Isaac Newton (1642 to 1727) Force. Newton s Three Law s of Motion. The First Law. The First Law. The First Law


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1 Isaac Newton (1642 to 1727) Force Chapter 4 Born 1642 (Galileo dies) Invented calculus Three laws of motion Principia Mathematica. Newton s Three Law s of Motion 1. All objects remain at rest or in uniform, straightline motion unless acted upon by an outside force. (inertia) mass  measure of inertia. Mass & inertia are directly proportional 2. Force = mass X acceleration 3. Every force has an equal and opposite force. Ball in jar example: The First Law The First Law The First Law Now stop pushing the jar (same as not wearing a seatbelt) Direction of jar Marble keeps going Jar gets stopped 1
2 School bus example The First Law The First Law Bee in a car example 1. Bee is in air when car starts 2. Bee is on the seat when car starts This is the way you want to keep going The bus has turned, so you feel pulled to the side. Does it take less force to push the elephant (ignore friction) on earth or on the moon? Does it take less force to move the elephant if he is weightless in space? Inertial Reference Frames Nonaccelerating (constant velocity) reference frame All laws of physics are identical Cannot tell if you are moving in an Inertial Reference Frame Speed of light question The Second Law Force = mass X acceleration SF = ma Sum of all the forces acting on a body 2
3 The Second Law The Second Law Situation One: Constant Velocity Objects Sitting objects still have forces Constant velocity objects SF = 0 Force of the material of the rock Force of gravity Situation Two: Accelerating Object SF = ma ma = F pedaling F air  F friction F F pedaling air F friction The Second Law Unit of Force = the Newton SF=ma SF = (kg)(m/s 2 ) 1 N = 1 kg m/s 2 1 Newton accelerates a 1 kg object from rest to 1 m/s in 1 s. A 60.0 kg bike and rider accelerates at m/s 2. How much extra force did the rider s legs have to provide? A force of 5.00 Newtons can accelerate a watermelon 2.50 m/s 2. What is the mass of the watermelon? SF = ma = (60.0 kg)(0.500 m/s 2 ) SF = 30.0 kg m/s 2 = 30 N (No actual watermelons were harmed in the production of this example problem) 3
4 What Force is needed to accelerate a 5.00 kg bowling ball from 0.00 to 20.0 m/s over a time period of 2.00 seconds? Calculate the net force required to stop a kg car from a speed of km/h within a distance of 55.0 m. (10,500 N) The 3 rd Law SF = ma = (1500 kg)(7.03 m/s 2 ) = 10,500 N (negative sign tells us that the force is the in the opposite direction of motion) For every force, there is an equal and opposite force. F brakes Direction of original motion Runner example: Does the runner push on the earth? Why does the runner move more? Does the earth move at all? F forward F gases Backward force for the earth Forward force for the runner 4
5 Sled of bricks on Ice: Would the sled move? ICE Mass vs. Weight Mass (kg) Amount of matter in an object Independent of gravity (the # of atoms in an object does not change) English unit is a slug Weight (N) Force from gravity pulling on an object Weight = mg English unit is a pound 1 lb = 4.45 N 2.2 lb equivalent to 1 kg What is the weight of a 60.0 kg man? If a freshman weighs 500 N, what is her mass? A 60.0 kg person weighs N on the moon. What is the acceleration of gravity on the moon? Weight = mg moon g moon = Weight/m g moon = N/60.0 kg = 1.67 m/s 2 (Note that this is almost exactly 1/6 th of earth s gravity) Calculate the weight of a 56 kg person: a) On earth (549 N) b) On the moon (g=1.7 m/s 2 ) (95.2 N) c) Suppose that person is taken to a different planet and has a weight of 672 N. Calculate the acceleration of gravity on that planet. (12.0 m/s 2 ) d) If the person dropped a ball from a height of 1.80 m on that planet, calculate the speed that it would hit the ground. (6.57 m/s) 5
6 The Normal Force Normal Force The force exerted by a surface perpendicular to the contact Normal Force: Example 1a A 10.0 kg present is sitting on a table. Calculate the weight and the normal force. F N F N F N F g = W F g = W F g = W Since the box is not moving SF = 0 SF = F N W 0 = F N 98.0 N F N = N Normal Force: Example 1b Suppose Mr. Saba leans on the box, adding an additional 40.0 N of force. Calculate the normal force. Again, the box is not moving so SF = 0 SF = F N W 40.0 N 0 = F N 98.0 N 40.0 N F N = N Normal Force: Example 1c Now Mr. Saba lifts up with a string (but does not lift the box off the table). Calculate the normal force. 6
7 Normal Force: Example 1d Again, the box is not moving so SF = 0 SF = F N N W 0 = F N N 98.0 N F N = N What happens when Mr. Saba pulls upward with a force of 100 N? F p = N F g = mg = 98.0 N A 2.0 kg ball is thrown into the air with a force of 30.0 N. a) Draw a free body diagram while the ball is being thrown. b) Draw a free body diagram while the ball is above the thrower. c) Calculate the acceleration. (5.2 m/s 2 ) d) Calculate speed at the release point, 2.00 m. (4.56 m/s) Calculate the sum of the two forces acting on the boat shown below. (53.3 N, o ) A basketball player throws a basketball through 1.2 m from rest to a speed of 1.50 m/s. The basketball has a mass of 0.56 kg and travels vertically. a) Calculate the acceleration of the basketball. (0.938 m/s 2 ) b) Draw a free body diagram of the ball while being thrown. c) Calculate the force the player exerted on the ball. (6.01 N) d) Draw a free body diagram of the ball in the air. e) Calculate how long it will take the ball to reach the peak height from when it leaves the player s hands. (0.153 s) A hockey puck slides at constant velocity across the ice. Which of the following is the correct freebody diagram? 7
8 A person drags the box (10.0 kg) at an angle as shown below. a. Calculate the acceleration of the box (3.46 m/s 2 ) b. Calculate the normal force. (78.0 N) F p = 40.0 N 30 o Here is the free body diagram F px = (40N)(cos30 o )= 34.6N F py = (40N)(sin30 o ) = 20.0 N F p = 40.0 N 30 o F N F N mg mg Your lawn mower has a mass of 18.0 kg and the handle makes 60.0 o angle with the ground. You push with a force of 9.00 N. a) Calculate acceleration. (0.25 m/s 2 ) b) Calculate the normal force. (184 N) c) How far will the lawn mower have travelled in 2.50 s. (78.1 cm) You pull a child and sled (50.0 kg) with a rope at an angle of 35.0 o with the horizontal. The child and sled experience an acceleration of 1.64 m/s 2. in the horizontal direction. a) Calculate the horizontal force on the sled. (82 N) b) Calculate the Force of the pull (the hypotenuse) (100N) c) Calculate the vertical force of the pull on the sled. (57.4 N) d) Calculate the normal force on the sled. (433 N) A man pushes a stroller (20.0 kg) across the floor. He pushes at an angle of 65.0 o, with a horizontal acceleration of 0.50 m/s 2. a) Calculate the x component of the push. (10.0N) b) Calculate the actual force of the push (23.7 N) c) Calculate the ycomponent of the force (21.4 N) d) Calculate the normal force on the stroller. (217.4 N) e) Calculate the speed of the stroller (from rest) if the stroller accelerates for 3.0 m (1.73 m/s) Example 3 A 10.0 kg box is placed next to a 5.00 kg box. The 10.0 kg box is pushed with a force of 20.0 N. a) Calculate the acceleration of the boxes (1.33 m/s 2 ) b) Calculate the contact force between the two boxes. (6.67 N) 20.0 N 10.0 kg 5.00 kg 8
9 Tension Flexible cord (can only pull) F T or T Neglect mass of the cord 12.0 kg 10.0 kg F p = 40.0 N A 5.00 kg box and a 12.0 kg box are connected by a cord. The 12.0 kg box is pulled horizontally with a force of 60.0 N. a) Calculate the acceleration of both boxes. (3.5 m/s 2 ) b) Calculate the force of tension in the cord. (17.6 N) A kg and a 5.00 kg box are connected by a cord. The boxes are not moving. a. Calculate the tension the cords (49N, 147 N) b. Suppose the boxes accelerate at 0.80 m/s2 upward. Now calculate the tensions. (53 N, 159 N) a) Calculate the acceleration. (3.68 m/s 2 ) b) Calculate the tension in the cord (91.9 N) c) Calculate how far the box will move in 1.50 s. (4.14 m) 25.0 kg 15.0 kg A 5.00 kg and a 2.00 kg box are connected by a cord. The top box (5.00 kg) is lifted by a second cord with an acceleration of m/s 2. a. Calculate the tension in the bottom, connecting cord. (21.1 N) b. Calculate the tension in the top cord. (73.85 N) 9
10 A 5.00 kg and a 2.00 kg box are connected by a cord. The top box (5.00 kg) is lifted by a second cord with a Force of 100 N. a. Calculate the acceleration. (4.5 m/s 2 ) b. Calculate the tension in the bottom, connecting cord. (28.6 N) Atwood s machine Calculate the acceleration of the elevator and the tension in the cable. (Atwood s Machine) Draw freebody diagrams for both the elevator and counterweight Ma = Mg T ma = T  mg M = 1150 kg m = 1000 kg (a = 0.68 m/s 2, F T = 10,500N) A sign is suspended by two cords. Assume the tensions in the two cords are equal at 30.0 N, and the angles are equal at 40.0 o. a. Calculate the mass of the sign. (3.94 kg) b. Now assume the mass of the sign is 10.0 kg. Calculate the new tensions in the cord. (76 N each) 10
11 Static Friction Friction that opposes motion before it moves coefficient of static friction = m s Generally greater than kinetic friction (m k ) f s = m s n Kinetic Friction Friction that opposes motion while it moves coefficient of kinetic friction (unitless) = m k Generally less than static friction f k = m k n Friction: Example 1 A 10.0 kg box is on a table. The coefficients of friction are m s =0.40, m k = What is the force of friction if the force pulling on the box is: a) 0 N b) 25N c) 38N f n d) 40N (calculate a) F p mg First let s calculate the maximum force from the static friction. A 450 g wooden box is pulled at a constant velocity with a force of 1.10 N. Calculate the value of m k. f max = m s n f max = m s mg f max = (0.40)(10.0 kg)(9.8 m/s 2 ) f max = 39 N (0.25) 11
12 A 2.50 kg sled is pulled at a constant velocity. What force is required if the m k is 0.11 on ice? Friction: Example 3 (Ans: 2.7 N) The coefficient of kinetic friction is a) Calculate acceleration (1.4 m/s 2 ) b) Calculate the force of tension in the cord. (16.8 N) m 1 =5.0 kg m 2 =2.0 kg The coefficient of kinetic friction is a) Calculate the acceleration and F T. b) Calculate how far the box will move in 0.50 s kg Two zombies (50.0 kg and kg) are pushed across a floor with a m k of The kg zombie is pushed with a force of N. (Draw a FDBD) a) Calculate the acceleration of the zombies. (1.08 m/s 2 ) b) Calculate the contact force between the zombies. (250 N) 15.0 kg N (2.14 m/s 2, 115 N, 26.8 cm) Your little sister wants a ride on her sled. Should you push or pull her? 12
13 You pull your sister and sled across the snow (m k = 0.200). The total mass is 50.0 kg and you pull at an angle of 60.0 o with a force of N a. Draw a free body diagram b. Calculate the normal force (273 N) c. Calculate the acceleration. (1.41 m/s 2 ) d. Calculate the frictional force (54.6 N) q Inclines mg n Inclines Inclines n mgsinq n = mgcosq f q mg q mgcosq q mgsinq A 75.0 kg sled (and rider) slides down a hill that has a 23.0 o degree incline. a) Calculate the force of acceleration. (287 N) b) Calculate the value of acceleration. (3.83m/s 2 ) c) Calculate the speed of the sled after it has gone 5.00 m. (6.2 m/s) d) Calculate the normal force on the sled. (677 N) A 20.0 kg child slides down a slide with a 45.0 o incline. a) Calculate the force of acceleration. (139 N) b) Calculate the value of acceleration. (6.93 m/s 2 ) c) Calculate the speed of the child after 1.50s (10.4 m/s) d) Calculate the normal force on the child. (139 N) 13
14 A duck slides down a 20.0 o bank. The bank is 2.5 m long and very smooth. a) Calculate the acceleration of the duck. (3.35 m/s 2 ) b) Calculate the duck s speed at the bottom of the bank. (4.1 m/s) c) The normal force of the duck is 9.2 N. Calculate the mass of the duck. (1.00 kg) a) What is the acceleration of the skier if snow has a m k of 0.10 (4.06 m/s 2 ) b) What is her speed after 4.0 s? Free Body Diagram First we need to resolve the force of gravity into x and y components: F Gy = mgcos30 o F Gx = mgsin30 o The pull down the hill is: F Gx = mgsin30 o The pull up the hill is: F fr = m k F N F fr = (0.10)(mgcos30 o ) SF = pull down pull up SF = mgsin30 o (0.10)(mgcos30 o ) ma = mgsin30 o (0.10)(mgcos30 o ) ma = mgsin30 o (0.10)(mgcos30 o ) a = gsin30 o (0.10)(gcos30 o ) a = 4.0 m/s 2 To find the speed after 4 seconds: v = v o + at v = 0 + (4.0 m/s 2 )(4.0 s) = 16 m/s (note that this is independent of the skier s mass) 14
15 Suppose the snow is slushy and the skier moves at a constant speed. Calculate m k A 1.00 kg block is placed on a 37.0 o incline and does not slide down. a) Draw a free body diagram b) Calculate the normal force (7.83 N) c) Calculate the force of acceleration (5.90 N) d) Calculate the minimum value of m s. (0.753) e) If the maximum value of m s is 0.800, calculate the angle at which it will slide? (38.7 o ) A kg filing cabinet slides down a ramp that has a coefficient of sliding friction (m k ) of The ramp has an angle of 25.0 o. a) Calculate the normal force. (888 N) b) Calculate the acceleration of the cabinet. (0.766 m/s 2 ) c) If the coefficient of static friction is (0.72), calculate the angle of the ramp at which it will start slipping. (35.7 o ) The acceleration of the following system is 1.00 m/s 2. The table is not smooth and has friction. a) Calculate the force of tension in the cord. (44.0 N) b) Calculate m k (0.35) c) How long will it take the boxes to move 1.50 m? (1.73s) 10.0 kg 5.0 kg Neglect friction for the following system. Assume the 50.0 kg block will slide down the incline. a. Calculate the tension in the cord (199 N) b. Calculate the acceleration of the blocks (0.157 m/s 2 ) For the following diagram, assume the m k = 0.18, and the ball falls toward the ground. a) Calculate the acceleration of the system (4.86 m/s 2 ) b) Calculate the tension in the cord. (494 N) 50.0 kg 50.0 kg 20.0 kg kg 20 o 25.0 o 15
16 Mr. Saba is on a 5.00 m waterslide. The angle of the slide is 25.0 o and it has a m k of a) Calculate his acceleration. (2.8 m/s 2 ) b) Calculate the time it will take him to reach the bottom of the slide. (1.9 s) Mr. Saba goes to a playground slide at an angle of 35.0 o. He slides down at a constant speed (no acceleration). a) Calculate the value of m k for this slide. (0.70) b) Suppose Mr. Saba puts on a few pounds. Will this now cause him to accelerate down the slide? A crate is on a dumptruck and m s is Calculate the angle at which the load will start to slip. Formula WrapUp SF = ma or SF = 0 Weight = mg F max = m s F N (before it moves) F fr = m k F N Inclined Plane n = mgcosq a = gsinq kg N 6. a) 196 N(both) b) 294 N c) 98.0 N N X 10 3 N in direction of ball s motion X 10 4 N X 10 4 N(max) 4.47 X 10 4 N (min) m/s 2 (down) 18.a) 7.4 m/s 2 (down) b) 1176 N ,000 N, 1.1 m 24. Southwest 26. a) F bat NE, mg S b) mg S 28. b) 62.2 N c) 204 N c) 100N 30. a) 29 N b) 34 N and 68 N 32. a) 320 N b) 0.98 m/s q = 22 o N, if m k =0 no force is required 40. a) static b) kinetic c) kinetic m ,000 N and 1.1 m 16
17 m 46. a) x min = v 2 /m s g b) 47 m c) 280 m 48. Bonnie s (slips if angle is > 8.5 o ) a) 3.67 m/s 2 b) 8.17 m/s 54. a) 1.85 m/s 2 b) 0.82 m up, 1.49s N, m k = a = m/s 2 time = 4.3 s ,000 N, 1.1 m 70. a) 16 m/s b) 12 m/s 74.a) 75.0 kg b) 75.0 kg c) 75.0 kg d) 98.0 kg e) 52.0 kg Force Table Lab Example Calculate the resultant vector (size and direction) of these three forces. 17
18 T 18
19 34 19
20 (0.27) 20
21 (11.6 kg sand) 21
22 A person pushes against a box, the box does not move until the Force reaches 100 N. Person pushes the box for a short while across the floor, and then lets the box slide. There is a friction. Draw a graph of Speed vs. time. You will be asked to draw your graph on the board and explain it. 22
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