3Change the direction of the force. What do Machines do?Do they allow one to do more work?Not really, at best they make completing a task easier.So then what do Machines do?Multiply the force.Multiply the distance.Change the direction of the force.
4Work = Force x Distance an object moves while the force is applied. W = F x dIn SI Units:Force is measured in newtons (N) distance is measured in meters (m)Work in N.m which is a joule (J).Named after James Prescott Joule
5Make something move faster. Lift something up. What does work do?Work causes a change in Energy. In other words, it can do any of the following:Make something move faster.Lift something up.Move something against friction.A combination of the above.
6Examples of Work: A cart is pushed to the right as illustrated. 50.0 Ndistance cart is moved 4.00 m.How much work is done on the cart?W = F x d = (50.0 N)(4.00 m) = 200. J
7Examples of Work: A box is lifted as illustrated. 80.0 NA box is lifted as illustrated.distance box is moved 20. m.How much work is done on the box?W = F x dW = (80.0 N)(20. m)W = 1600 J
8Now let’s apply this to some of our machines. The simplest is most likely levers.
9Class 1 Lever: load on one side of the fulcrum and the effort on the other side. The Fulcrum is the pivot point.The Load is what we are trying to lift or the output of the machine.EffortThe Effort is the force that is applied to lift the load or the input of the machine.LoadFulcrum
10Class 1 Lever: Load on one side of the fulcrum and the Effort on the other side.
11Class 1 Lever: More terminology Note that in lifting the load the Effort moved much farther than the Load.EffortWith a smaller Effort we could lift a Load that is heavier.LoadFulcrum
12Class 1 Lever: More terminology With an Effort of 300 N we were able to lift a Load of 900 N. We multiplied the input force by 3.Effort = 300 NThe Effort moved 60 cm while the Load moved only 20 cm. We moved 3 times farther than the LOAD.Load = 900NEffort distance, dE = 60 cmLoad distance, dL = 20 cmFulcrum
13Class 1 Lever: More terminology Work, W = F x dWIN = E x dE = (300 N)(0.60 m) = 180 JWOUT = L x dL = (900 N)(0.20 m) = 180 JEffort = 300 NNote: WorkIN = WorkOUTWe didn’t do more work, we just did it with less Effort than if I tried to lift it without the lever.Load = 900NEffort distance, dE = 60 cmLoad distance, dL = 20 cmFulcrum
14Class 1 Lever: More terminology We say that we have a Mechanical Advantage, MA.We can lift 3 times more than our input Effort.Effort = 300 NMA = Load/EffortMA = (900 N)/(300 N)MA = 3Load = 900NMA = dE/dLMA = (60 cm)/(20 cm)MA = 3Effort distance, dE = 60 cmLoad distance, dL = 20 cmFulcrum
15Class 1 Lever: More terminology We triple our Effort (input force) at the expense of moving the Load ⅓ as much.Effort = 300 NMA = Load/EffortMA = (900 N)/(300 N)MA = 3Load = 900NMA = dE/dLMA = (60 cm)/(20 cm)MA = 3Effort distance, dE = 60 cmLoad distance, dL = 20 cmFulcrum
16Class 1 Lever: More terminology We can also analyze the lever by measuring the distance from the Effort to the fulcrum (pivot point) and the distance from the Load to the fulcrum.This is called the lever arm or just arm and is often given the variable name “x.”Effort arm, xE = 3.0 mLoad arm, xL = 1.0 mFulcrum
17Class 1 Lever: More terminology This can also be used to calculate the Mechanical Advantage.MA = xE/xL = (3.00 m)/(1.00 m) = 3Looks familiar doesn’t it.Effort arm, xE = 3.0 mLoad arm, xL = 1.0 mFulcrum
18More terminology:Often we use the terms, Ideal Mechanical Advantage, IMA and Actual Mechanical Advantage, AMAIMA = xE/xL or dE/dLAMA = L/E with the load being just what you ultimately wanted to move, excluding anything else that may have to be moved with it.This will become clearer when we look at a 2nd class lever.
19More terminology:Often we use the terms, Ideal Mechanical Advantage, IMA and Actual Mechanical Advantage, AMAWe want to find out how well the particular machine does its work. This is called Efficiency, Eff.Efficiency, Eff = (AMA/IMA) x 100%
202nd class lever:Notice that the Effort and the Load are on the same side of the Fulcrum and the Load is between the Effort and the Fulcrum.EffortLoadFulcrum
212nd class lever:Again the Effort moves much farther than the Load. We are getting more force out than what we put in, but the load only moves a short distance. We are also lifting the lever along with the load.LoadEffortLoadEffortFulcrum
22IMA = xE/xL = (2.8 m)/(0.35 m) = 8 2nd class lever: The load is 900 N and since the Effort has to lift the Load and the lever, let’s say that the Effort is 150 N.Load arm, xL = 0.35 mEffortEffort arm, xE = 2.8 mLoadFulcrum
26There is a lab part of the competition There is a lab part of the competition. Let’s look at some of the basic concepts for a lever that is in static equilibrium.
27The easiest lever to analyze is the first class lever (seesaw), that is balanced by itself. The center of gravity of the lever is on the fulcrum.c.g.
28If a lever is not moving (rotating) then it is said to be at static equilibrium. When an object is at static equilibrium the following is true:ΣF = 0, that is netF = 0, no unbalanced forces.Στ = 0, that is there are no unbalanced torques.If you place a seesaw so that its center of gravity is on the fulcrum, it will balance. That is, the left side balances the right side.
29Torque is the tendency of a force to cause an object to rotate around an axis. In the case of a lever, the axis is the fulcrum.ForceIn this case the force would make the left side of the lever go down or rotate the lever counterclockwise, ccw.
30Torque is the tendency of a force to cause an object to rotate around an axis. In the case of a lever, the axis is the fulcrum.In this case the force would make the left side of the lever go down or rotate the lever counterclockwise, CCW.Force
31What if the force is 24 N, what torque is applied? Earlier we talked about the lever arm or arm being the distance from the fulcrum (axis) to the force. We will use the letter “x” as the symbol for lever arm.The symbol for torque is the Greek letter tau, τF = 24 Nτ = (F)(x) = (24 N)(1.2 m)τ = 28.8 N.m = 29 N.mx = 1.2 mA torque of 29 N.m will rotate the lever CCW.
32The two balance each other and it does not rotate. The weight of the seesaw on the left creates a torque that tries to rotate the seesaw counter-clockwise, CCW, so that the left side would go down.The weight of the seesaw on the right creates a torque that tries to make it rotate clockwise,CW, so that the right side would go down.FFThe two balance each other and it does not rotate.
33Note: The center of gravity may not be at the geometric center. Another way to look at this is that we can place all the weight of the seesaw ( FL ) at its center of gravity.The center of gravity of the seesaw is at the axis of rotation (fulcrum) so the value of the lever arm is zero and the force creates no torque.c.g.Note: The center of gravity may not be at the geometric center.Especially when using wooden meter sticks!FL
34How do we analyze this situation? Sample problem: Two identical 40.0 kg twin girls are sitting on opposite ends of a seesaw that is 4.0 m long and weighs 700 N, so that the center of gravity of the seesaw is on the fulcrum.c.g.How do we analyze this situation?
35FN = 1500 N x1 = 2.0 m x2 = 2.0 m F1 = 400 N FL = 700 N F2 = 400 N First, we need to draw a torque diagram of the seesaw. This is a free body diagram which includes the lever arms.FN = 1500 NNext, we define the axis of rotation (circle with a dot in the middle) and the lever arms.We place all the forces at their proper location.x1 = 2.0 mx2 = 2.0 mc.g..F1 = 400 NFL = 700 NF2 = 400 N
36Στ = 0 or ΣτCCW = ΣτCW F1x1 = F2x2 FN = 1500 N x1 = 2.0 m x2 = 2.0 m FL and FN both act through the axis of rotation, so their lever arm is zero, making their torque 0.F1x1 = F2x2FN = 1500 Nx1 = 2.0 mx2 = 2.0 mFL = 700 Nc.g..F1 = 400 NF2 = 400 N(400 N)(2.0 m) = (400 N)(2.0 m)The torques balance so the seesaw can be in static equilibrium.800 Nm = 800 Nm
37It is important that the seesaw be level, so that the force applied by each of the girls is acting downward and is perpendicular to the lever arm. If the Force and the Lever Arm are not perpendicular, then the equation for the torque becomes complex. It is better that we avoid that situation.
38So, what do you do to balance the seesaw if the two people are not the same weight (mass)? One 400 N girl sits on one end of a seesaw that is centered on the fulcrum, is 4.0 m long, and weighs 700 N. Where must her 650 N brother sit in order for the seesaw to be in static equilibrium??c.g.Option #1, move the heavier person closer to the fulcrum.
39Στ = 0 or ΣτCCW = ΣτCW FGxG = FBxB FN = 1750 N xG = 2.0 m xB = ?? m FL and FN both act through the axis of rotation, so their lever arms are zero.FGxG = FBxBFN = 1750 NxG = 2.0 mxB = ?? mFL = 700 Nc.g..FG = 400 NFB = 650 NxB = (800 Nm)/(650 N)xB = 1.23 m(400 N)(2.0 m) = (650 N)xB800 = 650xB
40Option #2, move the center of gravity of the seesaw so that more of the seesaw is on the side of the lighter person,One 400 N girl sits on one end of a 4.0 m long seesaw weighing 700 N That has moved the center of gravity of the lever 0.2 meters towards her. Where must her 650 N brother sit in order for the seesaw to be in static equilibrium??c.g.Now the weight of the seesaw creates a torque helping the girl.
41Στ = 0 or ΣτCCW = ΣτCW FGxG + FLxL = FBxB FN = 1750 N xG = 2.2 m FN acts through the axis of rotation, so its lever arm is zero.Στ = 0 or ΣτCCW = ΣτCWFGxG + FLxL = FBxBFN = 1750 NxG = 2.2 mxB = ?? mFL = 700 Nc.g..FG = 400 NFB = 650 N(400)(2.2) + (700)(0.2) = (650 N)xBxB = (1020 Nm)/(650 N)xB = 1.57 m= 650xBxL = 0.2 m
42For the Middle School (Division B) Competition, you will need to build a simple first class lever system. The lever may not be longer than 1.00 meter.
43Simple Machines. (Simple case.) Given small mass placed on one side.Given unknown large mass on the other.Unless the values are too extreme, you may be able to move the large mass close enough to the fulcrum.c.g.If this is the setup, you don’t have to worry about the weight of the lever.
44FN FS FB FL Simple Machines. (Simple case.) Big Mass Small Mass xS xB c.g.Big MassSmall MassFNxSxBc.g..FSFBFL
45FN FS FB FL Simple Machines. (Simple case.) In this case the torque equation is: τCCW = τCW(FS)(xS) = (FB)(xB) and you can solve for any value.FNxSxBc.g..FSFBFL
46So far we have been dealing with the force applied by the hanging mass So far we have been dealing with the force applied by the hanging mass. This force is known as the weight of the object or the force of gravity (Fg) acting on the object.The force of gravity acting on an object is the product of the mass of the object multiplied by the gravity constant on the planet Earth (9.8 N/kg).Fg = mg = m(9.8 N/kg) so, mass, m = Fg/(9.8 N/kg)This gets quite confusing because weight is measured in newtons and mass is measured in grams or kilograms (kg).You may have been told to weigh something but you actually measured its mass in grams.
47FN FS = mSg FB = mBg FL Simple Machines. (Simple case.) Knowing that Fg = mg(FS)(xS) = (FB)(xB) This equation can be written:(msg)(xS) = (mBg)(xB) Dividing by g we get:FN(msg)(xS)/g = (mBg)(xB)/g(ms)(xS) = (mB)(xB)xSxBc.g..FS = mSgFB = mBgFL
48(ms)(xS) = (mB)(xB) FN mS mB FL Simple Machines. (Simple case.) We can now solve for a mass using this equation and modify our torque diagram as shown:(ms)(xS) = (mB)(xB)FNxSxBc.g..mSmBFL
49FN mS = 125 g mB = ?? FL Simple Machines. (Simple case.) Suppose that you were given a small mass of 125 grams and an unknown large mass. You set up your lever so it balances as shown:FNxB = 10.0 cmxS = 47.6 cmc.g..mS = 125 gmB = ??FL
51Simple Machines (More realistic case) If the difference between the unknown mass and the known mass is large, move the fulcrum near one end of the lever.Place the unknown large mass on the short side.Then place the small mass on the long side.c.g.Now the lever helps balance the large weight.
52Στ = 0 or ΣτCCW = ΣτCW FSxS + FLxL = FBxB FN xS = cm xB = cm FL = N FN acts through the axis of rotation, so its lever arm is zero.Στ = 0 or ΣτCCW = ΣτCWFSxS + FLxL = FBxBFNxS = cmxB = cmFL = Nc.g..FS = NFB = ?? NxL = cm
53(mSg)xS + (mLg)xL = (mBg)xB FN acts through the axis of rotation, so its lever arm is zero.Στ = 0 or ΣτCCW = ΣτCWFSxS + FLxL = FBxBFNxS = cmxB = cmFL = mLgc.g..FS = mSgFB = mBgxL = cm(mSg)xS + (mLg)xL = (mBg)xBIf you divide through by “g” you get:mSxS + mLxL = mBxB
54Sample: Suppose that you have set up your 1 Sample: Suppose that you have set up your 1.00 meter long lever of mass 83.4 grams so that the center of gravity is 20.0 cm from the fulcrum. You have also determined that the big unknown mass will be placed 10.0 cm from the fulcrum.FNxS = cmxB = 10.0 cmFL = mLgc.g..FS = mSgFB = mBgxL = 20.0 cm
55You are given a known mass of 76. 2 grams and a big mass You are given a known mass of 76.2 grams and a big mass. Putting the masses on the lever, you find that it balances when the little mass is 62.8 centimeters from the fulcrum.FNxB = 10.0 cmmL = 83.4 g.mB = ??xL = 20.0 cmxS = 62.8 cmc.g.mS = 76.2 g
57As long as the levers are horizontal and in static equilibrium, you can use the equations with mass instead of force or weight. Your Physics teacher probably will not be too happy, but the equation is mathematically correct for the situation.
59First Class Lever Second Class Lever c.g.BFirst Class LeverSecond Class LeverFor the High School Competition you will need to build a compound lever system made up of a first class lever connected to a second class lever. Each lever may not be longer than 50.0 cm.
60mSxS + mL1xL1 = ExE1 xS = cm xE1 = cm mL1 = g mS = g E = ?? g xL1 = cm c.g.First part of lever systemmSxS + mL1xL1 = ExE1xS = cmxE1 = cmmL1 = gc.g..mS = gE = ?? gxL1 = cm
61ExE2 = mL2xL2 + mBxB xL2 = cm xE2 = cm E = g xB = cm mB = ?? g mL2 = g Second part of lever systemBc.g.ExE2 = mL2xL2 + mBxBxE2 = cmxB = cmmL2 = gc.g..E = gmB = ?? gxL2 = cm
62Sample: xE1 = 5.0 cm xE2 = 35.0 cm xB = 10.0 cm c.g.BxB = 10.0 cmSuppose that you built your lever system so that the fulcrum in the class 1 lever was 5.0 cm from the string connecting the levers.Also suppose that in the second class lever the string connecting the levers is 35.0 cm from its fulcrum and you set up the lever so that the unknown big mass is 10.0 cm from the fulcrum.
63Sample xE1 = 5.0 cm xE2 = 35.0 cm xS = 32.7 cm xB = 10.0 cm c.g.BxS = 32.7 cmxB = 10.0 cmYou are given a small mass of 86.0 grams and an unknown large mass that you place at the 10.0 cm mark.You slide the small mass along the class 1 lever and manage to get the levers to balance when the small mass is 32.7 cm from its fulcrum.
64xE1 = 5.0 cm xS = 32.7 cm xS = 32.7 cm xE1 = 5.0 cm mL1 = 27.4 g First part of lever systemYou also measured the mass of the lever as 27.4 g and arranged the lever so that its center of gravity is 16.0 cm from the fulcrum.c.g.xS = 32.7 cmxS = 32.7 cmxE1 = 5.0 cmmL1 = 27.4 gc.g..mS = 86.0 gE = ?? gxL1 = 16.0 cmmSxS + mL1xL1 = ExE1
65(86.0 g)(32.7 cm) + (27.4 g)(16.0 cm) = E(5.0 cm) First part of lever systemmSxS + mL1xL1 = ExE1(86.0 g)(32.7 cm) + (27.4 g)(16.0 cm) = E(5.0 cm)E = [(86.0 g)(32.7 cm) + (27.4 g)(16.0 cm)]/(5.0 cm)E = 650 gxS = 32.7 cmxE1 = 5.0 cmmL1 = 27.4 gc.g..mS = 86.0 gE = ?? gxL1 = 16.0 cm
66xE2 = 35.0 cm xB = 10.0 cm xL2 = 18.0 cm xE2 = 35.0 cm xB = 10.0 cm Second part of lever systemSuppose the mass of the lever is 31.4 g and its center of gravity is 18.0 cm from the fulcrum.Bc.g.xE2 = 35.0 cmxB = 10.0 cmmL2 = 31.4 gc.g..E = 650 gmB = ?? gxL2 = 18.0 cmExE2 = mL2xL2 + mBxB
67(650 g)(35.0 cm) = (31.4 g)(18.0 cm) + mB(10.0 cm) Second part of lever systemExE2 = mL2xL2 + mBxB(650 g)(35.0 cm) = (31.4 g)(18.0 cm) + mB(10.0 cm)mB = [(650 g)(35.0 cm) - (31.4 g)(18.0 cm)]/(10.0 cm)mB = 2218 g = kgxE2 = 35.0 cmxB = 10.0 cmmL2 = 31.4 gc.g..E = 650 gmB = ?? gxL2 = 18.0 cm
68Things to note:You must build your own lever system.You may want to have two set places to have your fulcrum depending on the given masses.You may want to have the unknown mass at a predetermined spot and thus notching the lever at that point.Make sure that you know the mass of your lever and have marked the location of its center of gravity.
70Fixed Pulley Effort Effort LoadEffortFixed PulleyA fixed pulley is basically a First Class Lever that can rotate. The mechanical advantage is 1. All a fixed pulley does is change the direction of the force.LoadEffortFulcrum
71Movable Pulley Effort Effort LoadEffortA movable pulley is basically a Second Class Lever that can rotate. The mechanical advantage is 2.LoadEffortFulcrum
72LoadEffortBlock & TackleA Block and Tackle is a combination of a movable pulley connected to a fixed pulley. In this case the mechanical advantage of the movable pulley is 2 and the MA of the fixed pulley is 1. Combined the mechanical advantage is 2.In order to calculate the Ideal Mechanical Advantage, IMA, of a Block and Tackle, you count the number of supporting ropes.
73LoadEffortBlock & TackleA Block and Tackle is a combination of a movable pulley connected to a fixed pulley. In this case the mechanical advantage of the movable pulley is 2 and the MA of the fixed pulley is 1. Combined the mechanical advantage is 2.Work is done to lift the Load, but you also must lift the movable pulley with it. The AMA will be less, not only because of friction in the system but because the weight of the movable pulley also has to be lifted by the Effort.
74Wheel and Axle Effort Effort Load dE dL A Wheel and Axle is two different diameter cylinders on the same shaft. This also is a first class lever that can rotate.LoadEffortThe Ideal Mechanical Advantage is the ratio of the diameters. IMA = dE/dL.FulcrumEffortLoad
76Inclined PlanexhAn inclined plane is basically a ramp that is stationary. A load is pushed up the ramp instead of being lifted straight up. The Ideal Mechanical Advantage of a ramp is the ratio of the length of the ramp (x) to the height of the ramp (h). IMA = x/h
77WedgeA wedge is like an inclined plane, but instead of being stationary the wedge is driven into something or between things.