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)