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## Force

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**Force**Chapter 4**Isaac Newton (1642 to 1727)**• Born 1642 (Galileo dies) • Invented calculus • Three laws of motion • Principia Mathematica**Newton’s Three Law’s of Motion**• All objects remain at rest or in uniform, straight-line motion unless acted upon by an outside force. (inertia) • Force = mass X acceleration • Every force has an equal and opposite force.**The First Law**Inertia –tendency of an object to remain at rest or in constant motion. mass - measure of inertia. Mass & inertia are directly proportional**The First Law**Ball in jar example:**The First Law**Direction of jar**The First Law**Now stop pushing the jar (same as not wearing a seatbelt) Jar gets stopped Marble keeps going**The First Law**School bus example This is the way you want to keep going The bus has turned, so you feel pulled to the side.**The First Law**Bee in a car example • Bee is in air when car starts • Bee is on the seat when car starts**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**• Non-accelerating (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**The Second Law**Situation One: Non-moving Object • Still has forces Force of the material of the rock Force of gravity**The Second Law**Situation Two: Moving Object SF = ma ma = Fpedalling – Fair - Ffriction Fpedalling Fair Ffriction**The Second Law**• Unit of Force = the Newton SF=ma SF = (kg)(m/s2) 1 N = 1 kg-m/s2 • A force of 1 Newton can accelerate a 1 kg object from rest to 1 m/s in 1 s.**The Second Law: Example 1**A 60.0 kg bike and rider accelerates at 0.5 m/s2. How much extra force did the rider’s legs have to provide? SF = ma = (60.0 kg)(0.5 m/s2) SF = 30 kg m/s2 = 30 N**The Second Law: Example 2**A force of 5 Newtons can accelerate a watermelon 2.5 m/s2. What is the mass of the watermelon? (No actual watermelons were harmed in the production of this example problem)**The Second Law: Example 3**What Force is needed to accelerate a 5 kg bowling ball from 0 to 20 m/s over a time period of 2 seconds?**The Second Law: Example 4**Calculate the net force required to stop a 1500 kg car from a speed of 100 km/h within a distance of 55 m. 100 km/h = 28 m/s v2 = vo2 + 2a(x-xo) a = (v2 - vo2)/2(x-xo) a = 02 – (28 m/s)2/2(55m) = -7.1 m/s2**SF = ma = (1500 kg)(-7.1 m/s2) = -1.1 X 104 N**(negative sign tells us that the force is the in the opposite direction of motion) Direction of original motion Fbrakes**The 3rd Law**“For every force, there is an equal and opposite force.”**The 3rd Law**Runner example: • Does the runner push on the earth? • Why does the runner move more? • Does the earth move at all? Backward force for the earth Forward force for the runner**The 3rd Law**Fforward Fgases**The 3rd Law**Sled of bricks on Ice: • Would the sled move? ICE**The 3rd Law**Why would Cyclops be in trouble?**Mass vs. Weight**Mass (kg) • Amount of matter in an object • Independent of gravity (the # of atoms in an object does not change) Weight (N) • Force from gravity pulling on an object • Weight = mg • English unit is a pound**Cereal Boxes**• Metric unit of mass = kilogram • English unit of mass = slug 1 lb = 4.45 N 2.2 lb equivalent to 1 kg**Weight Example 1**What is the weight of a 60.0 kg man?**Weight Example 2**If a freshman weighs 500 N, what is her mass?**Weight Example 4**A 60.0 kg person weighs 100.0 N on the moon. What is the acceleration of gravity on the moon? Weight = mgmoon gmoon = Weight/m gmoon = 100.0 N/60.0 kg = 1.67 m/s2 (Note that this is almost exactly 1/6th of earth’s gravity)**Calculate the weight of a 56 kg person:**• On earth (549 N) • On the moon (g=1.7 m/s2) (95.2 N) • Suppose that person is taken to a different planet and has a weight of 672 N. Calculate the acceleration of gravity on that planet. • 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)**The Normal Force**Normal Force – The force exerted by a surface perpendicular to the contact FN FN Fg = W Fg = W**Normal Force: Example 1a**A 10.0 kg present is sitting on a table. Calculate the weight and the normal force. FN Fg = W**Since the box is not moving SF = 0**SF = FN – W 0 = FN – 98.0 N FN = +98.0 N**Normal Force: Example 1b**Suppose someone 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 = FN – W – 40.0 N 0 = FN – 98.0 N – 40.0 N FN = +138.0 N**Normal Force: Example 1c**Now your friend lifts up with a string (but does not lift the box off the table). Calculate the normal force.**Again, the box is not moving so SF = 0**SF = FN+40.0 N – W 0 = FN + 40.0 N – 98.0 N FN = +58.0 N**Normal Force: Example 1d**What happen when the person pulls upward with a force of 100 N? SF = FN+ Fp – mg SF = 0 +100.0N – 98N = 2.0N ma = 2N a = 2N/10.0 kg = 0.2 m/s2 Fp = 100.0 N Fg = mg = 98.0 N**Example**A 2.0 kg ball is thrown into the air with a force of 30.0 N. • Calculate the acceleration. (5.2 m/s2) • Calculate speed at the top of the throw, 2.00 m. (4.56 m/s) • Draw a free body diagram while the ball is being tossed. • Draw a free body diagram while the ball is above the thrower.**Free Body Diagrams: Ex. 1**Calculate the sum of the two forces acting on the boat shown below. (53.3 N, +11.0o)**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. • Calculate the acceleration of the basketball. (0.938 m/s2) • Calculate the force the player exerted on the ball. (6.01 N) • Draw a free body diagram of the ball while being thrown. • Draw a free body diagram of the ball in the air. • Calculate how long it will take the ball to reach the peak height from when it leaves the player’s hands. (0.153 s)**Free Body Diagrams: Ex. 2**A hockey puck slides at constant velocity across the ice. Which of the following is the correct free-body diagram?**A person drags the box (10.0 kg) at an angle as shown below.**• Calculate the acceleration of the box (3.46 m/s2) • Calculate the normal force. (78.0 N) Fp = 40.0 N 30o FN mg**Here is the free body diagram**Fpx = (40N)(cos30o)= 34.6N Fpy = (40N)(sin30o) = 20.0 N Fp = 40.0 N 30o FN mg**Your lawnmower has a mass of 18.0 kg and the handle makes**60.0o angle with the ground. You push with a force of 9.00 N. • Calculate acceleration. (0.25 m/s2) • Calculate the normal force. (184 N) • How far will the lawnmower 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 35.0o with the horizontal. The child and sled experience an acceleration of 1.64 m/s2. in the horizontal direction. • Calculate the horizontal force on the sled. (82 N) • Calculate the Force of the pull (the hypotenuse) (100N) • Calculate the vertical force of the pull on the sled. (57.4 N) • Calculate the normal force on the sled. (433 N)