For every action force there is an equal and opposite

reaction force!!

For every action force there is an equal and opposite

reaction force!!

For every action force there is an equal and opposite

reaction force!!

•These Two forces are known as These Two forces are known as an action/reaction pair.an action/reaction pair.

•Gravity acts on an object and Gravity acts on an object and the object acts with the object acts with equalequal force. force.

Newton’s Third LawNewton’s Third Law

Action-Reaction PairsAction-Reaction Pairs

The hammer exerts a The hammer exerts a force on the nail to force on the nail to the right.the right.

The nail exerts an The nail exerts an equal but opposite equal but opposite force on the hammer force on the hammer to the left. to the left.

Acting and Reacting Acting and Reacting ForcesForces

• Use the words Use the words byby and and onon to study to study action/reaction forces below as action/reaction forces below as they relate to the they relate to the handhand and the and the barbar::

• Use the words Use the words byby and and onon to study to study action/reaction forces below as action/reaction forces below as they relate to the they relate to the handhand and the and the barbar::

The The actionaction force is exerted force is exerted by by the _____ the _____ onon the _____. the _____.

The reaction force is The reaction force is exerted exerted by by the _____ the _____ onon the _____.the _____.

barbar

handhandss

barbar

handhandss

Action

Reaction

A A 60-kg60-kg athlete exerts a force on a athlete exerts a force on a 10-10-kgkg skateboard. If she receives an skateboard. If she receives an acceleration of acceleration of 4 m/s4 m/s22, what is the , what is the acceleration of the skateboard?acceleration of the skateboard?

Force on runner = -(Force on Force on runner = -(Force on board)board)

mmrr aarr = -m = -mbb aabb

(60 kg)(4 m/s(60 kg)(4 m/s22) = -(10 kg) ) = -(10 kg) aabb

a = - 24 m/s2a = - 24 m/s2

Force on Force on RunnerRunner

Force on Force on BoardBoard

2(60 kg)(4 m/s)24 m/s

-(10 kg)a

Applying Newton’s LawApplying Newton’s Law

• Read, draw, and label problem.Read, draw, and label problem.

• Draw free-body diagram for each body.Draw free-body diagram for each body.

• Choose x or y-axis along motion and Choose x or y-axis along motion and choose direction of motion as positive.choose direction of motion as positive.

• Write Newton’s law for both axes:Write Newton’s law for both axes:

FFxx = m = m axx FFyy = m = m ayy

• Solve for unknown quantities. Solve for unknown quantities.

• Read, draw, and label problem.Read, draw, and label problem.

• Draw free-body diagram for each body.Draw free-body diagram for each body.

• Choose x or y-axis along motion and Choose x or y-axis along motion and choose direction of motion as positive.choose direction of motion as positive.

• Write Newton’s law for both axes:Write Newton’s law for both axes:

FFxx = m = m axx FFyy = m = m ayy

• Solve for unknown quantities. Solve for unknown quantities.

What is the tension What is the tension TT in the rope in the rope below if the block accelerates upward below if the block accelerates upward at at 4 m/s4 m/s22? (Draw sketch and free-? (Draw sketch and free-body.)body.)

1010 kgkg

aa = = +4 m/s+4 m/s22

TT aa

TT

mgmg

++

FFxx = m = m aaxx = 0 (No = 0 (No Motion)Motion)FFyy = m = m aayy = m = m

aaT - mg = m T - mg = m aa

mg = mg = (10 kg)(9.8 m/s) = 98 (10 kg)(9.8 m/s) = 98 NN

m m aa= = (10 kg)(4 m/s) = 40 N(10 kg)(4 m/s) = 40 N

T - - 98 N98 N = = 40 40 NN

T = 138 N

T = 138 N

Two-Body Problem: Find tension in the Two-Body Problem: Find tension in the connecting rope if there is no friction on the connecting rope if there is no friction on the surfaces.surfaces.

2 2 kgkg 4 kg4 kg

12 N12 N Find acceleration Find acceleration of system and of system and tension in tension in connecting cord.connecting cord.

First apply First apply F = mF = maa to entire system ( to entire system (both both massesmasses).).

12 N12 Nnn

((mm22 + + mm44))gg

FFxx = = ((mm22 + m + m44) ) aa

12 N = (6 kg) 12 N = (6 kg) aa

aa = =1212 NN

6 kg6 kga = 2 m/s2

a = 2 m/s2

The two-body problem.The two-body problem.

2 kg

4 kg12 N Now find tension Now find tension

T in connecting T in connecting cord.cord.

Apply F = m Apply F = m aa to the to the 22 kgkg mass where mass where aa = = 2 2 m/sm/s22..

TTnn

mm22 g g

FFxx = = mm2 2 aa

TT = (2 kg)(2 m/s = (2 kg)(2 m/s22))

T = 4 NT = 4 N

The two-body problem.The two-body problem.

2 2 kgkg 4 kg4 kg

12 N12 N Same answer for Same answer for TT results from results from focusing on focusing on 4-kg4-kg by by itself.itself.

Apply Apply F = m F = m aa to the to the 4 kg4 kg mass where mass where aa = 2 = 2 m/sm/s22..

FFxx = = mm4 4 aa

12 N - 12 N - TT = (4 kg)(2 m/s = (4 kg)(2 m/s22))

T = 4 NT = 4 N

1212 NNnn

mm22 g g

TT

Find acceleration of system and tension Find acceleration of system and tension in cord for the arrangement shown.in cord for the arrangement shown.

First apply First apply F = m F = m aa to entire to entire system system alongalong the line of the line of

motion.motion.FFxx = = ((mm22 + m + m44) ) aa

a = 6.53 m/s2

a = 6.53 m/s2

nn

mm22 gg

TT

mm44 gg

TT

+ + aa

Note Note mm22gg is balanced by is balanced by

nn..mm44g = g = ((mm22 + m + m44) )

aa(4 kg)(9.8 (4 kg)(9.8 m/sm/s22))

2 2 kgkg + + 4 kg4 kgaa = = = =

mm44gg

mm22 + m + m44

22 kgkg

44 kgkg

Now find the tension Now find the tension TT given that the given that the acceleration is acceleration is aa = 6.53 m/s= 6.53 m/s22..

To find To find TT, apply , apply F = m F = m aa to to just the just the 2 kg2 kg mass, ignoring 4 mass, ignoring 4

kg.kg.

T = T = (2 kg)(6.53 m/s(2 kg)(6.53 m/s22))

T = 13.1 NT = 13.1 N

Same answer if using 4 Same answer if using 4 kg.kg.

mm44g - T = mg - T = m4 4 aaT = mT = m44(g - (g - aa)) = = 13.1 N13.1 N

nn

mm22 gg

TT

mm44 gg

TT

+ + aa

22 kgkg

4 kg4 kg2 2 or xF m a T m a

Vector ComponentsVector Components• So we have a So we have a

person pulling a person pulling a sled 30sled 30oo with with respect to the respect to the horizontal at a horizontal at a force of 50 N.force of 50 N.

• We need to think of We need to think of it like the sled it like the sled being pulled being pulled vertically and vertically and horizontally at the horizontally at the same time, giving it same time, giving it both components.both components.

Θ=30o

F=50NFy

Fx

Vector ComponentsVector Components

• To solve for FTo solve for Fxx, we will use , we will use cosine because it is the cosine because it is the adjacent side and we have the adjacent side and we have the hypotenuse.hypotenuse.

• To solve for FTo solve for Fyy, use the same , use the same process but with sine.process but with sine.

Θ=30o

F=50m/s

Fy

Fx

θFFF

Fθ

hyp

oppθ

y

y

sin

sin

sin

cos

cos

cos

FFF

F

hyp

adj

x

x

Angled Free-body DiagramAngled Free-body Diagram

300 600

4 kg

A ABB

W = mg

300 600

Bx

By

Ax

Ay

1. Draw and label sketch.1. Draw and label sketch.

2. Draw and label vector force diagram.2. Draw and label vector force diagram.

3. Dot in rectangles and label x and y 3. Dot in rectangles and label x and y compo-nents opposite and adjacent to compo-nents opposite and adjacent to angles.angles.

In the absence of friction, what is the In the absence of friction, what is the acceleration down the acceleration down the 303000 incline? incline?

303000

mgmg

nn

303000

nn

WW

mg mg sin 30sin 3000mg mg cos 30cos 3000

++

FFxx = m = m aaxx

mg mg sin 30sin 3000 = m = m aa

aa = g = g sin 30sin 3000

aa = = (9.8 m/s(9.8 m/s22) sin ) sin 303000

a = 4.9 m/s2

a = 4.9 m/s2

SummarySummary

Newton’s Second Law: A resultant force produces an acceleration in the direction of the force that is directly proportional to the force and inversely proportional to the mass.

Newton’s Second Law: A resultant force produces an acceleration in the direction of the force that is directly proportional to the force and inversely proportional to the mass.

Newton’s First Law:Newton’s First Law: An object at rest or an An object at rest or an object in motion at constant speed will object in motion at constant speed will remain at rest or at constant speed in the remain at rest or at constant speed in the absence of a resultant force.absence of a resultant force.

Newton’s First Law:Newton’s First Law: An object at rest or an An object at rest or an object in motion at constant speed will object in motion at constant speed will remain at rest or at constant speed in the remain at rest or at constant speed in the absence of a resultant force.absence of a resultant force.

Newton’s Third Law:Newton’s Third Law: For every action For every action force, there must be an equal and opposite force, there must be an equal and opposite reaction force. Forces occur in pairs.reaction force. Forces occur in pairs.

Newton’s Third Law:Newton’s Third Law: For every action For every action force, there must be an equal and opposite force, there must be an equal and opposite reaction force. Forces occur in pairs.reaction force. Forces occur in pairs.