Chapter 4: Newton's Laws of p Motion
Chapter p 4: Newton’s Laws of Motion We will study classical motion: • No q quantum mechanics • No relativity We introduce the concept of force and define it in i terms t off the th acceleration l ti off a “standard “t d d body”
Intuitively, i i l we know k that h force f is i a “push” “ h” or “pull”. Forces come in different classes (types): Contact Field (or action at a distance)
Force F is a vector quantity
F
In 1686, 1686 Newton presented The Laws of Motion: Newton’s First Law Every body continues in its state of rest or of p in a straight g line as longg as uniform speed no net force acts on it.
Examples 1. An object that is moving and that continues to move with constant velocity when the summation of the forces acting on it is zero. 2. An object at rest that remains at rest. 3 What about pushing a chair? 3. 4. What happens when you turn a corner quickly in your car?
We know from experience that different objects resist a change in motion differently. Example: • push a door • push a train ⇒ Not N t th the same response!!
Mass The tendency of an object to resist a change in its velocity is called inertia. The measure of inertia is mass. – SI units of mass is the kilogram (kg).
Newton’s First Law tells us about motion if F = 0. 0 What if F ≠ 0?
Newton’ss Second Law Newton The acceleration of an object is directly proportional to the h resultant l force acting i on it i andd inversely i l proportional to its mass. The direction of the acceleration is the direction of the resultant force. force
G G F = ma
The resultant force acting on the mass m.
The Vector Nature of Forces In the formula F = ma, F is the total (net or resultant) lt t) force f acting ti on the th object. bj t We W mustt consider the vector sum of all forces acting on an object. bj t We W can also l consider id eachh dimension separately: ∑ Fx
= ma x
∑ Fy
= ma y
∑ Fz
= ma z
Example An object of mass 5 kg undergoes an acceleration of 8 m/s2 ŷ. What is the force on that object? F = ma = (5 kg)(8 m/s2) = 40 kg⋅m/s2 ŷ The force is in the same direction as the acceleration.
U it Units The SI unit of Force is the Newton defined as: 1 N = 1 kg⋅m/s2 For comparison: 1 lb. = 4.448 N
Example p A catcher stops a 92 mi/h pitch in his glove, glove bringing it to rest in 0.15 m. If the force exerted by the catcher is 803 N,, what is the mass of the ball?
Newton’s Third Law action and re-action If object 1 exerts a force F on object 2, 2 then object 2 exerts a force –F on object 1. –F Forces come iin pairs. i – The force pairs act on different objects. – The Th forces f have h the h same magnitude i d andd opposite direction. Example: I push on the wall with a force of 20 N. The wall pushes back on me with a force of 20 N in the opposite direction.
Weight The weight of any object on the Earth is the gravitational force exerted on it by the Earth: W = mg Note: Weight is a force (and therefore a vector). Weight is not equivalent to mass, but it is proportional. person’s weight g be zero? Can a p When we say we want to “lose weight”, what do we really mean?
G G W = mg Mass “m”
W = mg
Notice the direction of the weight!
Problem Solving 1. Draw a sketch of the situation. 2. Consider only one body at a time and draw a free body diagram for it, showing all of the forces acting on it. 3. Resolve all of the vectors into x and y components. 4. Apply Newton’s laws to each component separately. For example, evaluate the net force in the x direction. 5. Solve the equation for the unknown.
Free-Body Diagrams Drawing a picture always helps us understand goingg on! what is g Consider a tug of war between two companies of rats Al h Alpha
E h Echo
What is the net force on the rope? FE
Fr
Since FE is larger than FA, Echo company wins.
FA
Free-Body Diagrams Continued
What about a box sitting on the floor? The normal force, FN, is the force acting perpendicular to the surface to the surface the box is sitting on. FN
∑ Fx = W + FN = − mg + FN = 0 FN = mg
W=mg
Example What is the net force on a 10 kg g box that is being g pushed across the floor at a constant velocity of 2.5 m/s? Draw a free-body diagram.
What about more complicated systems? Consider two forces y acting on a ball. F1 F2
x
These arrows represent force vectors. All you have to do is add the vectors. Graphically, y x
F1 F2
FNET
In the picture we just drew, assume that force 1 has a magnitude of 10 N directed at 230o relative to the x axis and force 2 has a magnitude of 2 N at an angle of 0o relative to the x axis. Calculated the resultant vector. vector
F1 F2
x
Calculate the vector components, F1x = 10N cos 230o = -6.4N F1y = 10N sin 230o = -7.7N F2x = 2N cos 0o = 2N F2y = 2N sin 0o = 0N So the resultant vector is
Fr = −4.4 4 4 N iˆ − 7.7 7 7 N ˆj Magnitude, g , Fr = (−4.4 N ) 2 + (−7.7 N ) 2 = 8.9 N
Direction, ⎛ Fry ⎞ −1 ⎛ − 7.7 N ⎞ o ⎜ ⎟ θ = tan ⎜ ⎟ = tan ⎜ ⎟ = 60 ⎝ − 4 .4 N ⎠ ⎝ Frx ⎠
-x
θ
−1
-y
Example a pe 34. A 7500 kg helicopter accelerates upward at 0.52 34 0 52 m/s2 while lifting a 1200 kg car. (a) What is the lift force exerted by the rotors on the air? (b) What is the tension in the cable that h connects the h car to the h helicopter? h li ?
Example A box of mass m slides down a frictionless inclined plane with an angle θ relative to the horizontal. (a) Draw a free body diagram (b) What is the acceleration of the box down the ramp? diagram. (c) Does the acceleration depend on mass? (d) If θ = 45° , what is the acceleration of the box?
Example: A farm tractor tows a 4300-kg trailer up a 26° incline at a steady speed p of 3.0 m/s. What force does the tractor exert on the trailer? (Ignore friction.)
Since the trailer is not accelerating, the sum of the forces along g the incline = zero. Ftractor - mg sin θ = 0 Ftractor = 4300 × 9.81× sin 26D =18491.8 N ≈ 18 kN
Example 19. The cable supporting 19 s pporting a 2100 kg elevator ele ator has a maximum ma im m strength of 21,750 N. What maximum upward acceleration can it give the elevator without breaking?
Example 23. An exceptional standing jump would raise a person 0.80 23 0 80 m off the ground. To do this, what force must a 61 kg person exert against the ground? Assume the person crouches a di t distance off 0.20 0 20 m prior i to t jumping, j i andd thus th the th upwardd force f has this distance to act over before he leaves the ground.
Example 11. A fisherman yyanks a fish out of the water with an acceleration of 3.5 m/s2 using very light fishing line that has a breaking strength of 25 N. The fisherman unfortunately loses the fish as the line snaps. snaps What can you say about the mass of the fish?
Example 31. The two forces F1 and F2 shown in the figures below act on a 29.0 29 0 kg object on a frictionless tabletop. tabletop If F1=20.2N =20 2N and F2=26.0N, find the net force on the object and its acceleration for each situation a and b.
Example 6. What average force is required to stop a 1050 kg car in 7.0 s if it is traveling at 90 km/h.
Example 41. A block is given an initial speed of 4.0 m/s up the 22° plane shown h iin the th figure fi below. b l (a) ( ) How H far f up the th plane l will ill it go?? (b) how much time will have elapsed before it returns to its starting position? Ignore friction.
Intuitively, i i l we know k that h force f is i a “push” “ h” or “pull”. Forces come in different classes (types): Contact Field (or action at a distance)
Force F is a vector quantity
F
In 1686, 1686 Newton presented The Laws of Motion: Newton’s First Law Every body continues in its state of rest or of p in a straight g line as longg as uniform speed no net force acts on it.
Examples 1. An object that is moving and that continues to move with constant velocity when the summation of the forces acting on it is zero. 2. An object at rest that remains at rest. 3 What about pushing a chair? 3. 4. What happens when you turn a corner quickly in your car?
We know from experience that different objects resist a change in motion differently. Example: • push a door • push a train ⇒ Not N t th the same response!!
Mass The tendency of an object to resist a change in its velocity is called inertia. The measure of inertia is mass. – SI units of mass is the kilogram (kg).
Newton’s First Law tells us about motion if F = 0. 0 What if F ≠ 0?
Newton’ss Second Law Newton The acceleration of an object is directly proportional to the h resultant l force acting i on it i andd inversely i l proportional to its mass. The direction of the acceleration is the direction of the resultant force. force
G G F = ma
The resultant force acting on the mass m.
The Vector Nature of Forces In the formula F = ma, F is the total (net or resultant) lt t) force f acting ti on the th object. bj t We W mustt consider the vector sum of all forces acting on an object. bj t We W can also l consider id eachh dimension separately: ∑ Fx
= ma x
∑ Fy
= ma y
∑ Fz
= ma z
Example An object of mass 5 kg undergoes an acceleration of 8 m/s2 ŷ. What is the force on that object? F = ma = (5 kg)(8 m/s2) = 40 kg⋅m/s2 ŷ The force is in the same direction as the acceleration.
U it Units The SI unit of Force is the Newton defined as: 1 N = 1 kg⋅m/s2 For comparison: 1 lb. = 4.448 N
Example p A catcher stops a 92 mi/h pitch in his glove, glove bringing it to rest in 0.15 m. If the force exerted by the catcher is 803 N,, what is the mass of the ball?
Newton’s Third Law action and re-action If object 1 exerts a force F on object 2, 2 then object 2 exerts a force –F on object 1. –F Forces come iin pairs. i – The force pairs act on different objects. – The Th forces f have h the h same magnitude i d andd opposite direction. Example: I push on the wall with a force of 20 N. The wall pushes back on me with a force of 20 N in the opposite direction.
Weight The weight of any object on the Earth is the gravitational force exerted on it by the Earth: W = mg Note: Weight is a force (and therefore a vector). Weight is not equivalent to mass, but it is proportional. person’s weight g be zero? Can a p When we say we want to “lose weight”, what do we really mean?
G G W = mg Mass “m”
W = mg
Notice the direction of the weight!
Problem Solving 1. Draw a sketch of the situation. 2. Consider only one body at a time and draw a free body diagram for it, showing all of the forces acting on it. 3. Resolve all of the vectors into x and y components. 4. Apply Newton’s laws to each component separately. For example, evaluate the net force in the x direction. 5. Solve the equation for the unknown.
Free-Body Diagrams Drawing a picture always helps us understand goingg on! what is g Consider a tug of war between two companies of rats Al h Alpha
E h Echo
What is the net force on the rope? FE
Fr
Since FE is larger than FA, Echo company wins.
FA
Free-Body Diagrams Continued
What about a box sitting on the floor? The normal force, FN, is the force acting perpendicular to the surface to the surface the box is sitting on. FN
∑ Fx = W + FN = − mg + FN = 0 FN = mg
W=mg
Example What is the net force on a 10 kg g box that is being g pushed across the floor at a constant velocity of 2.5 m/s? Draw a free-body diagram.
What about more complicated systems? Consider two forces y acting on a ball. F1 F2
x
These arrows represent force vectors. All you have to do is add the vectors. Graphically, y x
F1 F2
FNET
In the picture we just drew, assume that force 1 has a magnitude of 10 N directed at 230o relative to the x axis and force 2 has a magnitude of 2 N at an angle of 0o relative to the x axis. Calculated the resultant vector. vector
F1 F2
x
Calculate the vector components, F1x = 10N cos 230o = -6.4N F1y = 10N sin 230o = -7.7N F2x = 2N cos 0o = 2N F2y = 2N sin 0o = 0N So the resultant vector is
Fr = −4.4 4 4 N iˆ − 7.7 7 7 N ˆj Magnitude, g , Fr = (−4.4 N ) 2 + (−7.7 N ) 2 = 8.9 N
Direction, ⎛ Fry ⎞ −1 ⎛ − 7.7 N ⎞ o ⎜ ⎟ θ = tan ⎜ ⎟ = tan ⎜ ⎟ = 60 ⎝ − 4 .4 N ⎠ ⎝ Frx ⎠
-x
θ
−1
-y
Example a pe 34. A 7500 kg helicopter accelerates upward at 0.52 34 0 52 m/s2 while lifting a 1200 kg car. (a) What is the lift force exerted by the rotors on the air? (b) What is the tension in the cable that h connects the h car to the h helicopter? h li ?
Example A box of mass m slides down a frictionless inclined plane with an angle θ relative to the horizontal. (a) Draw a free body diagram (b) What is the acceleration of the box down the ramp? diagram. (c) Does the acceleration depend on mass? (d) If θ = 45° , what is the acceleration of the box?
Example: A farm tractor tows a 4300-kg trailer up a 26° incline at a steady speed p of 3.0 m/s. What force does the tractor exert on the trailer? (Ignore friction.)
Since the trailer is not accelerating, the sum of the forces along g the incline = zero. Ftractor - mg sin θ = 0 Ftractor = 4300 × 9.81× sin 26D =18491.8 N ≈ 18 kN
Example 19. The cable supporting 19 s pporting a 2100 kg elevator ele ator has a maximum ma im m strength of 21,750 N. What maximum upward acceleration can it give the elevator without breaking?
Example 23. An exceptional standing jump would raise a person 0.80 23 0 80 m off the ground. To do this, what force must a 61 kg person exert against the ground? Assume the person crouches a di t distance off 0.20 0 20 m prior i to t jumping, j i andd thus th the th upwardd force f has this distance to act over before he leaves the ground.
Example 11. A fisherman yyanks a fish out of the water with an acceleration of 3.5 m/s2 using very light fishing line that has a breaking strength of 25 N. The fisherman unfortunately loses the fish as the line snaps. snaps What can you say about the mass of the fish?
Example 31. The two forces F1 and F2 shown in the figures below act on a 29.0 29 0 kg object on a frictionless tabletop. tabletop If F1=20.2N =20 2N and F2=26.0N, find the net force on the object and its acceleration for each situation a and b.
Example 6. What average force is required to stop a 1050 kg car in 7.0 s if it is traveling at 90 km/h.
Example 41. A block is given an initial speed of 4.0 m/s up the 22° plane shown h iin the th figure fi below. b l (a) ( ) How H far f up the th plane l will ill it go?? (b) how much time will have elapsed before it returns to its starting position? Ignore friction.
