PHY10T3FCE&NSLM

NEWTON’S LAW OF MOTION

FORCE (F)Described as either push or pull that can cause a mass (body) to accelerate (cause of motion)

A vector quantity that is the product of mass (m) & acceleration (a).

F = m aIn MKS it is in unit : Newtons (N) , 1 N = 1 kg-m/s2

In CGS it is in unit : dynes , 1 dyne = 1 g-cm/s2

In English Units : Pound (₤ or lbs) , Pound-force (lbf)

CONVERSION :

1 dyne = 1x10-5 N = 10-5 N1 lbs = 4.448 N

COMPONENTS

m

F = 5 N

θ

m Fx = 5 N cos θ

θ

Fy = 5 N sin θ

=

FORCE (F)

1. Force due to Gravity / WeightFGRAV = W = mg, where (for earth) : g = 9.8 m/s2 or 32 ft/s2

m

W = mg

m

W = mgθ

Weight is ALWAYS directed TOWARDS (attractive to) the earth, even if the surface is at an angle.

Weight is ALWAYS acting on the body

KINDS OF FORCES

Weight being a force is a VECTOR. Mass is SCALAR.

It is possible to have zero weight , but mass is not zero. (i.e. objects in outer space)

2. Longitudinal ForcesForces acting along the length of an object

Common in ropes, cables, solid cylindersTENSION (T)

Pull Force on an object. Its end effect is to STRETCH an object.

COMPRESSION (C)

Push Force on an object. Its end effect is to FLATEN an object.

- Compression usually is due to normal forces between two objects in contact

C C

TT

Length (L)

Length (L)

KINDS OF FORCES

3. Contact Forces

Forces due to interaction between different surfacesFRICTIONAL FORCE / FRICTION (f)

Force that oppose motion of an object.

Always parallel to the contact surface & directed opposite the motion of the object

NORMAL FORCE (Ŋ or N)

Reaction Force due to Weight of the object(s) in contact

Always Perpendicular to the contact surface

m

W = mgN

f

m

W = mgθ N

f

KINDS OF FORCES

3. Contact Forcesf α N

f = μ N

μ – Coefficient of Friction

μS – Coefficient of Static Friction

μK – Coefficient of Kinetic FrictionμS = tan θf , (This θ MUST be the Angle of friction or repose)

KINDS OF FORCES

3. Contact Forces

Equations for Normal & Frictional Forces

m

W = mg

m

W = mg

θ

N

N

m

W = mg

β

N

θ

F

N = W = mg N = Wy = W cosθ

N = mgcosθ

N = Wy – Fy

N = mgcosβ – Fsinθ

F

KINDS OF FORCES

f = μN = μmg

f = μN = μmg cosθ f = μN

f = μ(mgcosβ – F sinθ)

FIRST LAW : Law of Inertia

“A body acted by NO net force either stays motionless or moves, but with constant velocity and zero acceleration”


SECOND LAW : Law of Acceleration

“A body requires a net force to accelerate” “The acceleration is directly proportional to the net force but inversely proportional to the body’s mass”

“The direction of the net force is the same as the direction of the acceleration”

THIRD LAW : Law of Action – Reaction

“To every action there is always opposed an equal reaction, same in magnitude but opposite in direction.”


m

W = mg

N m

W = mg

T

EQUILIBRIUM – The effects of all forces acting on a single point on the body cancel one another. There is no change in motion

An object or body is at the state of Equilibrium, when :

(1)The body stays at rest

OR

(2)The body moves in a straight line but in constant or uniform velocity (No acceleration)

For Both :ΣFx = 0 & ΣFy = 0

Therefore : R = 0 or Fnet = 0

FIRST CONDITION OF EQUILIBRIUM (FCE)

FREE BODY DIAGRAM (FBD)Used to analyze forces acting on a body by isolating the body with all forces acting on it.

Steps

1. A. Represent the object as a point mass

B. If it is a system (multiple objects), a certain point on it will be indicated for reference, use this as the “point mass”.

2. Draw ALL forces ACTING ON the body from this point

Internal – WeightExternal – Applied Force, Friction Normal Force (Coming from Other Body in contact), Tension (Always away from the body)

3. Don’t forget to show also the angle of the force (if angled).

WRONG FBD CORRECT FBD

W = mg W = mg

f f

N

N

θ

F

m

W = mgN

f

θ

F

θ

F

TIPS ON FBD DON’T Draw Vectors GOING towards the point mass (even though it may show in the figure) draw it away from the point mass, it does the same thing.

EXAMPLE 1

TIPS ON FBD EXAMPLE 2

m

W = mgθ N

f

WRONG FBD * CORRECT FBD *

W = mg W = mg

f f

Nθ θ

N

* Using the inclined surface as the x-axis

TIPS ON FBD PULLEYS & WeightsPulleys are analyzed as frictionless & of negligible weight.For FCE : Tension of rope or cable passing through the pulley is equal to the weights HANGING from them.

T = Wθ

2 kg

W = mg

T = W

T = W T = W

2 kg

W1 = mg

T = W1

T = W1

T = W1

T = W1

Using the inclined surface as the x-axis (rotate of axis)

m

W = mgθ N

f

1000 lbs

45°

60°

O

Must have at least 1 pair of perpendicular forces and 90° angle is visible.

ALLOWED since f & N are Perpendicular

NOT ALLOWED :No perpendicular forces or angles with respect to point O, thus not practical to use any inclined as x-axis.

TIPS ON FBD

If a system consists of two or more objects. Multiple FBD’s may be required :

A

BP = ? AB

TIPS ON FBD

FIRST CONDITION OF EQUILIBRIUM (FCE)Sample Problems :

1. A 5 kg block will start to slide down at constant speed from a surface when it is inclined at 40° with the horizontal. Determine the Frictional force, Normal force and the coefficient of static friction.

m = 5kg

40°

Solution : Draw the forces acting on the body

fN

Draw the FBD

W = mg

fN

40°m = 5kg

40° W = mg

Since we have perpendicular forces along the inclined (f & N) we can use the inclined as our x-axis

Re-draw the FBD

W = mg

f

N

40°

Draw the component vectors of angled vectors

W

f

N

40°

Wx

Wy


W

f

N

40°

Wx

Wy

ΣFx = 0 → (+)

Fnet = 0

− Wx + f = 0f = Wx = Wsinθ

ΣFy = 0↑ (+)

+ N – Wy = 0N = Wy = Wcosθ

N = mg cos θN = (5kg)(9.8m/s2)(cos 40°)

N = 37.54 N

Friction and Normal force and μS

f = mg sinθf = (5kg)(9.8m/s2)(sin 40°)

f = 31.5 Nf = μ N

μ = f/N μS = f/N = 31.5 N / 37.54 N μS = 0.839

or μS = tan θf = tan 40° μS = 0.839


2. A woman at an airport is pulling a 15 kg suitcase (with wheels) at constant speed of 2 m/s by pulling on the handle attached to the bag (this makes an angle θ above the horizontal). She pulls with a 56 N force, and the frictional force is 20N. What is the angle “θ”, the normal force & the coefficient of kinetic friction?

F = 56 N

m = 15 kg

f = 20 N

θ N

W = mg

F = 56 N

θf = 20 N

N

W = mg

Fy

θf = 20 N

Fx


ΣFx = 0 → (+)

Fnet = 0

− f + Fx = 0f = Fx

ΣFy = 0↑ (+)+ N + Fy – W = 0

N = W – Fy

N = mg – F sin θN = (15kg)(9.8m/s2) – (56N)(sin 69.08°)

N = 94.69 N

Solving for θ

f = F cosθ20 N = 56 N(cos θ)

θ = 69.08°

f = μ N μ = f/N

μK = f/N = 20 N / 94.69 N μK = 0.211

N

W = mg

Fy

θf = 20 N

Fx

,Normal force and μK


3. A wet shirt weighs 4 N. It is hanged to dry on a metal clothesline. The shirt is placed at the very center of the length of the clothesline, and the angle formed with respect to the horizontal due to the weight of the shirt on either side are equal. What are the tensions on each side of the clothesline?

θθ

W = 4 N

T2

θθ

T1

W = 4 N

T2y

θθ

T1y

T1x T2x

2 m

0.4 m

θ1 m

0.4 mtan θ = (0.4 m)/(1 m)θ = 21.8°


ΣFx = 0 → (+)

Fnet = 0

− T1x + T2x = 0T2x = T1x

ΣFy = 0↑ (+)+ T1y + T2y – W = 0T1 sinθ + T2 sinθ = WT1 sinθ + T1 sinθ = W

Solving for Tensions

T2 cosθ = T1cosθT2 = T1

W = 4 N

T2y

θθ

T1y

T1x T2x

2(T1 sinθ) = WT1 = W/ (2sinθ)

T1 = [(4N)/[2sin(21.8°)]

T1 =5.385N T2 = 5.385N


Q : What relates FCE to Newton’s First Law of Motion (NFLM)?

A : FCE is an application of NFLM. Where the object is and must be at

equilibrium.

SECOND LAW

Recall : NEWTON’S LAW OF MOTION

“A body requires a net force to accelerate”

“The acceleration is directly proportional to the net force but inversely proportional to

the body’s mass”

“The direction of the net force is the same as the direction of the acceleration”

m

Body of mass “m” at rest on a frictionless surface

m

Due to net force “Fnet” going to the left the object will accelerate also to the left

Fnet

aFnet = ma

mF

a

W = mgN

a) ΣFx = max & ΣFy = may

b) ΣFx = max & ΣFy = 0

c) ΣFx = 0 & ΣFy = may

If Fnet = maPossibilities :

Purely Horizontal Movement

Purely Vertical Movement

NEWTON’S SECOND LAW OF MOTION (NSLM)

Sample Problems :

1. A 5 kg block slides down a plane inclined at 40° to the horizontal. Find the acceleration of the block

a) If the plane is frictionlessb) If the coefficient of kinetic friction is

0.20

m = 5kg

40°

a

Solution : Draw the forces acting on the body

fN

Draw the FBD

W = mg

fN

40°am = 5kg

40°

a

W = mg


Since we have perpendicular forces along the inclined (f & N) we can use the inclined as our x-axis

Re-draw the FBD

W = mg

f

N

40°

a

Draw the component vectors of angled vectors

W

f

N

40°

a

Wx

Wy


ΣFx = max → (+)

Fnet = ma

− Wx + f = − mama = Wx – f

ΣFy = 0↑ (+)

+ N – Wy = 0N = Wy = Wcosθ

N = mg cos θma = mg sinθ – f f = μ N

W

f

N

40°

Wx

Wy

a

ma = mg sinθ - μNma = mg sinθ – μmg cosθma = mg (sin θ – μ cosθ)a = g (sin θ – μ cosθ)


1. A 5 kg block slides down a plane inclined at 40° to the horizontal. Find the acceleration of the block

a) If the plane is frictionlessb) If the coefficient of kinetic friction is 0.20

(a) a = ? If f = 0, hence μ = 0a = g [ (sin 40°) − μ ( cos 40°) ]a = (9.8 m/s2) [ (sin 40°) − 0 ( cos 40°) ]

a = 6.3 m/s2

(b) a = ? If μ = 0.2a = g [ (sin 40°) − μ ( cos 40°) ]a = (9.8 m/s2) [ (sin 40°) − 0.2 ( cos 40°) ]

a = 4.8 m/s2


2. A car (2,000 kg) is traveling at 28.7 m/s when the driver locks the breaks to stop the car. What will be the shortest distance ( from the point where the breaks were locked up to the full stopping point), if the coefficient of kinetic friction between the tires and pavement (road) is 0.8?

VO = 28.7 m/s

s

VF = 0a

Using Kinematics Eq’n (3)

VF2 = VO

2 + 2as

s = (VF2 − VO

2)/(2a)


f

W = mg

N


af

W = mg

N

a

Using NSLM to determine the acceleration FBD :



f

W = mg

N

a ΣFx = max → (+)Fnet = ma

− f = + maΣFy = 0↑ (+)+ N – W = 0

N = WN = mg

f = μ N – μN = +ma – μmg = +ma – μg = +a

a = – μg = − (0.8)(9.8 m/s2) = − 7.84 m/s2

a = 7.84 m/s2, deceleration



VO = 28.7 m/s

s

VF = 0a

Using Kinematics Eq’n (3)

VF2 = VO

2 + 2as

s = (VF2 − VO

2)/(2a)

s = [(02 − (28.7m/s)2]/[(2)(−7.84 m/s2)]

s = 52.53 m

a = 7.84 m/s2, deceleration

s = (−823.69 m2/s2)/(−15.68 m/s2)


3. A 3.5 kg pail is dropped into a 15 m empty deep well, starting from rest at the top. The tension in the rope is constant at 14.8 N as the pail drops. What is the time to reach the bottom of the well?

h = 15 m

VO = 0Using Kinematics Eq’n (2)

h = VOt + ½ at2

Where a ≠ g



Using NSLM to determine the acceleration FBD :

T = 14.8 N

W = mg

a

W = mg

T = 14.8 N

a



W = mg

T = 14.8 N

a

Fnet = maΣFx = 0 → (+) ΣFy = may↑ (+)

+ T – W = – ma ma = mg – T

a = ( mg – T ) / m

a = [(3.5kg)(9.8 m/s2) – 14.8 N]/(3.5kg) a = 5.57 m/s2 , (downward)



h = 15 m

VO = 0Using Kinematics Eq’n (2)

h = VOt + ½ at2

Where a ≠ g

a = 5.57 m/s2 , (downward)

– 15m = (0)t + ½(– 5.57 m/s2)t2

– 15m = (– 2.785 m/s2)t2

t2 = (15m/ 2.785 m/s2)t2 = 5.386s2

t = 2.32 s


Q : What relates FCE to NSLM?

A : FCE analysis is similar to NSLM, but the system’s acceleration (a) is

equal to zero during FCE.

Prob 4 : Given :

System is released from rest. Determine the acceleration of the system, when it is already in motion

m1 = 2 kg

m2

μS = 0.65 μK = 0.2

a = ?

FBD of m1

W1 = m1g

N1

TfFBD of m2

W2 = m2g

T

a

a

Derive first the equation for the accelerationUse NSLM on the first FBD (m1)

ΣFx = max → (+)Fnet = ma

− f + T = + m1aΣFy = 0↑ (+)

+ N1 – W1 = 0N1 = W1

N1 = m1g f = μ N

– μN1 + T = m1a – μm1g + T = m1a We solve for “T” because it is the common force between m1 & m2

T = m1a + μm1g (eq’n 1)

NSLM


m1 = 2 kg

m2

μS = 0.65 μK = 0.2

a = ?

FBD of m1

W1 = m1g

N1

TfFBD of m2

W2 = m2g

T

a

a

Use NSLM on the second FBD (m2)

ΣFx = 0 → (+)

Fnet = ma

ΣFy = may↑ (+)

T – W2 = – m2a

T = W2 – m2a

T = m2g – m2a (eq’n 2)

Prob 4 : Given : NSLM


m1 = 2 kg

m2

μS = 0.65 μK = 0.2

a = ?

FBD of m1

W1 = m1g

N1

TfFBD of m2

W2 = m2g

T

a

a

m2g – m2a

(eq’n 1 = eq’n 2)

m1a + μm1g = m1a + m2a = m2g – μm1g a (m1 + m2) = g (m2 – μm1)

a = g (m2 – μm1)

(m1 + m2)



m1 = 2 kg

m2

μS = 0.65 μK = 0.2

a = ?

FBD of m1

W1 = m1g

N1

TfFBD of m2

W2 = m2g

T

a

a

a = g (m2 – μm1)

(m1 + m2) Solve first for the value of m2

When the object is about to move the frictional coefficient is static and the acceleration is zero

0 = g (m2 – μSm1)

(m1 + m2)

0 = m2 – μSm1 The equation is reduced to :

m2 = μSm1 = (0.65)(2kg) m2 = 1.3 kg



m1 = 2 kg

m2

μS = 0.65 μK = 0.2

a = ?

FBD of m1

W1 = m1g

N1

TfFBD of m2

W2 = m2g

T

a

a

a = g (m2 – μKm1)

(m1 + m2)

Next we solve for acceleration using µK and m2

a = 9.8m/s2 [1.3kg – (0.20)(2kg)]

(2kg + 1.3kg)

a = 2.673 m/s2


A

B

Consider the figure shown below. Block A weighs 50N and block B weighs 27N . Once block B is set into downward motion, it descends at a constant speed.a)Calculate the coefficient of kinetic friction between block A and the table top.b)A cat, of weight 50N , jumps on top of block B. If block B is now set in to downward motion, what is its acceleration?

WA = 50N

WB = 27N

This has the very same procedure in derivation for acceleration using NSLM as problem 1

a = g (mB – μmA)

(mA + mB)

The working equation is still


A

B


WA = 50N

WB = 27N

a) Solve for µK

a = g (mB – μKmA)

(mA + mB)

Condition : Block B (& the system) moves at constant speed (a = 0) downward

Prob 5 : Given :

0 = g (mB – μKmA)

(mA + mB)

0 = mB – μKmA µK = mB/mA

µK = (WB/g)/(WA/g) = (WB/(WA) µK = (27N/50N)

µK = 0.54

NSLM

A

B


WA = 50N

WB = 27N+50N

b) Solve for “a”

a = g (mB – μKmA)

(mA + mB)

Condition : Block B now has additional weight due to the cat, the system will now accelerate

Prob 5 : Given :

mA = WA/g = (50N/9.8m/s2)

mB = WB/g = (27N+50N)/(9.8m/s2)

a = 9.8m/s2 [7.857kg – (0.54)(5.102kg)]

(5.102kg + 7.857kg)

mB = 7.857 kg

a = 3.86 m/s2

mA = 5.102 kg

a

NSLM

BA

Consider the figure shown below. Block A is 1kg and block B is 2kg. The inclined of block A is 50° with the horizontal while that of block B is 25°. In what direction will the system go and what is the acceleration? Assume that the surface is frictionless.

mA = 1 kg

mB = 2 kg

θA = 50° θB = 25°FBD of mA

WA = mAg

NA Ta

θA = 50°

WA = mAg

NA

T

a

θA = 50°WAY = WAcosθA

NA

T

a

WAX = WAsinθA

a

θA = 50°

Assume direction is to the left. (This is just an assumed or guess direction)

Prob 6: Given : NSLM

BA


mA = 1 kg

mB = 2 kg

FBD of mA

WAY = WAcosθ

NA

T

a

WAX = WAsinθ

a


ΣFy = 0↑ (+) – WAX + T = – mAa – WAsinθA + T = – mAa

T = WAsinθA – mAa (eq’n 1)

θA = 50° θB = 25°

θA = 50°


BA


mA = 1 kg

mB = 2 kg

FBD of mB

NB

T

a

θB = 25°

WB = mBg

NB

T

a

θB = 25°WBY = WBcosθB

NB

T

a

WBX = WBsinθB

a

θA = 50° θB = 25°

θB = 25°

Prob 6: Given : NSLM

BA


mA = 1 kg

mB = 2 kg

FBD of mB

WBY = WBcosθB

NB

T

a

WBX = WBsinθB

a

θA = 50° θB = 25°

θB = 25°


ΣFy = 0↑ (+) – T + WBX = – mBa − T + WBsinθB = – mBa

T = WBsinθB + mBa (eq’n 2)


BA


mA = 1 kg

mB = 2 kg

a

θA = 50° θB = 25°

T = WBsinθB + mBa (eq’n 2)T = WAsinθA – mAa (eq’n 1)

(mAg)sinθA – mAa = (mBg)sinθB + mBa

(eq’n 1) = (eq’n2)

(mAg)sinθA – (mBg)sinθB = mAa + mBa

g(mAsinθA – mBsinθB) = a(mA+ mB)

g(mAsinθA – mBsinθB)(mA+ mB)a =

Substitute all given values :

a = – 0.259m/s2 The negative sign means that the assumed direction of “a” was wrong. Hence it should be to the right :

a = 0.259m/s2 to the right


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PHY10T3FCE&NSLM

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