Forces and Newton's Three Laws

Force

quantitative measure of a push or pull

vector (magnitude and direction)

causes a change in motion

two objects must be involved (force requires an agent)

two types:

1) contact

2)noncontact

labeling:

FAB (force of A on B)

Fbox, hand (force of box by hand)

Contact force

forces that touch an object

normal, friction, spring, tension

Noncontact force

long range force

gravity, electromagnetism, etc

Net Force (Fnet)

all the forces acting on the system added together (vector addition)

Newton's 1st Law

an object remains in state of motion or at rest unless acted upon by a net force (Galileo's law of inertia)

If Fnet = 0, v is constant or vice versa

v is constant -> delta v is 0 -> a is 0 so Fnet is 0

Newton's 2nd Law

Fnet = ma = m x (delta v/delta t) with units of Kg x m/s^2 or N (Newton)

If Fnet = 0, a = 0 so delta v is 0

Fnet is in the direction of a

In 2D:

1) Fnet x = max

2) Fnet y = may

Newton's 3rd Law

when two objects interact, thy have magnitudes that are equal and have opposite signs

FAB = -FBA

-forces always come in pairs

criteria:

1) same type of force

2) subscripts switch FAB -> -FBA

3) force pairs have different FBD (not on same FBD)

Free Body Diagram

1. represent a system as a point particle

2. put tails of forces on object

3. only draw forces acting on system (forces touching system or gravity)

4. label forces by magnitudes (no signs)

5. choose a coordinate system

6. do not add arrows for Fnet or a on FBD

Weight/Gravitational Force (Fg/W/Fmg)

force done by gravity

Earth is the agent carrying out the force

m and g are both constant in motion or at rest

g = 9.8 m/s^2

Free fall: gravity is only force for any object , a = g

Normal Force (Fn/N)

perpendicular to surface

when force pushes on ground/wall, atoms kind of act like a spring and push back

doesn't have a constant value so it is often found when we have net force and other forces acting on system

ex. N = mg sometimes or N have other values

Spring Force (Fs)

Fs = -k (x-xo)

-opposite direction of displacement when spring from equilibrium (restoring force)

Spring constant k = N/m (Newton/meters)

Compressed spring - pushing force on object, x < 0 (push spring to left, displacement is negative and spring pushes right (positive force) to return to equilibrium (original length)

Stretched spring - pulling force on object, x > 0 (push spring to right, displacement is positive and spring push left (negative force) to return to equilibrium (original length)

Tension Force (T)

rope-like object is inextensible (does not stretch) and have noticeable mass (ignore m, ex T1 = t2)

direction is equal to the direction rope is pulled

Friction Force (Ff)

direction is against direction of motion

two types: Kinetic (fk) and Static (fs)

Kinetic friction (fk) - when object slides across a surface, opposes motion, opposite direction from velocity, constant value

fk = muk (coefficient of fk)*N (normal)

Static friction (fs) - keeps object at rest at a surface, in direction that prevents motion , larger than fk, not constant

fs < or = to mus (coefficient of fs) * N (normal)

fs, max = mus (coefficient of fs) * N (normal)

Drag Force (D)/Air resistance

opposes motion in gas/liquid

direction that opposes motion

Force pairs/action-reaction pairs

are internal forces

At equlibirum,

when all forces cancel each other out, Fnet is 0

Pulley problems (Atwood Machine)

ideal:

1) pullet: no mass, friction at top

2) string: no mass, no friction, no stretch

ex. masses (mB > mA)

same acceleration since objects are moving together, tension is same

a↑|O| ↓a

| ▢B

A□

An air-track glider slides at constant speed. There is no friction and air resistant is assumed to be negligible. What can you say about the net force (total force) on the glider?

a) Fnet = 0

b) Fnet does not equal 0, to the right

c) Fnet does not equal 0, to the left

d) Fnet does not equal 0, perpendicular to the motion

correct answer:

Fnet = 0 because v is constant -> delta v is 0 -> a = 0

A sailboat is blown straight across the sea at a constant velocity. What is the direction of the net force on the boat?

a) left

b) right

c) down

d) the net force is 0

correct answer:

The net force is 0 since v is constant -> delta v is 0 -> a= 0

Two forces act on the same object: vector F1 an vector F2 both have the same magnitude F but are at right angles to each other. What is the magnitude of the net force (total force) acting on the object? ↑ →

a) F

b) 2F

c) something between F and 2F

d) more than F

e) not enough information

correct answer:

something between F and 2F

Fnet = F1 + F2

Fnet^2 = F1^2 + F2^2 = 2F^2

Fnet = square root of 2F = 1.41 F

A moving van collides with a small car in a high speed head on collision (everyone survives). During the impact, the van exerts a force Fvan on the car and the car exerts a force Fcar on the van. Which statement about these forces is true?

a) Fvan = Fcar

b) Fvan > Fcar

c) Fvan < Fcar

d) not enough information

correct answer:

Fvan = Fcar

according to Newton's third law:

FAB = -FBA

The car and van exerts equal force on each other but the car accelerates more since it is lighter than the van

A heavy duty truck pulling a light cart is accelerating forward. How does the magnitude of the force on the truck from the cart compare to the magnitude of the force on the cart from the truck (ignore rope and assume contact)

a) F on truck by car > F on cart by truck

b) F on truck by cart < F on cart by truck

c) F on truck by cart = F on cart by truck

correct answer:

F on truck by cart = F on cart by truck

according to Newton's third law:

FAB = -FBA

so the magnitudes of both forces must be equal and opposite in direction

An object is being lowered on a cord at constant speed. Assume no air resistance. How does the magnitude of tension T in the cord compared to the magnitude of the weight mg of the object? There is a constant downward velocity applied.

|

0 v↓

a) T = mg

b) T > mg

c) T < mg

d) not enough information

correct answer:

T = mg

if we draw a FBD and set y+ upwards,

we can see that T and W have equal magnitudes

Then if we apply Newton's 2nd Law:

Fnet = ma = 0

0 = +T - W

T = W

T = mg

An object is being lowered on a cord at a speed which is decreasing with time (Assume only forces acting are weight and tension) What is the direction of acceleration? Decreasing speed is applied.

|

0 v↓

a) up

b) down

c) a = 0

d) not enough information

correct answer:

up

To have decreasing speed, acceleration needs to have a sign opposite of velocity. Velocity is pointing down in this scenario, so the direction of acceleration should be up.

Homeyra (H) and a Sumo Wrestler (W) are having a tug of war. So far, no one is winning. What is the direction of the force of friction on Homeyra's feet by the ground

H --- W

a) right

b) left

correct answer:

left

if we draw a FBD, the Ff on Homeyra by ground is pointing to the left which is opposite to the direction of motion (which is tension pulled by Sumo wrestler, the tension force is pulling Homerya to the right)

Ho large is the force of friction fs on Steve's feet compared to the force of friction fw on the Wrestler's feet? (no one is winning)

a) fsteve > fw

b) fsteve = fw

c) fsteve < fw

correct answer:

fsteve = fw

if no one is winning, then the force of friction for both the Wrestler and Steve must be equal in magnitude to the Forces applied by the rope on Steve and the Wrestler

You're on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by your seat is equal to your weight mg. How does N change at the tip of the Ferris wheel when you are in motion.

a) N remains equal to mg

b) N is smaller than mg

c) N is larger than mg

d) none of the above

correct answer:

N is smaller than mg

At the very top, Gravity (mg) is pointing downward towards the center and Normal force (N) is pointing upward away from the center

If we apply Newton's 2nd law and y+ as down,

Fnet = ma

Fnet y = +m(v^2/R)

mg - N = +m (v^2/R)

N = mg - m (v^2/R) < mg

Conical Pendulum: Consider the force of gravity W = mg and the tension T that both act on the pendulum. Pendulum moves in a flat circle and doesn't move up and down. Y+ is up and x+ is to the right. Which force is larger?

a) W > T

b) T > W

C) W = T

correct answer:

T > W

ay = 0

Fnet = ma

Fnet y = 0

since Fnet y = 0

Ty = W

A component of T cancels out W, so T > W

An Atwood's machine is a pulley with two masse connected by a string as shown. The mass of object A, mA, is less than the mass of object B, mB. The tension T in the string on the left, above mass A is?

a↑|O| ↓a

| ▢B

A□

a) T = mAg

b) T = mBg

c) neither of these

d) both A and B are true

e) ???

correct answer: neither of these

A is accelerating up and B is accelerating down

y+ is set as up

FBD for A: T and WA have equal magnitudes

Newton's 2nd Law on A:

Fnet A = mAa

Fnet Ay = mAay

T - mAg = mAa

T = mA (a + g)

Fnet A is up not 0 so T can't equal mAg

Fnet B is down so it is not 0 so T can't equal mBg

FBD for B: T > WB in regards to magnitude

Newton's 2nd Law on B:

Fnet B = mBaB

T - mBg = mb (+/- a)

T - mBg = mb (- a)

T = mB (-a + g)

a = g (mB - mA)/(mB + mA)