Newton's Laws of Motion Dr. Robert MacKay Clark College, Physics

Introduction Newtons 3 laws of motion 1. Law of inertia 2. Net Force = mass x acceleration ( F = M A ) 3. Action Reaction Newtons Universal Law of Gravity

Isaac Newton 1689 Isaac Newton 1702 Isaac Newton 1726 Isaac Newton Knighted by Queen Anne 1705

Other topics Why do objects accelerate? Why do objects not accelerate? Forces in Balance (Equilibrium) Forces out of Balance Friction Air resistance Terminal Velocity

Law of inertia (1st Law) Every object continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed upon it. acceleration = 0.0 unless the objected is acted on by an unbalanced force

Law of inertia (1st Law) Inertia (The intrinsic tendency of an object to resist changes in motion) Mass is a measure of an objects inertia Mass is also a measure of the amount of an objects matter content. (i.e. protons, neutrons, and electrons) Weight is the force upon an object due to gravity

Newtons 2nd Law Net Force = Mass x Acceleration F = M A

Newtons Law of Action Reaction (3rd Law) You can not touch without being touched For every action force there is and equal and oppositely directed reaction force

Newtons 2nd Law Net Force = Mass x Acceleration F = M A M=2.0 kgF=? A= 6.0 m/s 2 F=M A = 2.0 kg x 6.0 m/s 2 =12.0 Newtons = 12.0 N

An object experiences a net force and exhibits an acceleration in response. Which of the following statements is always true? (a) The object moves in the direction of the force. (b) The acceleration is in the same direction as the velocity. (c) The acceleration is in the same direction as the force. (d) The velocity of the object increases.

Newtons 2nd Law Net Force = Mass x Acceleration F = M A M=2.0 kgF=6.0 N A= ? A=F / M = ? m/s 2

Newtons 2nd Law Net Force = Mass x Acceleration F = M A M=?F= 10.0 N A= 20.0 m/s 2 M= F/A = ? kg

Newtons 2nd Law Net Force = Mass x Acceleration F = M A M= 8.0 kgF= ? N A= 10.0 m/s 2 F=M A = ? N

Newtons 2nd Law Net Force = Mass x Acceleration F = M A F= ? N F= weight A= 10.0 m/s 2 F=M A = ? N M= 8.0 kg

Weight W = m g g = 10 m/s 2 weight is the force due to the gravitational attraction between a body and its planet W= ? m= 6.0 kg

Question 1: A force of 45 N pulls horizontally on a 15 kg crate resting on a level frictionless surface. (Actually the crate has real good tiny wheels) What is the acceleration of the crate? m/s/s m/s/s m/s/s m/s/s

Question 2: A 15.0 kg crate is in contact with a 30.0 kg crate on a level frictionless surface as shown. If the 15.0 kg mass is pushed with a force of 45.0 N what is the acceleration of the two masses? A. 1.0 m/s/s B. 1.5 m/s/s C. 2.0 m/s/s D. 3.0 m/s/s

Question 3: A 15.0 kg crate is in contact with a 30.0 kg crate on a level frictionless surface as shown. If the 15.0 kg mass is pushed with a force of 45.0 N what is the force that the 15.0 kg mass exerts on the 30.0 kg mass? 15 N 6 28% 20 N 2 9% 25 N 1 4% 30 N 12 57%

Weight W = m g W= ? 8.0 kg

A baseball of mass m is thrown upward with some initial speed. A gravitational force is exerted on the ball (a) at all points in its motion (b) at all points in its motion except at the highest point (c) at no points in its motion

Newtons 2nd Law Net Force = Mass x Acceleration F = M A M= 5.0 kg F= N A= ? m/s 2 Net Force = ? D =120.0 N

Newtons 2nd Law Net Force = Mass x Acceleration F = M A M= 5.0 kg F= N A= 20.0 m/s 2 Net Force = ? D=?

Newtons 2nd Law Net Force = Mass x Acceleration F = M A M= 5.0 kg F= N A= 0.0 m/s 2 Net Force = ? D=?

Which of the following statements is most correct? (a) It is possible for an object to have motion in the absence of forces on the object. (b) It is possible to have forces on an object in the absence of motion of the object. (c) Neither (a) nor (b) is correct. (d) Both (a) and (b) are correct.

An object experiences no acceleration. Which of the following cannot be true for the object? (a) A single force acts on the object. (b) No forces act on the object. (c) Forces act on the object, but the forces cancel.

On Earth, where gravity is present, an experiment is performed on a puck on an air hockey table, with negligible friction. A constant horizontal force is applied to the puck and its acceleration is measured. The experiment is performed on the same puck in the far reaches of outer space where both friction and gravity are negligible. The same constant force is applied to the puck and its acceleration is measured. The pucks acceleration in outer space will be a) greater than its acceleration on Earth, b) less than its acceleration on Earth, c) exactly the same as its acceleration on Earth, d) infinite since neither friction nor gravity are holding it back? (end of section 5.5) QUICK QUIZ 5.3

Given : 1 =60 degrees 2 =45 degrees T 3 =200 N (45 lbs) Find: T 1 and T 2 X Y T1 -T 1 cos( 1 ) T 1 sin( 1 ) T2 +T 2 cos( 2 ) T 2 sin( 2 ) T N R 0 0

30 kg 20 kg F = 200 N Find: A & T T

30 kg 20 kg Find: a & T T

30 kg 20 kg Find: a & T T Friction=80 N ( k = 0. 27)

Newtons 2nd Law Friction depends on surfaces in contact (roughness) contact force pushing surfaces together M= 5.0 F= N A= 0.0 m/s 2 Net Force = ? Friction ?

f s,max Applied Force=Static frictional force F=fs Static friction kinetic friction F f s,max = s N f K = K N (sliding friction) f K = K N F net =P-f K P f

Air ResistanceForce Depends on: velocity Air density Shape and aerodynamics of object

Terminal Velocity When air resistance force balances an objects weight w Air Drag Acceleration= 0.0 ===> Terminal velocity

Terminal Velocity w Air Drag Acceleration = 0.0 ===>Terminal velocity w Air Drag 80 kg10 kg which has the greatest force of air resistance?

Not Terminal Velocity Acceleration = ? W = ? Air Drag = 240N 80 kg which has the greatest force of air resistance?

Not Terminal Velocity Acceleration = ? W = 800 N Air Drag = 240N 80 kg +x

Not Terminal Velocity Acceleration = ? W = 800 N Air Drag = 240N 80 kg +x F = m a +800N - 240N= 80kg a +560N=80kg a a=7.0 m/s2 Down

1.A falling elephant, or 2.A falling feather? Which encounters the greater force of air resistance

1. A falling elephant, or 2.A falling feather? Which encounters the greater force of air resistance

Newtons 2nd Law Free Body diagrams A rock in Free Fall

Newtons 2nd Law Free Body diagrams A rock in Free Fall w

Free Body diagrams A rock At the top of its parabolic trajectory (no air resistance)

Free Body diagrams A rock At the top of its parabolic trajectory (no air resistance) w

Newtons 2nd Law Free Body diagrams A rock in Free Fall w

Newtons 2nd Law Free Body diagrams A rock Falling with air resistance

Newtons 2nd Law Free Body diagrams A rock Falling with air resistance w R

Newtons 2nd Law Free Body diagrams A rock on a string

Newtons 2nd Law Free Body diagrams A rock on a String w T For the FBD is acceleration, a, up or down ?

Newtons 2nd Law Free Body diagrams A rock sliding without friction Vo

Newtons 2nd Law Free Body diagrams A rock sliding without friction w N

Newtons 2nd Law Free Body diagrams A rock sliding with friction

Newtons 2nd Law Free Body diagrams A rock sliding with friction wW=mg N FkFk +x +y

A rock or car sliding with friction wW=mg N FkFk +x +y

A rock or car sliding with friction wW=mg N FkFk +x +y

1. Sketch the problem 2. Draw a FBD with all forces 3. Label x and y axes 4. Resolve vectors into components 5. Use: to help solve the problem 6. Think

Elevator

#50. A boy pulls on a box of mass 30 kg with a Force of 25 N in the direction shown. (a) What is the acceleration of the box? (b) What is the normal force? (ignore friction)

#50. A boy pulls on a box of mass 30 kg with a Force of 25 N in the direction shown. (a) What is the acceleration of the box? (b) What is the normal force? (ignore friction) w N Fy=25sin(30)=12.5N Fx=25cos(30)=21.7N

#50. A boy pulls on a box of mass 30 kg with a Force of 25 N in the direction shown. (a) What is the acceleration of the box? (b) What is the normal force? (sliding WITH friction) w N Fy=25sin(30)=12.5N Fx=25cos(30)=21.7N f= k N

A car coasting without friction

w=mg N x y W=mg

A car coasting without friction w=mg N x y W=mg wxwx wywy W x =Wsin W y =Wcos

A car coasting without friction w=mg N x y W=mg wxwx wywy W x =Wsin W y =Wcos

A rock or car sliding without friction w=mg N x y W=mg wxwx wywy W x =Wsin W y =Wcos

A rock sliding with friction

w=mg N x y W=mg FkFk

A rock or car sliding without friction w=mg N x y W=mg wxwx wywy W x =Wsin W y =Wcos FkFk

A rock or car sliding without friction w=mg N x y W=mg wxwx wywy W x =Wsin W y =Wcos FkFk

A rock or car sliding without friction w=mg N x y wxwx wywy W x =Wsin W y =Wcos FkFk

A rock or car sliding without friction w=mg N x y wxwx wywy W x =Wsin FkFk

1. Sketch the problem 2. Draw a FBD with all forces 3. Label x and y axes 4. Resolve vectors into components Use: to help solve the problem 5. Think

Atwoods Machine m1 m2

Atwoods Machine m1 m2 +m 1 g -m 2 g -x +x T T

a 1 x g If m 1 =m 2 =m a=0 T=2m 2 /(2m) T=mg

F=80 N m1=4.0 kgm2 = 6.0 kg µ k =0.0 Find the acceleration a And the tension T in the cord connecting the two blocks +x +y

F=80 N m1=4.0 kgm2 = 6.0 kg µ k =0.0 Find the acceleration a And the tension T in the cord connecting the two blocks +x +y F=80 N N1 N2 w2w1 TT

F=80 N m1=4.0 kgm2 = 6.0 kg µ k =0.30 Find the acceleration a And the tension T in the cord connecting the two blocks +x +y

F=80 N m1=4.0 kgm2 = 6.0 kg µ k =0.30 Find the acceleration a And the tension T in the cord connecting the two blocks N1 N2 w2w1 TT F k2 F k1 N 2 =m 2 g F k2 = 0.3m 2 g N 1 =m 1 g F k1 = 0.3m 1 g