a push or a pull -we could name hundreds… Stretch, squeeze,press,twist,crumple,bend,punch, etc etc

Net Force The cause of an acceleration or the change in an object’s velocity. Net Forces can cause objects to… 1.Start moving 2.Stop moving 3.Change their direction

For years physicists have been grouping forces. Eventually they would like to have all forces in one unified theory For years physicists have been grouping forces. Eventually they would like to have all forces in one unified theory GUT’s (Grand Unification Theory) TOE’s (Theory Of Everything)

For now– 4 groups of forces For now– 4 groups of forces – Gravitational : exists between all things with mass – Electromagnetic : forces giving materials strength, bending squeezing, shattering. (object interaction) – Strong nuclear: holds particles of atomic nucleus together **strongest of four forces! – Weak Nuclear : involved in decay of atomic nuclei *

Forces –a force is a vector quantity (magnitude & direction)  Aristotle, Copernicus & Galileo all explained force and motion.  Isaac Newton (1642 – 1727) explained what we study today  Einstein extended our knowledge of gravity and forces – You need to take a higher level of Physics to learn about it.

Forces can act through contact or at a distance Contact Forces – forces resulting from physical contact between 2 objects. Field Forces – do not involve physical contact. An example could be gravity, the earth exerts a force even though its not necessarily contacting the object. Also magnets and electricity have forces without touching.

Unit of force is named after Isaac Newton Newton Symbol = N N= Kg m/s 2

Pounds and Newtons measure Force 1 pound = Newtons 1 Newton = pounds Remember that a Newton is a small thing. If you weigh 150lbs, you also weigh 667N

Free Body Diagram; FBD Free Body Diagrams isolate an object and the forces acting on it. We draw a simple picture of the object and then draw arrows (vectors) to represent the forces on the object. Be sure to label them. Force diagrams are similar but they have multiple objects.

The Net Force causes acceleration There are many forces that act on objects…. –Weight –Normal –Friction –Applied –Tension –Centripetal

A pictorial representation of forces complete with labels. W 1,F g 1 or m 1 g Weight(mg) – Always drawn from the center, straight down Force Normal(F N ) – A surface force always drawn perpendicular to a surface. Tension(T or F T ) – force in ropes and always drawn AWAY from object. Friction(Ff)- Always drawn opposing the motion. m2gm2g T T FNFN FfFf

mg FNFN FfFf

Holt worksheet FBD

On page 127 in the Text book Lets look at the picture of this red car getting towed and do a sample FBD together.

Do you see 4 forces acting on the car? Weight 14700N, Ground 13690N, Truck hitch 5800N, Friction 775N Draw a FBD for this scene.

Weight p.141 in text Is the magnitude of the force of gravity acting on an object. Weight depends on location – if you change the value of g, your weight will change. (I weigh 220lbs on Earth : I would weigh 37lbs on the moon: 80lbs on Mars).

Mass Weight Mass Weight - is measured on - is measured on a balance scale a spring scale - is the amount of - is the force of gravity matter in a body pulling on the mass SCALAR - VECTOR - is the same - Depends on where wherever you go you are (planets, levels) - Has the units of… - Has the units of… Kg metric Newton Metric Kg metric Newton Metric slug imperial Pound imperial - Is a fundamental unit - derived unit N= kg* m/s 2 unit N= kg* m/s 2 we must know the difference

however, mass and weight are relatedhowever, mass and weight are related Force = mass acceleration Weight(force) = mass gravity N= kg m/s 2 W= mg g= W/m m= W/g Look at units Mass = Weight/ GravityMass = Weight/ Gravity Mass = N/ m/s 2Mass = N/ m/s 2 M= kg m/s 2 / m/s 2M= kg m/s 2 / m/s 2 M= KgM= Kg

The Normal Force p.141 in text A force exerted by one object on another in a direction perpendicular to the surface of contact. Many times the normal force will be the contact force an object has with the ground. If the object is on a slope(ramp), the normal will be perpendicular with the slope.

Normal Force t.v. table Draw a FBD for this scene, the object is the t.v.

Normal Force on a ramp Normal Force can be calculated. It’s the perpendicular component of the weight. The angle with the weight vector and perpendicular component is the same as the ramp angle.

any force opposing motionany force opposing motion occurs between any two surfaces that are touchingoccurs between any two surfaces that are touching –No surface is perfectly smooth 2 Kinds of Friction - Static: occurs when no motion is present between two surfaces - Kinetic: (dynamic) occurs when objects on surfaces slide past one another (sliding) STATIC is the greater of the two p. 142 text

The origin of friction: on a microscopic scale, most surfaces are rough.

 Coefficient of friction   is a unit less number used to describe how much friction is present for certain material  lesser the friction, lower the number  greater the friction, greater the number

 hot car tire  car tire  sand paper  tile floor  soapy water  ice  Teflon

µ is the symbol for the coefficient of friction (mu or mew)µ is the symbol for the coefficient of friction (mu or mew) W FnFn

Friction, µ, and normal force are related this way µ = F f F n F n F n = F f µ F f = µ F n F f = µ W F f = µ m g

many forces act on a moving object many forces act on a moving object F normal F applied F weight F friction

Balanced and Unbalanced Forces Sometimes all the forces on an object will cancel out. We say that object is in equilibrium or a state of balance Sometimes all the forces will not cancel out and we say there is a Net Force acting on the object. The Net Force is the Force that causes acceleration.

Consider the Forces on this box, are they balanced or not?

**over head bucket problem sometimes F net = Zero when f net = 0 -- No acceleration will occur Remember, even without acceleration, you can still have velocity When F net = 0 for a moving object, that object is moving at CONSTANT VELOCITY

for the case of constant velocityfor the case of constant velocity friction force = applied force F fr = F app this occurs when you are on cruise control at highway speed, going in a straight path. -- no acceleration but you still have velocity

Free Fall all objects fall at the same accelerating rate, with out air resistanceall objects fall at the same accelerating rate, with out air resistance This is ONLY true when air resistance is not presentThis is ONLY true when air resistance is not present Galileo showed this is true, but he couldn’t explain whyGalileo showed this is true, but he couldn’t explain why

Here is why its true:Here is why its true: A= F/m F/ m = F/m NEWTON”S 2 nd LAW

lets compare a 1 kg ball to a 10 kg balllets compare a 1 kg ball to a 10 kg ball Weight = 10N = 100N 10 N/ 1 KG = 10 m/s 2 100N / 10 Kg= 10 m/s 2 You cannot just look at mass or weight alone, you must consider them together A big mass has a big force A small mass has a small force

Falling With Air air resistance is a frictional force, air resistance causes objects to slow, even to where no acceleration occurs.air resistance is a frictional force, air resistance causes objects to slow, even to where no acceleration occurs. When F net = 0 No Acceleration When air resistance = weight of object Fnet= 0 N

terminal velocity occurs when –Acceleration = 0 –Fnet = 0 –Weight - resistance = 0 A = Fnet/ mass = Weight – Resistance/ Mass = 0 / Mass = 0 m/s 2 Falling ants Sky Divers Flying squirrels

Terminal Velocity As an object moves through the air, it collides w/ air molecules that exert a force on it. (drag, F fr, air resistance) The force depends on the size and shape of the object, the density of air, and the speed of motion.As an object moves through the air, it collides w/ air molecules that exert a force on it. (drag, F fr, air resistance) The force depends on the size and shape of the object, the density of air, and the speed of motion. The faster you go, the more air resistance you get, at some point drag force will equal the force of gravityThe faster you go, the more air resistance you get, at some point drag force will equal the force of gravity ** overhead plane model**

An object at rest has a tendency to stay at rest. An objects in motion has a tendency to stay in motion, unless acted on by an outside force An object at rest has a tendency to stay at rest. An objects in motion has a tendency to stay in motion, unless acted on by an outside force 1 st Law!

coin cup democoin cup demo magician table cloth pullmagician table cloth pull roll a ball, does it keep going? why?!roll a ball, does it keep going? why?! flip a coin in deep space, what happens?flip a coin in deep space, what happens? tug of wartug of war voyager space craftvoyager space craft Inertia- the more mass, the greater the tendency to remain

Newton’s 1 st law  An object with no force on it remains at rest or moves with a constant velocity in a straight line.  An object at rest has a tendency to stay at rest; an object in motion has a tendency to remain in motion; unless acted on by another force. The tendency to resist motion change is called I N E R T I A

 roll ball only friction stops it  coin, cup, card  magician-table cloth  cruising airplane constant velocity jets overcome friction

 The acceleration of a body is directly proportional to the net force on it and inversely proportional to its mass 2 nd law can be summarized a= f net /m more commonly f net = ma F net = net force m= mass a= acceleration

for a constant mass, a bigger force makes a faster acceleration.for a constant mass, a bigger force makes a faster acceleration. for a constant force, a bigger mass makes a smaller acceleration.for a constant force, a bigger mass makes a smaller acceleration. Newton’s second law describes the unit of force

Net force is the force that causes acceleration F net is the sum of ALL forces acting on a body F F net 10 kg 100N 20N 100N 10N f = 100+(-100)+20+(-10) F net = +10N this is the force that will accelerate the block

A= F net /m a= 10N/10kg a= 1m/s 2

Net Force You have to find the net external force to calculate the acceleration or mass of an object. Remember the F net is the sum of all forces acting on an object.

 a force that causes a mass of 1 kg to accelerate at a rate of 1 m/s 2 is defined as 1 Newton (N) Newtons are derived units Fundamentally they are F = ma Kg m/s 2

 whenever one object exerts a force on a second, the second object exerts an equal and opposite force on the first  For every action there is an equal and opposite reaction

 forces occur in pairs  ACTiON/ REACTION pairs  They are equal in size and opposite in direction. (The one we label as action doesn’t matter, the point is both forces are a co-part of a single interaction)

Consider the bucketConsider the bucket To raise the bucket more upward force than downward force will cause the bucket to rise Earth

push on a wall – wall pushes on youpush on a wall – wall pushes on you hammer- nailhammer- nail walk on icewalk on ice tires- roadtires- road rocket – exhaust vs. liftrocket – exhaust vs. lift falling object -earth pulls ball, ball pulls earthfalling object -earth pulls ball, ball pulls earth Gun recoilGun recoil

recoil – this is the result of Newton’s 3 rd lawrecoil – this is the result of Newton’s 3 rd law –when the powder explodes, it produces a force –The action force is the bullet leaving the gun –the reaction force is the gun pushing back on you The forces are equal but the acceleration of each are NOT The reason is mass- The gun has large mass, the bullet has small mass a= f/m

Horse pulling a cart problem

Soccer ball kick example Each force acts on a different object. your kick acts on the ballyour kick acts on the ball your kick is the f net of the ballyour kick is the f net of the ball your kick accelerates the ballyour kick accelerates the ball The reaction force acts on your leg, it doesn’t act on the ball, the reaction force slows your leg down

A 10-kg box is being pulled across the table to the right at a constant speed with a force of 50N. a)Calculate the Force of Friction b)Calculate the Force Normal mg FNFN FaFa FfFf

Suppose the same box is now pulled at an angle of 30 degrees above the horizontal. a)Calculate the Force of Friction b)Calculate the Force Normal mg FNFN FaFa FfFf 30 F ax F ay

If an object is NOT at rest or moving at a constant speed, that means the FORCES are UNBALANCED. One force(s) in a certain direction over power the others. THE OBJECT WILL THEN ACCELERATE.

The acceleration of an object is directly proportional to the NET FORCE and inversely proportional to the mass. Tips: Draw an FBD Resolve vectors into components Write equations of motion by adding and subtracting vectors to find the NET FORCE. Always write larger force – smaller force. Solve for any unknowns

A 10-kg box is being pulled across the table to the right by a rope with an applied force of 50N. Calculate the acceleration of the box if a 12 N frictional force acts upon it. mg FNFN FaFa FfFf In which direction, is this object accelerating? The X direction! So N.S.L. is worked out using the forces in the “x” direction only

m1gm1g m2gm2g T T FNFN A mass, m 1 = 3.00kg, is resting on a frictionless horizontal table is connected to a cable that passes over a pulley and then is fastened to a hanging mass, m 2 = 11.0 kg as shown below. Find the acceleration of each mass and the tension in the cable.

2) 8N

4.1 A 5.0-kg object is to be given an upward acceleration of 0.3 m/s 2 by a rope pulling straight upward on it. What must be the tension in the rope? FTFT FGFG a (+) m = 5 kg a = 0.3 m/s 2 ΣF y = F T - F G = ma F T = m(a + g) = 5( ) = 50.5 N 5 kg N2L

4.2 A cord passing over a frictionless pulley has a 7.0 kg mass hanging from one end and a 9.0-kg mass hanging from the other. (This arrangement is called Atwood's machine). a. Find the acceleration of the masses. FTFT FTFT F G1 F G2 a (+) m 1 = 7 kg m 2 = 9 kg 7kg 9 kg N2L

FTFT FTFT F G1 F G2 a (+) ΣF = F T - F G1 - F T + F G2 = m TOTAL a = 1.22 m/s 2

FTFT FTFT F G1 F G2 a (+) b. Find the tension of the cord Using either side of the pulley yields the same answer! F T – F G1 = m 1 a F T = m 1 a + F G1 = m 1 (a + g) = 7( ) = 77.1 N

APPARENT WEIGHT The actual weight of a body is the gravitational force that acts on it. The body's apparent weight is the force the body exerts on whatever it rests on. Apparent weight can be thought of as the reading on a scale a body is placed on. scale

4.3 What will a spring scale read for the weight of a 75 kg man in an elevator that moves: m = 75 kg F G = 75(9.8) = 735 N

ΣF y = F N - F G = 0 F N = F G = 735 N FNFN FGFG a. With constant upward speed of 5 m/s m = 75 kg F G = 735 N b. With constant downward speed of 5 m/s F N = 735 N N1L

ΣF = F N - F G = ma F N = ma + F G = 75 (2.45) = 919 N FNFN FGFG b. Going up with an acceleration of 0.25 g a = 2.45 m/s 2 N2L

ΣF = F G - F N = ma F N = F G - ma = (2.45) = 551 N FNFN FGFG c. Going down with an acceleration of 0.25 g a = 2.45 m/s 2 N2L

For free fall the only force acting is F G so F N = 0 N FGFG e. In free fall? N2L

“Weightlessness” More properly, this effect is called apparent weightlessness, because the gravitational force still exists. It can be experienced on Earth, but only briefly:

FRICTION: STATIC AND KINETIC FRICTION The kinetic friction (or sliding friction) that occurs afterward is usually less than the starting friction, so less force is needed to keep the book moving than to start it moving.

COEFFICIENT OF FRICTION The frictional force between two surfaces depends on the normal force F N pressing them together and on the natures of the surfaces. The latter factor is expressed quantitatively in the coefficient of friction  (mu) whose value depends on the materials in contact. The frictional force is experimentally found to be: Static friction Kinetic friction

4.4 A horizontal force of 140 N is needed to pull a 60.0 kg box across the horizontal floor at constant speed. What is the coefficient of friction between floor and box? F a = 140 N m = 60 kg ΣF y = F N – F G = 0 F N = F G =mg = 60(9.8) = 588 N FNFN FGFG FfFf FaFa ΣF x = F a – F f = 0 F a = F f = μF N = 0.24 N1L

4.5 A 70-kg box is slid along the floor by a horizontal 400-N force. Find the acceleration of the box if the value of the coefficient of friction between the box and the floor is m = 70 kg F a = 400 N μ = 0.5 ΣF y = F N – F G = 0 F N = F G =mg = 70(9.8) = 686 N ΣF x = F a – F f = ma F f = μF N = (0.5)(686) = 343 N FNFN FaFa FGFG FfFf N2L

m = 70 kg F a = 400 N μ = 0.5 ΣF x = F a – F f = ma = 0.81 m/s 2 FNFN FaFa FGFG FfFf

4.6 A 70-kg box is pulled by a rope with a 400-N force at an angle of 30  to the horizontal. Find the acceleration of the box if the coefficient of friction is 0.50 m = 70 kg F a = 400 N, 30  μ = 0.5 FGFG FaFa FfFf FNFN F ax = 400 cos 30  = N F ay = 400 sin 30  = 200 N ΣF y = F N + F ay - F g = 0 F N = F G - F ay = 70(9.8) = 486 N F f = μF N = (0.5)(486) = 243 N N2L

FGFG FaFa FfFf FNFN ΣF x = F ax – F f = ma = 1.47 m/s 2 m = 70 kg F a = 400 N, 30  μ = 0.5

FNFN FfFf FGFG F Gy F Gx θ 4.7 A box sits on an incline that makes an angle of 30˚ with the horizontal. Find the acceleration of the box down the incline if the coefficient of friction is θ = 30  μ = 0.3 N2L ΣF y = F N - F Gy = 0 F N = F Gy ΣF x = F Gx – F f = ma F f = μF N ΣF x = F Gx – μ F Gy = ma

FNFN FfFf FGFG F Gy F Gx θ ΣFx = F Gx – μ F Gy = ma mg sin θ - μ mg cos θ = ma = 2.35 m/s 2

4.8 Two blocks m 1 (300 g) and m 2 (500 g), are pushed by a force F. If the coefficient of friction is a. What must be the value of F if the blocks are to have an acceleration of 200 cm/s 2 ? m 1 = 0.3 kg m 2 = 0.5 kg μ = 0.4 a = 2 m/s 2 F G1 = F N1 F G2 = F N2 F G1 F G2 F N1 F N2 F F f2 F f1 ΣF x = F - F f1 - F f2 = m T a F = m T a + F f1 + F f2 = m T a + μ F NT = m T a + μ m T g = 0.8(2) + (0.4) 0.8 (9.8) = 4.7 N N2L

b. How large a force does m 1 then exert on m 2 ? m 2 alone ΣF x = m 2 a F 12 - F f2 = m 2 a F 12 = F f2 + m 2 a = μ m 2 g + m 2 a = 0.4(0.5) (9.8) (2) = 2.96 N F f2 F 12 F N2 F G2

4.9 An object m A = 25 kg rests on a tabletop. A rope attached to it passes over a light frictionless pulley and is attached to a mass m B = 15 kg. The coefficient of friction between the table and block A is a. What is the acceleration of the 25 kg block? m A = 25 kg m B = 15 kg μ = 0.2 F N = F GA = 25(9.8) = 245 N F fA = μF N = 0.2 (245) = 49 N FNFN F GA FfFf FTFT FTFT F GB 25 kg 15 kg N2L

FNFN F GA FfFf FTFT FTFT F GB ΣF = F T - F fA + F GB - F T = m T a = 2.45 m/s 2 m A = 25 kg m B = 15 kg μ = 0.2

FNFN F GA FfFf FTFT FTFT F GB ΣF = F T - F f = m A a F T = m A a + F f = 25 (2.45) + 49 = N b. What is the tension on the string?

4.10 Blocks 1 and 2 of masses m l and m 2, respectively, are connected by a light string, as shown above. These blocks are further connected to a block of mass M by another light string that passes over a pulley of negligible mass and friction. Blocks l and 2 move with a constant velocity v down the inclined plane, which makes an angle  with the horizontal. The kinetic frictional force on block 1 is f and that on block 2 is 2f.

a. On the figure below, draw and label all the forces on block m l. FGFG FNFN FTFT FfFf

Express your answers to each of the following in terms of m l, m 2, g, , and f. b. Determine the coefficient of kinetic friction between the inclined plane and block 1.

c. Determine the value of the suspended mass M that allows blocks 1 and 2 to move with constant velocity down the plane. Mg m 1 g m 2 g f 2f FTFT FTFT FTFT F N1 F N2 N1L FTFT

m 1 g f F N1 d. The string between blocks 1 and 2 is now cut. Determine the acceleration of block 1 while it is on the inclined plane. N2L

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

Summary: Procedure Read and write data with units.Read and write data with units. Draw free-body diagram for each object.Draw free-body diagram for each object. Choose x or y-axis along motion and choose direction of motion as positive.Choose x or y-axis along motion and choose direction of motion as positive. Write Newton’s law for both axes:Write Newton’s law for both axes:  F x = m x  F y = m y  F x = m a x  F y = m a y Solve for unknown quantities.Solve for unknown quantities. Read and write data with units.Read and write data with units. Draw free-body diagram for each object.Draw free-body diagram for each object. Choose x or y-axis along motion and choose direction of motion as positive.Choose x or y-axis along motion and choose direction of motion as positive. Write Newton’s law for both axes:Write Newton’s law for both axes:  F x = m x  F y = m y  F x = m a x  F y = m a y Solve for unknown quantities.Solve for unknown quantities. N = (kg)(m/s 2 )

Translational Equilibrium An object is said to be in Translational Equilibrium if and only if there is no resultant force. This means that the sum of all acting forces is zero. In the example, the resultant of the three forces A, B, and C acting on the ring must be zero. A C B

Visualization of Forces Force diagrams are necessary for studying objects in equilibrium. Don’t confuse action forces with reaction forces. Equilibrium: The action forces are each ON the ring. A B C Force A: By ceiling on ring. Force B: By ceiling on ring. Force C: By weight on ring.

Visualization of Forces (Cont.) Now let’s look at the Reaction Forces for the same arrangement. They will be equal, but opposite, and they act on different objects. Reaction forces: Reaction forces are each exerted: BY the ring. ArAr BrBr CrCr Force A r : By ring on ceiling. Force B r : By ring on ceiling. Force C r : By ring on weight.

Vector Sum of Forces An object is said to be in Translational Equilibrium if and only if there is no resultant force. The vector sum of all forces acting on the ring is zero in this case. W 40 0 A B C Vector sum:  F = A + B + C = 0

Vector Force Diagram W 40 0 A B C W A B C AxAx AyAy A free-body diagram is a force diagram AyAy showing all the elements in this diagram: axes, vectors, components, and angles.

Free-body Diagrams: Read problem; draw and label sketch. Isolate a common point where all forces are acting. Construct force diagram at origin of x,y axes. Dot in rectangles and label x and y components opposite and adjacent to angles. Label all given information and state what forces or angles are to be found. Read problem; draw and label sketch. Isolate a common point where all forces are acting. Construct force diagram at origin of x,y axes. Dot in rectangles and label x and y components opposite and adjacent to angles. Label all given information and state what forces or angles are to be found.

Look Again at Previous Arrangement W 40 0 A B C 1.Isolate point. 2. Draw x,y axes. 3. Draw vectors. 4. Label components. 5. Show all given information. A 40 0 W AyAy B C AyAy AxAx

Example 1. Draw a free-body diagram for the arrangement shown on left. The pole is light and of negligible weight. W 30 0 A B C 700 N Careful: The pole can only push or pull since it has no weight. The force B is the force exerted on the rope by the pole. Don’t confuse it with the reaction force exerted by the rope on the pole. B 30 0 A C 700 N AyAy AxAx Isolate the rope at the end of the boom. All forces must act ON the rope! On rope B

Translational Equilibrium The First Condition for Equilibrium is that there be no resultant force. This means that the sum of all acting forces is zero.

Example 2. Find the tensions in ropes A and B for the arrangement shown. 200 N 40 0 A B C The Resultant Force on the ring is zero: R =  F = 0 R x = A x + B x + C x = 0 R y = A y + B y + C y = N 40 0 A B C AxAx AyAy AyAy

Example 2. (Cont.) Finding components. Recall trigonometry to find components: The components of the vectors are found from the free- body diagram. 200 N 40 0 A B C AxAx AyAy BxBx CyCy C x = 0 C y = -200 N Opp = Hyp x sin Adj = Hyp x cos A x = A cos 40 0 A y = A sin 400 A B y = 0

Example 2. Continued... W 40 0 A B C A free-body diagram must represent all forces as components along x and y-axes. It must also show all given information. Components A x = A cos 40 0 A y = A sin 40 0 B x = B; B y = 0 C x = 0; C y = W AxAx AyAy AyAy

Example 2. Continued N 40 0 A B C 200 N 40 0 A B C AxAx AyAy AyAy  F x = 0  F y = 0 Components A x = A cos 40 0 A y = A sin 40 0 B x = B; B y = 0 C x = 0; C y = W

Example 2. Continued N 40 0 A B C AxAx AyAy AyAy Solve first for A Solve Next for B The tensions in A and B are A = 311 N; B = 238 N Two equations; two unknowns

Problem Solving Strategy 1.Draw a sketch and label all information. 2.Draw a free-body diagram. 3.Find components of all forces (+ and -). 4.Apply First Condition for Equilibrium:  F x = 0 ;  F y = 0 5. Solve for unknown forces or angles.

Example 3. Find Tension in Ropes A and B A B 400 N A B 1. Draw free-body diagram. 2. Determine angles AyAy ByBy AxAx BxBx 3. Draw/label components. Next we will find components of each vector.

Example 3. Find the tension in ropes A and B.  F x = B x - A x = 0  F y = B y + A y - W = 0 B x = A x B y + A y = W A B W 400 N AyAy ByBy AxAx BxBx 4. Apply 1 st Condition for Equilibrium: First Condition for Equilibrium:  F x = 0 ;  F y = 0

Example 3. Find the tension in ropes A and B. B x = A x B y + A y = W A B W 400 N AyAy ByBy AxAx BxBx Using Trigonometry, the first condition yields: B cos 60 0 = A cos 30 0 A sin B sin 60 0 = 400 N A x = A cos 30 0 ; A y = A sin 30 0 B x = B cos 60 0 B y = B sin 60 0 W x = 0; W y = -400 N

Example 3 (Cont.) Find the tension in A and B. A B W 400 N AyAy ByBy AxAx BxBx B = A We will first solve the horizontal equation for B in terms of the unknown A: B cos 60 0 = A cos 30 0 A sin B sin 60 0 = 400 N We now solve for A and B: Two Equations and Two Unknowns.

Example 3 (Cont.) Find Tensions in A and B. A sin B sin 60 0 = 400 N B = A A sin (1.732 A) sin 60 0 = 400 N A A = 400 N A = 200 N A B 400 N AyAy ByBy AxAx BxBx B = A Now apply Trig to: A y + B y = 400 N A sin B sin 60 0 = 400 N

Example 3 (Cont.) Find B with A = 200 N. Rope tensions are: A = 200 N and B = 346 N This problem is made much simpler if you notice that the angle between vectors B and A is 90 0 and rotate the x and y axes (Continued) B = A A = 200 N B = 1.732(200 N) B = 346 N A B W 400 N AyAy ByBy AxAx BxBx

Example 4. Rotate axes for same example A B 400 N A B AyAy ByBy AxAx BxBx We recognize that A and B are at right angles, and choose the x-axis along B – not horizontally. The y-axis will then be along A—with W offset. x yW

Since A and B are perpendicular, we can find the new angle  from geometry. You should show that the angle  will be We now only work with components of W. x y A B N A B W =400 N x y 

Recall W = 400 N. Then we have: Apply the first condition for Equilibrium, and... A B x y 30 0 WxWxWxWx WyWyWyWy W x = (400 N) cos 30 0 W y = (400 N) sin 30 0 Thus, the components of the weight vector are: W x = 346 N; W y = 200 N B – W x = 0 and A – W y = N

Example 4 (Cont.) We Now Solve for A and B:  F x = B - W x = 0  F y = A - W y = 0 B = W x = (400 N) cos 30 0 B = 346 N A = W y = (400 N) sin 30 0 A = 200 N A B 400 N x y 30 0 WxWxWxWx WyWyWyWy Before working a problem, you might see if rotation of the axes helps.

Study forthe test!!Study forthe test!!Study forthe test!!Study forthe test!!