Newton s second law
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Newton’s Second Law. Chapter 6. The Second Law. Force = mass X acceleration S F = ma S F = 0 or S F = ma -Still object -Accelerating object -Obj. at constant velocity. Sum of all the forces acting on a body Vector quantity. The Second Law. Situation One:

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The second law
The Second Law

Force = mass X acceleration

SF = ma

SF = 0 or SF = ma

-Still object -Accelerating object

-Obj. at constant velocity

Sum of all the forces acting on a body

Vector quantity


The second law1
The Second Law

Situation One:

Non-moving Object

  • Still has forces

Force of the material of the rock

Force of gravity

http://alfa.ist.utl.pt/~vguerra/Other/Rodin/thinker.jpg


The second law2
The Second Law

Situation Two: Moving Object: Constant Velocity

SF = 0

Fpedalling = Fair + Ffriction

Fpedalling

Fair

Ffriction

http://2003.tour-de-france.cz/images/foto/05-07-2003/armstrong4.jpg


The second law3
The Second Law

Situation Two: Moving Object: Accelerating

SF = ma

ma = Fpedalling – Fair - Ffriction

Fpedalling

Fair

Ffriction

http://2003.tour-de-france.cz/images/foto/05-07-2003/armstrong4.jpg


The second law4
The Second Law

Unit of Force = the Newton

SF=ma

SF = (kg)(m/s2)

1 N = 1 kg-m/s2 (MKS)

1 Newton can accelerate a 1 kg object from rest to 1 m/s in 1 s.


Equilibrium
Equilibrium

  • No motion

  • Constant velocity

    BOTH INDICATE NO ACCELERATION

    SF=0


Three ropes are tied together for a wacky tug-of-war. One person pulls west with 100 N of force, another south with 200 N of force. Calculate the magnitude and direction of the third force.

?

100 N

200 N


A car with weight 15,000 N is being towed up a 20 person pulls west with 100 N of force, another south with 200 N of force. Calculate the magnitude and direction of the third force.o slope (smooth) at a constant velocity. The tow rope is rated at 6000 N. Will it break?


Accelerating systems
Accelerating Systems person pulls west with 100 N of force, another south with 200 N of force. Calculate the magnitude and direction of the third force.

  • SF=ma

  • Must add up all forces on the object


What Force is needed to accelerate a 5 kg bowling ball from 0 to 20 m/s over a time period of 2 seconds?


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


A 1500 kg car is pulled by a tow truck. The tension in the rope is 2500 N and the 200 N frictional force opposes the motion. The car starts from rest.

a. Calculate the net force on the car

b. Calculate the car’s speed after 5.0 s


A 500.0 gram model rocket (weight = 4.90 N) is launched straight up from rest by an engine that burns for 5 seconds at 20.0 N.

  • Calculate the net force on the rocket

  • Calculate the acceleration of the rocket

  • Calculate the height and velocity of the rocket after 5 s

  • Calculate the maximum height of the rocket even after the engine has burned out.



Mass vs weight
Mass vs. Weight below. (53.3 N, +11.0

Mass

  • The amount of matter in an object/INTRINSIC PROPERTY

  • Independent of gravity

  • Measured in kilograms

    Weight

  • Force that results from gravity pulling on an object

  • Weight = mg (g = 9.8 m/s2)


Mass vs weight1
Mass vs. Weight below. (53.3 N, +11.0

  • Weight = mg is really a re-write of F=ma.

    • Weight is a force

    • g is the acceleration (a) of gravity

  • Metric unit of weight is a Newton

  • English unit is a pound


A 60.0 kg person weighs 1554 N on Jupiter. What is the acceleration of gravity on Jupiter?


Elevator at constant velocity
Elevator at Constant Velocity acceleration of gravity on Jupiter?

a= 0

SF = FN – mg

ma = FN – mg

0 = FN – mg

FN = mg

Suppose Chewbacca has a mass of 102 kg:

FN = mg = (102kg)(9.8m/s2)

FN = 1000 N

FN

mg

a is zero


Elevator accelerating upward
Elevator Accelerating Upward acceleration of gravity on Jupiter?

a = 4.9 m/s2

SF = FN – mg

ma = FN – mg

FN = ma + mg

FN = m(a + g)

FN=(102kg)(4.9m/s2+9.8 m/s2)

FN = 1500 N

FN

mg

a is upward


Elevator accelerating downward
Elevator Accelerating Downward acceleration of gravity on Jupiter?

a = -4.9 m/s2

SF = FN – mg

ma = FN – mg

FN = ma + mg

FN = m(a + g)

FN=(102kg)(-4.9m/s2+9.8 m/s2)

FN = 500 N

FN

mg

a is down


At what acceleration will he feel weightless
At what acceleration will he feel weightless? acceleration of gravity on Jupiter?

FN = 0

SF = FN – mg

ma = FN – mg

ma = 0 – mg

ma = -mg

a = -9.8 m/s2

Apparent weightlessness occurs if a > g

FN

mg


A 10.o kg present is sitting on a table. Calculate the weight and the normal force.

FN

Fg = W


Suppose someone leans on the box, adding an additional 40.0 N of force. Calculate the normal force.


Now your friend lifts up with a string (but does not lift the box off the table). Calculate the normal force.


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


Free body diagrams ex 3
Free Body Diagrams: Ex. 3 N?

A person pulls on the box (10.0 kg) at an angle as shown below. Calculate the acceleration of the box and the normal force. (78.0 N)

Fp = 40.0 N

30o

FN

mg


Friction
Friction N?

  • Always opposes the direction of motion.

  • Proportional to the Normal Force (more massive objects have more friction)

FN

Ffr

Ffr

Fa

FN

Fa

mg

mg


Friction1
Friction N?

Static -opposes motion before it moves (ms)

  • Generally greater than kinetic friction

  • Fmax = Force needed to get an objct moving

    Fmax = msFN

    Kinetic - opposes motion while it moves (mk)

  • Generally less than static friction

    Ffr = mkFN


Friction and rolling wheels
Friction and Rolling Wheels N?

Rolling uses static friction

  • A new part of the wheel/tire is coming in contact with the road every instant

B

A


Braking uses N?kinetic friction

Point A gets drug across the surface

A


A 50.0 kg wooden box is pushed across a wooden floor ( N?mk=0.20) at a steady speed of 2.0 m/s.

  • How much force does she exert? (98 N)

  • If she stops pushing, calculate the acceleration. (-1.96 m/s2)

  • Calculate how far the box slides until it stops. (1.00 m)


A 100 kg box is on the back of a truck ( N?ms = 0.40). The box is 50 cm X 50 cm X 50 cm.

a. Calculate the maximum acceleration of the truck before the box starts to slip.



Inclines
Inclines push or pull her?

What trigonometric function does this resemble?


Inclines1
Inclines push or pull her?

FN

mg

q


Inclines2
Inclines push or pull her?

FN

q

mgcosq

mg

q

mgsinq


Inclines3
Inclines push or pull her?

FN

Ffr

mgsinq

mgcosq

q


A 50.0 kg file cabinet is in the back of a dump truck ( push or pull her?ms = 0.800).

  • Calculate the magnitude of the static friction on the cabinet when the bed of the truck is tilted at 20.0o (170 N)

  • Calculate the angle at which the cabinet will start to slide. (39o)


Given the following drawing: push or pull her?

  • Calculate the acceleration of the skier. (snow has a mk of 0.10) (4.0 m/s2)

  • Calculate her speed after 4.0 s? (16 m/s)



F components:Gy = mgcos30o

FGx = mgsin30o

The pull down the hill is:

FGx = mgsin30o

The pull up the hill is:

Ffr= mkFN

Ffr= (0.10)(mgcos30o)


S components:F = pull down – pull up

SF = mgsin30o– (0.10)(mgcos30o)

ma = mgsin30o– (0.10)(mgcos30o)

ma = mgsin30o– (0.10)(mgcos30o)

a = gsin30o– (0.10)(gcos30o)

a = 4.0 m/s2

(note that this is independent of the skier’s mass)


To find the speed after 4 seconds: components:

v = vo + at

v = 0 + (4.0 m/s2)(4.0 s) = 16 m/s


Suppose the snow is slushy and the skier moves at a constant speed. Calculate mk

SF = pull down – pull up

ma = mgsin30o– (mk)(mgcos30o)

ma = mgsin30o– (mk)(mgcos30o)

a = gsin30o– (mk)(gcos30o)


Since the speed is constant, acceleration =0 speed. Calculate

0 = gsin30o– (mk)(gcos30o)

(mk)(gcos30o) = gsin30o

mk= gsin30o= sin30o = tan30o =0.577

gcos30o cos30o


Drag speed. Calculate

D ≈ ¼Av2

D = drag force

A = Area

V = velocity

Fails for

  • Very small particles (dust)

  • Very fast (airplanes)

  • Water and dense fluids


Finding acceleration with drag
Finding Acceleration with Drag speed. Calculate

Derive the formula for the acceleration of a freefalling object including the drag force.


Terminal speed
Terminal Speed speed. Calculate

  • Find the formula for terminal speed (a=0) for a freefalling body

  • Calculate the terminal velocity of a person who is 1.8 m tall, 0.40 m wide, and 75 kg. (64 m/s)


A 1500 kg car is travelling at 30 m/s when the driver slams on the brakes (mk = 0.800). Calculate the stopping distance:

  • On a level road. (57.0 m)

  • Up a 10.0o incline (48.0 m)

  • Down a 10.0o incline (75.0 m)


A dogsled has a mass of 200 kg. The sled reaches cruising speed, 5.0 m/s in 15 m. Two ropes are attached to the sled at 10.0o, one on each side connected to the dogs. (mk = 0.060)

  • Calculate the acceleration of the sled. (0.833 m/s)

  • Calculate T1 and T2 during the acceleration period. (140 N)


Formula wrap up
Formula Wrap-Up speed, 5.0 m/s in 15 m. Two ropes are attached to the sled at 10.0o, one on each side connected to the dogs. (

SF=ma

Weight = mg (g = 9.8 m/s2)

Fmax = msFN

Ffr = mkFN