mass on the top (m1) is chosen, the solution is as follows: Apply Newton’s 2nd Law and solve for the tension force. Let the positive direction be upwards. T1 m1 T2 m1g

42. Q4.1 Motor An elevator is being lifted at a constant speed by a steel cable attached to an electric motor. There is no air resistance, nor is there any friction between the elevator and the walls of the elevator shaft. The upward force exerted on the elevator by the cable is Cable v Elevator A. greater than the downward force of gravity. B. equal to the force of gravity. C. less than the force of gravity. D. any of the above, depending on the speed of the elevator.

43. A4.1 Motor An elevator is being lifted at a constant speed by a steel cable attached to an electric motor. There is no air resistance, nor is there any friction between the elevator and the walls of the elevator shaft. The upward force exerted on the elevator by the cable is Cable v Elevator A. greater than the downward force of gravity. B. equal to the force of gravity. C. less than the force of gravity. D. any of the above, depending on the speed of the elevator.

44. Q4.9 A woman pulls on a 6.00-kg crate, which in turn is connected to a 4.00-kg crate by a light rope. The light rope remains taut. Compared to the 6.00–kg crate, the lighter 4.00-kg crate A. is subjected to the same net force and has the same acceleration. B. is subjected to a smaller net force and has the same acceleration. C. is subjected to the same net force and has a smaller acceleration. D. is subjected to a smaller net force and has a smaller acceleration. E. none of the above

45. A4.9 A woman pulls on a 6.00-kg crate, which in turn is connected to a 4.00-kg crate by a light rope. The light rope remains taut. Compared to the 6.00–kg crate, the lighter 4.00-kg crate A. is subjected to the same net force and has the same acceleration. B. is subjected to a smaller net force and has the same acceleration. C. is subjected to the same net force and has a smaller acceleration. D. is subjected to a smaller net force and has a smaller acceleration. E. none of the above

46. 2 – Dimensional Forces Homework: Handout #3, all problems.

47. a T3 T2 T1 3m 2m m ConcepTest 4.17 Three Blocks Three blocks of mass 3m, 2m, and m are connected by strings and pulled with constant acceleration a. What is the relationship between the tension in each of the strings? 1) T1 > T2 > T3 2) T1 < T2 < T3 3) T1 = T2 = T3 4) all tensions are zero 5) tensions are random

48. a T3 T2 T1 3m 2m m ConcepTest 4.17 Three Blocks T1 pulls the wholeset of blocks along, so it must be the largest. T2 pulls the last two masses, but T3 only pulls the last mass. 1) T1 > T2 > T3 2) T1 < T2 < T3 3) T1 = T2 = T3 4) all tensions are zero 5) tensions are random

49. vector magnitude Forces are _________ quantities: they have both a _____________ and a ____________. direction For Newton’s 2nd Law, both the net force and the acceleration are vectors. It is easier to resolve the vectors into components to solve 2 – dimensional problems. On any given problem, draw in a coordinate set and label which directions are positive . Example #1: A 40.0 kg block rests on a smooth surface. A force of 400 N is applied to the block at an angle of 36.9o above the horizontal. a. Label all forces acting on the object, and resolve these forces into horizontal and vertical components. Fa n = normal q = 36.9o m mg = weight

50. Fa n = normal q = 36.9o +y m +x mg = weight The normal force has only a y – component: nx = 0, ny = +n The weight has only a y – component: wx = 0, wy = –mg The x – component of the applied force, Fa, is: Fa,x = +Facosq The y – component of the applied force, Fa, is: Fa,y = +Fasinq