2Things to Remember from Last Year Newton's Three Laws of MotionInertial Reference FramesMass vs. WeightForces we studied: weight / gravity normal force tension friction (kinetic and static)Drawing Free Body DiagramsProblem Solving
3Newton's Laws of Motion ΣF = ma 1. An object maintains its velocity (both speed anddirection) unless acted upon by a nonzero net force.2. Newton’s second law is the relation betweenacceleration and force.3. Whenever one object exerts a force on a secondobject, the second object exerts an equal force inthe opposite direction on the first object.ΣF = ma
4Which of Newton's laws best explains why motorists should buckle-up? 1Which of Newton's laws best explains why motorists should buckle-up?AFirst LawBSecond LawCThird LawDLaw of Gravitation
5speed remained the same, but it is turning to the right. 2You are standing in a moving bus, facing forward, and you suddenly fall forward. You can infer from this that the bus'sAvelocity decreased.Bvelocity increased.Cspeed remained the same, but it is turning to the right.Dspeed remained the same, but it is turning to the left.
6normal force due to your contact with the floor of the bus 3You are standing in a moving bus, facing forward, and you suddenly fall forward as the bus comes to an immediate stop. What force caused you to fall forward?AgravityBnormal force due to your contact with the floor of the busCforce due to friction between you and the floor of the busDThere is not a force leading to your fall.
7slowly slow down, then stop. 4When the rocket engines on the spacecraft are suddenly turned off, while traveling in empty space, the starship willAstop immediately.Bslowly slow down, then stop.Cgo faster.Dmove with constant speed
85A net force F accelerates a mass m with an acceleration a. If the same net force is applied to mass 2m, then the acceleration will beA4aB2aCa/2Da/4
96A net force F acts on a mass m and produces an acceleration a. What acceleration results if a net force 2F acts on mass 4m?Aa/2B8aC4aD2a
10The table pushing up on the object with force mg 7An object of mass m sits on a flat table. The Earth pulls on this object with force mg, which we will call the action force. What is the reaction force?AThe table pushing up on the object with force mgBThe object pushing down on the table with force mgCThe table pushing down on the floor with force mgDThe object pulling upward on the Earth with force mg
11the force on the truck is greater then the force on the car 8A 20-ton truck collides with a 1500-lb car and causes a lot of damage to the car. Since a lot of damage is done on the carAthe force on the truck is greater then the force on the carBthe force on the truck is equal to the force on the carCthe force on the truck is smaller than the force on the carDthe truck did not slow down during the collision
12Inertial Reference Frames Newton's laws are only valid in inertial reference frames:An inertial reference frame is one in which Newton’s first law is valid. This excludes rotating and accelerating frames.
13Mass and WeightMASS is the measure of the inertia of an object, the resistance of an object to accelerate.WEIGHT is the force exerted on that object by gravity. Close to the surface of the Earth, where the gravitational force is nearly constant, the weight is:Mass is measured in kilograms, weight Newtons.FG = mg
14Normal Force and Weight FNmgThe Normal Force, FN, is ALWAYS perpendicular to the surface.Weight, mg, is ALWAYS directed downward.
15Mass is less, weight is same 9The acceleration due to gravity is lower on the Moon than on Earth. Which of the following is true about the mass and weight of an astronaut on the Moon's surface, compared to Earth?AMass is less, weight is sameBMass is same, weight is lessCBoth mass and weight are lessDBoth mass and weight are the same
1610A 14 N brick is sitting on a table. What is the normal force supplied by the table?A14 N upwardsB28 N upwardsC14 N downwardsD28 N downwards
17Kinetic FrictionFriction forces are ALWAYS parallel to the surface exerting them.Kinetic friction is always directed opposite to direction the object is sliding and has magnitude:fK = μkFNfKv
1811A 4.0kg brick is sliding on a surface. The coefficient of kinetic friction between the surfaces is What it the size of the force of friction?A8.0B8.8C9.0D9.8
19A brick is sliding to the right on a horizontal surface. 12A brick is sliding to the right on a horizontal surface.What are the directions of the two surface forces: the friction force and the normal force?Aright, downBright, upCleft, downDleft, up
20Static FrictionStatic friction is always equal and opposite the Net Applied Force acting on the object (not including friction).Its magnitude is:fS ≤ μSFNfSFAPP
2113A 4.0 kg brick is sitting on a table. The coefficient of static friction between the surfaces is What is the largest force that can be applied horizontally to the brick before it begins to slide?A16.33B17.64C17.98D18.12
2214A 4.0kg brick is sitting on a table. The coefficient of static friction between the surfaces is If a 10 N horizontal force is applied to the brick, what will be the force of friction and will the brick move?A16.12, noB17.64, noC16.12, yesD17.64, yes
23Tension ForceFTmgaWhen a cord or rope pulls on an object, it is said to be under tension, and the force it exerts is called a tension force, FT.
2415A crane is lifting a 60 kg load at a constant velocity. Determine the tension force in the cable.A568 NB578 NC504 ND600 N
2516A system of two blocks is accelerated by an applied force of magnitude F on the frictionless horizontal surface. The tension in the string between the blocks is:A6F6 kg4 kgFB4FC3/5 FD1/6 FE1/4 F
27Adding Orthogonal Forces Since forces are vectors, they add as vectors.The simplest case is if the forces are perpendicular (orthogonal) with another.Let's add a 50 N force at 0o with a 40 N force at 90o.
28Adding Orthogonal Forces 1. Draw the first force vector, 50 N at 0o, beginning at the origin.180o90o270o0o
29Adding Orthogonal Forces 1. Draw the first force vector, 50 N at 0o, beginning at the origin.2. Draw the second force vector, 40 N at 90o, with its tail at the tip of the first vector.180o90o270o0o
30Adding Orthogonal Forces 1. Draw the first force vector, 50 N at 0o, beginning at the origin.2. Draw the second force vector, 40 N at 90o, with its tail at the tip of the first vector.3. Draw Fnet from the tail of the first force to the tip of the last force.180o90o270o0o
31Adding Orthogonal Forces We get the magnitude of the resultant from the Pythagorean Theorem.c2 = a2 + b2Fnet2 = (50N)2 + (40N)2= 2500N N2= 4100N2Fnet = (4100N2)1/2= 64 N180o90o270o0o
32Adding Orthogonal Forces We get the direction of the net Force from the inverse tangent.tan(θ) = opp / adjtan(θ) = (40N) / (50N)tan(θ) = 4/5tan(θ) = 0.8θ = tan-1(0.80) = 39oFnet = 64 N at 39o180o90o270o0o
33Decomposing Forces into Orthogonal Components Let's decompose a 120 N force at 34o into its x and y components.Fθ0o
34Decomposing Forces into Orthogonal Components We have Fnet, which is the hypotenuse, so let's first find the adjacent side by usingcosθ = adj / hypcosθ = Fx / FnetFx = Fnet cosθ= 120N cos34o= 100 NFθ0oFx
35Decomposing Forces into Orthogonal Components Now let's find the opposite side by usingsinθ = adj / hypsinθ = Fy / FnetFy = Fnet sinθ= 120N sin34o= 67 NFyFθ0oFx
36Adding non-Orthogonal Forces Now that we know how to add orthogonal forces.And we know now to break forces into orthogonal components, so we can make any force into two orthogonal forces.We combine those two steps to add any number of forces together at any different angles.
37Adding non-Orthogonal Forces Let's add together these three forces:F1 = 8.0 N at 50oF2 = 6.5 N at 75oF3 = 8.4 N at 30oFirst, we will do it graphically, to show the principle for how we will do it analytically.Then, we will do it analytically, which is easy once you see why it works that way.
38Adding non-Orthogonal Forces First, here are the vectors all drawn from the origin.F1 = 4.0 N at 50oF2 = 6.5 N at 75oF3 = 8.4 N at 30o
39Adding non-Orthogonal Forces Now we arrange them tail to tip.F3F2F1F1 = 4.0 N at 50oF2 = 6.5 N at 75oF3 = 8.4 N at 30o
40Adding non-Orthogonal Forces Then we draw in the Resultant, the sum of the vectors.F3F2FFF1F2F3F1F1 = 4.0 N at 50oF2 = 6.5 N at 75oF3 = 8.4 N at 30o
41Adding non-Orthogonal Forces F1xF2xF3xF1yF2yF3yFNow, we graphically break each vector into its orthogonal components.F1 = 4.0 N at 50oF2 = 6.5 N at 75oF3 = 8.4 N at 30o
42Adding non-Orthogonal Forces Then, we remove the original vectors and note that the sum of the components is the same as the sum of the vectors.F3yF3xFF2yF2xF1yF1 = 4.0 N at 50oF2 = 6.5 N at 75oF3 = 8.4 N at 30oF1x
43Adding non-Orthogonal Forces This is the key result:The sum of the x- components of the vectors equals the x- component of the Resultant.As it is for the y- components.F3yFF2yFyF1yF1xF2xF3xF1 = 4.0 N at 50oF2 = 6.5 N at 75oF3 = 8.4 N at 30oFx
44Adding non-Orthogonal Forces This explains how we will add vectors analytically.1. Write the magnitude and direction (from 0 to 360 degrees) of each vector.2. Compute the values of the x and y components.3. Add the x components to find the x component of the Resultant.4. Repeat for the y-components.5. Use Pythagorean Theorem to find the magnitude of the Resultant.6. Use Inverse Tangent to find the Resultant's direction.
45Adding non-Orthogonal Forces First make a table and enter what you know.F1 = 8.0 N at 50oF2 = 6.5 N at 75oF3 = 8.4 N at 30oForceMagnitudeDirectionx-comp. (Fcosθ)y-comp. (Fsinθ)F1F2F3Fnet
46Adding non-Orthogonal Forces Then calculate the x and y components.F1 = 8.0 N at 50oF2 = 6.5 N at 75oF3 = 8.4 N at 30oForceMagnitudeDirectionx-comp. (Fcosθ)y-comp. (Fsinθ)F18.0 N50oF26.5 N75oF38.4 N30oFnet
47Adding non-Orthogonal Forces Then add up the x and y components.F1 = 8.0 N at 50oF2 = 6.5 N at 75oF3 = 8.4 N at 30oForceMagnitudeDirectionx-comp. (Fcosθ)y-comp. (Fsinθ)F18.0 N50o5.14 N6.13 NF26.5 N75o1.68 N6.28 NF38.4 N30o7.27 N4.2 NFnet
48Adding non-Orthogonal Forces Then use Pythagorean Theorem and tan-1 to determine Fnet.F1 = 8.0 N at 50oF2 = 6.5 N at 75oF3 = 8.4 N at 30oForceMagnitudeDirectionx-comp. (Fcosθ)y-comp. (Fsinθ)F18.0 N50o5.14 N6.13 NF26.5 N75o1.68 N6.28 NF38.4 N30o7.27 N4.20 NFnet14.09 N16.61 N
49Adding non-Orthogonal Forces Let's add together these three forces:F1 = 8.0 N at 50oF2 = 6.5 N at 75oF3 = 8.4 N at 30oForceMagnitudeDirectionx-comp. (Fcosθ)y-comp. (Fsinθ)F18.0 N50o5.14 N6.13 NF26.5 N75o1.68 N6.28 NF38.4 N30o7.27 N4.20 NFnet22.9 N14.09 N16.61 N
50Adding non-Orthogonal Forces Now use the same procedure to add these forces.DO NOT IGNORE NEGATIVE SIGNSF1 = 21 N at 60oF2 = 65 N at 150oF3 = 8.4 N at 330o
51Adding non-Orthogonal Forces F1 = 21 N at 60oF2 = 65 N at 150oF3 = 8.4 N at 330oForceMagnitudeDirectionx-comp. (Fcosθ)y-comp. (Fsinθ)F121 N60oF265 N150oF345 N330oFnet
52Answers Force F1 F2 F3 Fnet Magnitude Direction 21 N 60o 65 N 150o F1 = 21 N at 60oF2 = 65 N at 150oF3 = 8.4 N at 330oForceMagnitudeDirectionx-comp. (Fcosθ)y-comp. (Fsinθ)F121 N60o10.50 N8.66 NF265 N150oN32.50 NF345 N330o38.97 NNFnet131 N290o-6.82 N18.66 N
53I. The vector sum of the three forces must equal zero. 17Three forces act on an object. Which of the following is true in order to keep the object in translational equilibrium?I. The vector sum of the three forces must equal zero.II. The magnitude of the three forces must be equal.III. All three forces must be parallel.AI onlyBII onlyCI and III onlyDII and III onlyEI, II, and III
54Force and friction acting on an object Previously, we solved problems with multiple forces, but they were either parallel or perpendicular.For instance, draw the free body diagram of the case where a box is being pulled along a surface, with friction, at constant speed.
55Force and friction acting on an object FNmgFAPPfNow find the acceleration given that the applied force is 20 N, the box has a mass of 3.0kg, and the coefficient of kinetic friction is 0.20.
56Force and friction acting on an object FNmgFAPPfx - axisy - axisΣF = ma FAPP - fk = maFAPP - μkFN = maFAPP - μkmg = maa = (FAPP - μkmg)/ma = (20N - (0.20)(30N))/3.0kga = (20N - 6.0N)/3.0kg a = (14N)/3.0kga = 4.7 m/s2ΣF = maFN - mg = 0FN = mgFN = (3.0kg)(10m/s2)FN = 30NFAPP = 20Nm = 3.0kgμk = 0.20
57Forces at angles acting on an object Now we will solve problems where the forces act at an angle so that it is not parallel or perpendicular with one another.First we do a free body diagram, just as we did previously.
58Forces at angles acting on an object The next, critical, step is to choose axes.Previously, we always used vertical and horizontal axes, since one axis lined up with the forces...and the acceleration.FNmgFAPPfNow, we must choose axes so that all the acceleration is along one axis, and there is no acceleration along the other.You always have to ask, "In which direction could this object accelerate?" Then make one axis along that direction, and the other perpendicular to that.What's the answer in this case?
59Forces at angles acting on an object This time vertical and horizontal axes still work...since we assume the box will slide along the surface.However, if this assumption is wrong, we will get answers that do not make sense, and we will have to reconsider our choice.FNmgFAPPfyX
60Forces at angles acting on an object yXNow we have to break any forces that do not line up with our axes into components that do.In this case, FAPP, must be broken intoFx and FyFNFAPPfmg
61Forces at angles acting on an object FNmgFAPPfyXFyFxOnce we do that, we can now proceed as we did previously, using each component appropriately.
62Force and friction acting on an object FNmgfyXFAPPFyFxLet's use our work to find the acceleration if the applied force is 20 N at 37o above horizontal, the box has a mass of 3.0kg, and the coefficient of kinetic friction is 0.20.
63Force and friction acting on an object FAPP = 20N at 37om = 3.0kgμk = 0.20x - axisy - axisΣF = ma Fx - fk = maFx - μkFN = maFcosθ - μkmg = maa = (Fcosθ - μkFN)/ma = (20N cos37o- (0.20)(18N))/3.0kga = (16N - 3.6N)/3.0kg a = (12.4N)/3.0kga = 4.1 m/s2ΣF = maFN + Fy - mg = 0FN = mg - FyFN = mg - FsinθFN = (3.0kg)(10m/s2)- (20N)(sin37o)FN = 30N - 12NFN = 18NNote that FN is lower due to the force helping to support the objectFNmgfyXFAPPFAPPyFAPPx
64Normal Force and Friction Friction was reduced because the Normal Force was reduced; the box's weight, mg, was supported by the y- component of the force plus the Normal Force...so the Normal Force was lowered...lowering friction.mgFAPPFAPPyFNJust looking at the y-axisΣF = maFN + Fy - mg = 0FN = mg - FyFN = mg - FsinθFNFymg
65Normal Force and Friction yXFNWhat would happen with both the Normal Force and Friction in the case that the object is being pushed along the floor by a downward angled force?fFAPPmg
66Normal Force and Friction yXFNIn this case the pushing force is also pushing the box into the surface, increasing the Normal Force as well as friction.fFxFyFAPPmgJust looking at the y-axisΣF = maFN - Fy - mg = 0FN = mg + FyFN = mg + FsinθFNmgFy
6718 A mg B mg sin(θ) C mg cos(θ) D mg + F sin(θ) E mg – F sin(θ) The normal force on the box is:FappθAmgBmg sin(θ)Cmg cos(θ)Dmg + F sin(θ)Emg – F sin(θ)
6819 A B μ(mg - Fsin(θ)) C μ(mg + Fcos(θ)) D E μmg The frictional force on the box is:Aμ(mg + Fsin(θ))FappθBμ(mg - Fsin(θ))Cμ(mg + Fcos(θ))Dμ(mg - Fcos(θ))Eμmg
6920 v m A mg - Fapp cosθ B C mg D mg + Fapp sinθ E mg + Fapp cosθ A block of mass m is pulled along a horizontal surface at constant speed v by a force Fapp , which acts at an angle of θ with the horizontal. The coefficient of kinetic friction between the block and the surface is μ.The normal force exerted on the block by the surface is:FappθvmAmg - Fapp cosθBmg - Fapp sinθCmgDmg + Fapp sinθEmg + Fapp cosθ
70The friction force on the block is: 21A block of mass m is pulled along a horizontal surface at constant speed v by a force Fapp , which acts at an angle of θ with the horizontal. The coefficient of kinetic friction between the block and the surface is μ.The friction force on the block is:FappθvmAμ(mg - Fapp cosθ)Bμ(mg - Fapp sinθ)CμmgDμ(mg + Fapp sinθ)Eμ(mg + Fapp cosθ)
71Normal Force and Weight FNmgPreviously we dealt mostly with horizontal (or, rarely, vertical surfaces). In that case FN, and mg were always along the same axis.Now we will look at the more general case.
72Inclined PlaneOn the picture, draw the free body diagram for the block.Show the weight and the normal force.
73Inclined Plane FN is ALWAYS perpendicular to the surface. mg is ALWAYS directed downward.But now, they are neither parallel or perpendicular to one another.FNmg
74Choosing AxesPreviously, we used vertical and horizontal axes. That worked because problems always resulted in an acceleration that was along one of those axes.We must choose axes along which all the acceleration is along one axis...and none along the other.FNmg
75Choose of AxesIn this case, the block can only accelerate along the surface of the plane.Even if there is no acceleration in a problem, this is the ONLY POSSIBLE direction of acceleration.So we rotate our x-y axes to line up with the surface of the plane.yXFNmga
76Inclined Plane Problems By the way, here's one way to see that θ is the both the angle of incline and the angle between mg and the new y-axis.Let's name the angle of the inclined plane θ and the angle between mg and the x-axis θ' and show thatθ = θ'Since the angles in a triangle add to 180o, and the bottom left angle is 90o, that means:α + θ = 90oy-axisFNmgαθ'θ
77Inclined Plane Problems Now look at the angles in the upper left corner of the triangle.The surface of the inclined plane forms an angle of 180o, so90o + θ' + α = 180oθ' + α = 90oBut we already showed thatα + θ = 90oθ' + α = 90o = α + θθ' = θy-axisFNmgαθ'θ
78FN Inclined Plane Problems a mg We need to resolve all forces which don't align with an axis into components that will.In this case, we resolve mg into its x and y components.yXFNmgmg sin θmg cos θθayXFNmgaθθ
79Now we just use Newton's Second Law, which is true for each axis. Inclined Plane ProblemsNow we just use Newton's Second Law, which is true for each axis. x - axisy - axisFNmgmg sin θmg cos θθayXΣFx = maxmgsinθ = maxa = gsinθdown the planeΣFy = mayFN - mgcosθ = 0FN = mgcosθ
80θ ΣFx = max mg sin θ = ma g sin θ = a a = g sin θ AnsweryxFNmgmg sin θmg cos θθΣFx = maxmg sin θ = mag sin θ = aa = g sin θa = 10 m/s2 sin (30o)a = 10 m/s2 (0.5)a = 5 m/s2θA 5 kg block slides down a frictionless incline at an angleof 30 degrees.a) Draw a free body diagram.b) Find its acceleration.(Use g = 10 m/s2)
81Inclined Plane Problems with Friction We can add kinetic friction to our inclined plane example. The kinetic friction points opposite the direction of motion.We now have a second vector along the x axis. fk points in the negative direction (recall fk = μk FN)yxFNmgmg sinθmg cosθθfkaFNmgθfka
83Inclined Plane Problems with Friction The general solution for objects sliding down an incline is:a = g(sinθ - μkcosθ)Note, that if there is no friction:μk = 0and we get our previous result for a frictionless plane:a = gsinθ
84If the object is sliding with constant velocity: a = 0 Inclined Plane Problems with Frictiona = g(sinθ - μkcosθ)If the object is sliding with constant velocity: a = 0yxFNa = g(sinθ - μkcosθ)0 = g(sinθ - μkcosθ)0 = sinθ - μkcosθμk cosθ = sinθμk = sinθ / cosθμk = tanθfka = 0mg cosθmgθmg sinθ
85Inclined Plane with Static Friction We just showed that for an object sliding with constant velocity down an inclined plane that:μk = tanθSimilarly, substituting μs for μk (at the maximum static force; the largest angle of incline before the object begins to slide) than:μs = tanθmaxBut this requires that the MAXIMUM ANGLE of incline, θmax, be used to determine μs.yxFNfsa = 0mg cosθmgθmg sinθ
86AnsweryxFNmgmg sin θmg cos θθfkθΣF = mamg sin θ - fK = 0mg sin θ = fKmg sin θ = μ FNmg sin θ = μ mg cos θμ = mg sin θ / mg cos θμ = tan θμ = tan 30 = 0.58A 5 kg block slides down a incline at an angle of 30o with a constant speed.a) Draw a free body diagram.b) Find the coefficient of frictionbetween the block and the incline.(Use g = 10 m/s2)
87AnswerFNmgmg sin θmg cos θθFappfKa = 0y-directionΣF = maFN - mg cos θ = 0FN = mg cos θθx-directionΣF = maFapp - mg sin θ - fk = maFapp - mg sin θ -μk FN = 0Fapp = mg sin θ + μk mg cos θFapp = 38 NA 5 kg block is pulled up an incline at an angle of 30o at a constant velocity The coefficient of friction between the block and the incline is 0.3.a) Draw a free body diagram.b) Find the applied force.(Use g = 10 m/s2)
88FN Fapp fK mg x-axis ΣF = ma Fapp - mg sin θ - fk = ma mg cos θθFappfKay-axisΣF = maFN - mg cos θ = 0FN = mg cos θx-axisΣF = maFapp - mg sin θ - fk = maFapp - mg sin θ - μk FN = maFapp - mg sin θ - μk mg cos θ = maa = (Fapp/m) - g sin θ - μk g cos θa = 0.4 m/s2A 5 kg block is pulled UP an incline at an angle of 30 degrees with a force of 40 N. The coefficient of friction between the block and the incline is 0.3.a) Draw a free body diagram.b) Find the block's acceleration.(Use g = 10 m/s2)
89Inclined Plane Problems If a mass, m, slides down a frictonless inclined plane, we have this general setup:yxFNmgθaIt is helpful to rotate our reference frame so that the +x axis is parallel to the inclined plane and the +y axis points in the direction of FN.
90FN fs mg a = 0 x-direction ΣF = ma fs = mg sin θ mg cos θθfsa = 0y-directionΣF = maFN - mg cos θ = 0FN = mg cos θx-directionΣF = ma fs = mg sin θμs mg cos θ = mg sin θμs cos θ = sin θμs = sin θ / cos θμs = tan θθ = tan-1 μs θ = 21.8oA 5 kg block remains stationary on an incline. The coefficients of static and kinetic friction are 0.4 and 0.3, respectively.a) Draw a free body diagram.b) Determine the angle that the block will start to move.(Use g = 10 m/s2)
9122A block of mass m slides down a rough incline as shown. Which free body diagram correctly shows the forces acting on the block? ABCDEfWNfWNfWNfWNfWN
9223A block with a mass of 15 kg slides down a 43° incline as shown above with an acceleration of 3 m/s2.What is the normal force N appliedby the inclined plane onthe block?A95 NB100 NC105 ND110 NE115 N
9324A block with a mass of 15 kg slides down a 43° incline as shown above with an acceleration of 3 m/s2.The magnitude of thefrictional force along theplane is nearly:A56 NB57 NC58 ND59 NE60 N
94Static EquilibriumThere is a whole field of problems called "Statics" that has to do with cases where no acceleration occurs, objects remain at rest.Anytime we construct something (bridges, buildings, houses, etc.) we want them to remain stationary, not accelerate. So this is a very important field.The two types of static equilibrium are with respect to linear and rotational acceleration, a balancing of force and of torque (forces that cause objects to rotate). We'll look at them in that order.
95Tension ForcePreviously, we did problems where a rope supporting an object exerted a vertical force straight upward, along the same axis as the force mg was pulling it down. That led to the simple case that if a = 0, then FT = mgFTmg
9625A uniform rope of weight 20 N hangs from a hook as shown above. A box of mass 60 kg is suspended from the rope. What is the tension in the rope?A20 N throughout the ropeB120 N throughout the ropeC200 N throughout the ropeD600 N throughout the ropeEIt varies from 600 N at the bottom of the rope to 620 N at the top.
97Tension ForceBut it is possible for two (or more) ropes to support an object (a = 0) by exerting forces at angles. In that case:The vertical components of the force exerted by each rope must add up to mg.And the horizontal components must add to zero.T2T1mg
98Tension ForceyXSo we need to break the forces into components that align with our axes.T2T1mg
99Tension Force y One result can be seen just from this diagram. mgT1xT2xyXT2yT1yOne result can be seen just from this diagram.In order for the forces along the x-axis to cancel out, the more vertical Tension must be much larger than the Tension which is at a greater angle to the vertical.In problems with two supporting ropes or wires, the more vertical has a greater tension.
100Tension Force50o20omgT1xT2xyXT2yT1yLet's calculate the tension in two ropes if T1 is at an angle of 50o from the vertical and T2 is at an angle of 20o from the vertical and they are supporting a 8.0 kg mass.Note, that in this case the horizontal component is given by Tsinθ, where before it was given by Tcosθ.To know which function to use you MUST draw the diagram and find the sides opposite and adjacent to the given angle.
101Tension Force y y - axis x - axis T2y T1y T1x T2x X mg ΣFx = max = 0 T1sinθ1 = T2sinθ2T1 = T2sinθ2/sinθ1T1 = T2 (sin20o/sin50o)T1 = T2 (0.34/0.77)T1 = 0.44 T2T1 = 0.44 (64N)T1 = 28NΣFy = may = 0T1y + T2y - mg = 0T1cosθ1 + T2cosθ2 = mg (0.44T2)cos(50o) + T2cos(20o) = mg(0.44T2)(0.64) + T2(0.94) =(8kg)(9.8m/s2)1.22 T2 = 78NT2 = 78N / 1.22T2 = 64N
102Tension ForceθθyXIf the ropes form equal angles to the vertical, the tension in each must also be equal, otherwise the x- components of Tension would not add to zero.TyTyθθTxTxmg
103Tension Force x - axis y - axis y Ty Ty Tx Tx X mg ΣFy = may = 0 θθyXΣFy = may = 0T1y + T2y - mg = 0Tcosθ + Tcosθ = mg 2Tcosθ = mg T = mg / (2cosθ)Note that the tension rises as cosθ becomes smaller...which occurs as θ approaches 90o.It goes to infinity at 90o, which shows that the ropes can never be perfectly horizontal.ΣFx = max = 0T1x - T2x = 0Tsinθ = Tsinθwhich just confirms that if the angles are equal, the tensions are equalTyTyθθTxTxmg
10426A lamp of mass m is suspended from two ropes of unequal length as shown above. Which of the following is true about the tensions T1 and T2 in the cables?T1T2AT1 > T2BT1 = T2CT1 < T2DT1 - T2 = mgET1 + T2 = mg
10527A large mass m is suspended from two massless strings of an equal length as shown below. The tension force in each string is:θmA½ mg cos(θ)B2 mg cos(θ)Cmg cos(θ)Dmg/cos(θ)Emg/2cos(θ)
106Torque and Rotational Equilibrium Forces causes objects to linearly accelerate.Torque causes objects to rotationally accelerate.Rotational dynamics is a major topic of AP Physics C: Mechanics. It isn't particularly difficult, but for AP B, we only need to understand the static case, where all the torques cancel and add to zero.First we need to know what torque is.
107Torque and Rotational Equilibrium Until now we've treated objects as points, we haven't been concerned with their shape or extension in space.We have assumed that any applied force acts through the center of the object and it is free to accelerate. That does not result in rotation, just linear acceleration.But if the force acts on an object so that it causes the object to rotate around its center of mass...or around a pivot point, that force has exerted a torque on the object.
108Torque and Rotational Equilibrium A good example is opening a door, making a door rotate. The door does not accelerate in a straight line, it rotates around its hinges.Think of the best direction and location to push on a heavy door to get it to rotate and you'll have a good sense of how torque works.Which force (blue arrow) placed at which location would create the most rotational acceleration of the green door about the black hinge?
109Torque and Rotational Equilibrium The maximum torque is obtained from:The largest forceAt the greatest distance from the pivotAt an angle to the line to the pivot thatis closest to perpendicularMathematically, this becomes:τ = Frsinθτ (tau) is the symbol for torque;F is the applied forcer is the distance from the pivotθ is the angle of the force to a line to the pivot
110Torque and Rotational Equilibrium τ = FrsinθWhen r decreases, so does the torque for a given force. When r = 0, τ = 0.When θ differs from 90o, the torque decreases. When θ = 0o or 180o, τ = 0.Fθr
111Two Physical Interpretations of Torque These are all equivalent. τ = F(rsinθ)The torque due to F acting at angle θ can be replaced by F acting at a smaller distance rsinθ.τ = r(Fsinθ)The torque due to F acting at angle θ can be replaced by a smaller perpendicular force Fsinθ acting at r.FsinθrsinθFThese are all equivalent.FθθrFFsinθFrθrrsinθ
112Rotational Equilibrium When the sum of the torques on an object is zero, the object is in rotational equilibrium.Define counter clockwise (CCW) as the positive direction for rotation and clockwise (CW) as the negative.For instance, what perpendicular force, F, must be applied at a distance of 7.0 m for the pivot to exactly offset a 20N force acting at a distance of 4.0m from the pivot of a door at an angle of 30o?
11612 kg15 kgA 12 kg load hangs from one end of a rope that passes over a small frictionless pulley. A 15 kg counterweight is suspended from the other end of the rope. The system is released from rest.a. Draw a free-body diagram for each object showing all applied forces in relative scale. Next to each diagram show the direction of the acceleration of that object.b. Find the acceleration each mass.c. What is the tension force in the rope?d. What distance does the 12 kg load move in the first 3 s?e. What is the velocity of 15 kg mass at the end of 5 s?
117m1 = 12 kg m2 = 15 kg c. FT = m1g + m1a or FT = m2g - m2a = 12kg(10 m/s2) + 12kg(1.11 m/s2)= 120 N N= NFT = m2g - m2a = 15kg(10 m/s2) - 15kg(1.11 m/s2) = 150 N N = Nb. ΣF = maΣF = FT - m1g = m1aFT = m1g + m1aΣF = FT - m2g = -m2a FT = m2g - m2aFT = FTm1g + m1a = m2g - m2am1a + m2a = m2g - m1ga(m1 + m2) = g(m2 - m1)a = (g(m2 - m1)) / (m1 + m2)= (10 m/s2 (15kg - 12kg)) / (12kg + 15kg)= 30 m/s2 * kg / 27kg a = 1.11 m/s2
118m1 = 12 kg m2 = 15 kg d. x = xo + vot + 1/2 at2 xo = 0 vo = 0 x = 1/2 at2= 1/2 (1.11 m/s2)(3s)2= 5 me. v = vo + atv = at= (1.11 m/s2)(5s)= 5.55 m/s
119500 g300 gA 500 g block lies on a horizontal tabletop. The coefficient of kinetic friction between the block and the surface is The block is connected by a massless string to the second block with a mass of 300 g. The string passes over a light frictionless pulley as shown above. The system is released from rest.a. Draw clearly labeled free-body diagrams for each of the 500 g and the 300g masses. Include all forces and draw them to relative scale. Draw the expected direction of acceleration next to each free-body diagram.b. Use Newton’s Second Law to write an equation for the 500 g mass.c. Use Newton’s Second Law to write an equation for the 300 g mass.d. Find the acceleration of the system by simultaneously solving the system of two equations.e. What is the tension force in the string?
121FA 10-kilogram block is pushed along a rough horizontal surface by a constant horizontal force F as shown above. At time t = 0, the velocity v of the block is 6.0 meters per second in the same direction as the force. The coefficient of sliding friction is 0.2. Assume g = 10 meters per second squared.a. Calculate the force F necessary to keep the velocity constant.
122a. Calculate the force F necessary to keep the velocity constant. m = 10 kgv = 6 m/sμ = 0.2g = 10 m/s2a. Calculate the force F necessary to keep the velocity constant.x yΣF = ma ΣF = FN - mg = 0Fapp - fk = 0 FN = mgFapp = fkFapp = μFNFapp = μmgFapp = (0.2)(10kg)(10m/s2)Fapp = 20 NFNfkFappmg
123A helicopter holding a 70-kilogram package suspended from a rope 5 A helicopter holding a 70-kilogram package suspended from a rope 5.0 meters long accelerates upward at a rate of 5 m/s2. Neglect air resistance on the package.a. On the diagram, draw and label all of the forces acting on the package.b. Determine the tension in the rope.c. When the upward velocity of the helicopter is 30 meters per second, the rope is cut and the helicopter continues to accelerate upward at 5 m/s2. Determine the distance between the helicopter and the package 2.0 seconds after the rope is cut.
124m = 70 kg a = 5 m/s2 x = 5 m ΣF = ma FT - mg = ma FT = mg + ma a. On the diagram, draw and label all of the forces acting on the package.b. Determine the tension in the rope.mgFTaΣF = maFT - mg = maFT = mg + maFT = m (g+a)FT = (70kg) (10 m/s2 + 5 m/s2)FT = (70kg) (15 m/s2)FT = 1050 N
125ahelicopterapackagem = 70 kga = 5 m/s2x = 5 mv = 30 m/sc. When the upward velocity of the helicopter is 30 meters per second, the rope is cut and the helicopter continues to accelerate upward at 5.2 m/s2. Determine the distance between the helicopter and the package 2 seconds after the rope is cut.Method 1Use the reference frame of the package after it is released.The relative acceleration of the helicopter is given by:arelative = ahelicopter - apackagearelative = 5 m/s2 - (-10 m/s2)arelative = 15 m/s2Their relative initial velocity is zero, since they are connected togetherThe helicopter is 5.0 m above the package when released, the length of the rope.y = y0 + v0 + 1/2at2y = 5m /2(15 m/s2)(2s)2y= 35m
126ahelicopterapackagem = 70 kga = 5 m/s2x = 5 mv = 30 m/sc. When the upward velocity of the helicopter is 30 meters per second, the rope is cut and the helicopter continues to accelerate upward at 5.2 m/s2. Determine the distance between the helicopter and the package 2 seconds after the rope is cut.Method 2Use the reference frame of the earth, but use the initial height of the package as zero.Package Helicopter y = y0 + v0t+ 1/2at2 y = y0 + v0t+ 1/2at2y = 0 + (30 m/s)(2s) + 1/2(-10 m/s2)(2s)2 y = 5m + (30 m/s)(2s) + 1/2(+5 m/s2)(2s)2y= 60 m - 20 m y= 5m + 60 m + 10 my= 40 m y= 75 mTheir separation is just the difference in their positionsΔy = yhelicopter - ypackageΔy = 75m - 40mΔy = 35m
127a. The acceleration of the 4-kilogram block 1 kg2 kg4 kgThree blocks of masses 1.0, 2.0, and 4.0 kilograms are connected by massless strings, one of which passes over a frictionless pulley of negligible mass, as shown above. Calculate each of the following.a. The acceleration of the 4-kilogram blockb. The tension in the string supporting the 4-kilogram blockc. The tension in the string connected to the l-kilogram block
129m1 = 1 kgm2 = 2 kgm3 = 4 kgg = 9.8 m/s2b. The tension in the string supporting the 4-kilogram blockFT2 = m1a + m2a FT2 = 1kg(5.6 m/s2) + 2kg(5.6 m/s2)FT2 = 16.8 NorFT2 = m3g - m3aFT2 = 4kg(9.8 m/s2) - 4kg(5.6 m/s2)c. The tension in the string connected to the l-kilogram blockFT1 = m1aFT1 = 1kg(5.6 m/s2)FT1 = 5.6 N
130a. Calculate the acceleration of the 10-kilogram block in case I. 10 kg5 kgCase IIA 10-kilogram block rests initially on a table as shown in cases I and II above. The coefficient of sliding friction between the block and the table is The block is connected to a cord of negligible mass, which hangs over a massless frictionless pulley. In case I a force of 50 newtons is applied to the cord. In case II an object of mass 5 kilograms is hung on the bottom of the cord. Use g = 10 meters per second squared.a. Calculate the acceleration of the 10-kilogram block in case I.b. On the diagrams below, draw and label all the forces acting on each block in case IIc. Calculate the acceleration of the 10-kilogram block in case II.d. Calculate the tension force in case II.
131a. Calculate the acceleration of the 10-kilogram block in case I. m1 = 10 kgm2 = 5 kgμ = 0.2FApp = 50 Ng = 10 m/s2a. Calculate the acceleration of the 10-kilogram block in case I.ΣF = maFApp - fK = maFApp - μmg = maa = (FApp - μmg)/ma = (50N - 0.2(10kg)(10m/s2))/10kga = 3 m/s2yΣF = ma ΣF = FN - mg = 0FN = mgfk = μFNfk = μmgb. On the diagrams below, draw and label all the forces acting on each block in case IIFNm1gfFTm2g5 kg10 kg
132m1 = 10 kgm2 = 5 kgμ = 0.2FApp = 50 Ng = 10 m/s2c. Calculate the acceleration of the 10-kilogram block in case II.ΣF = maFT - f = m1a FT - m2g = -m2aFT - μm1g = m1a FT = m2g - m2aFT = μm1g + m1aμm1g + m1a = m2g - m2am1a + m2a = m2g - μm1ga(m1 + m2) = g(m2 - μm1)a = g(m2 - μm1)/(m1 + m2) = 10 m/s2(5kg - 0.2*10kg)/(5kg +10kg)a = 2 m/s2d. Calculate the tension force in case II.FT - m2g = -m2aFT = m2g - m2aFT = m2(g - a) = 5kg (10 m/s2 - 2 m/s2) = 40 N
133c. The magnitude of the acceleration of M1 In the system shown above, the block of mass M1 is on a rough horizontal table. The string that attaches it to the block of mass M2 passes over a frictionless pulley of negligible mass. The coefficient of kinetic friction mk between M1 and the table is less than the coefficient of static friction msa. On the diagram below, draw and identify all the forces acting on the block of mass M1.b. In terms of M1 and M2 determine the minimum value of ms that will prevent the blocks from moving.The blocks are set in motion by giving M2 a momentary downward push. In terms of M1, M2, mk, and g, determine each of the following:c. The magnitude of the acceleration of M1d. The tension in the string.
134M1, M2, mk, ms, ga. On the diagram below, draw and identify all the forces acting on the block of mass M1.b. In terms of M1 and M2 determine the minimum value of ms that will prevent the blocks from moving.The pulley system is not moving, so acceleration is zero.ΣF = max-direction for M1: FT - f = 0 FT = fy-direction for M1: FN - M1g = 0 FN = M1gy-direction for M2: FT - M2g = 0 FT = M2gf = μsFN or msFN in this casef = msM1g = M2gms = M2 / M1M1gFNFTf
135The blocks are set in motion by giving M2 a momentary downward push The blocks are set in motion by giving M2 a momentary downward push. In terms of M1, M2, mk, and g, determine each of the following:c. The magnitude of the acceleration of M1ΣF = max-direction for M1: FT - f = M1a FT = f + M1ay-direction for M1: FN - M1g = 0 FN = M1gf = μkFN or mkFN in this casef = mkM1gy-direction for M2: FT - M2g = -M2a FT = M2g - M2aFT = f + M1a, FT = M2g - M2amkM1g + M1a = M2g - M2aM1a + M2a = M2g - mkM1ga(M1 + M2) = M2g - mkM1ga = M2g - mkM1g / (M1 + M2)d. The tension in the string.FT = f + M1a or FT = M2g - M2aFT = mkM1g + M1(M2g - mkM1g / (M1 + M2) ) orFT = M2g - M2(M2g - mkM1g / (M1 + M2) )
136a. What is the tension FT in the string. m1m22 m60oTwo 10-kilogram boxes are connected by a massless string that passes over a massless frictionless pulley as shown above. The boxes remain at rest, with the one on the right hanging vertically and the one on the left 2.0 meters from the bottom of an inclined plane that makes an angle of 60° with the horizontal. The coefficients of kinetic friction and static friction between the Ieft-hand box and the plane are 0.15 and 0.30, respectively. You may use g = 10 m/s2, sin 60° = 0.87, and cos 60° = 0.50.a. What is the tension FT in the string.b. On the diagram above, draw and label all the forces acting on the box that is on the plane.c. Determine the magnitude of the frictional force acting on the box on the plane.d. The string is then cut and the left-hand box slides down the inclined plane. What is the magnitude of its acceleration?
137m1 = 10 kgm2 = 10 kgx = 2 mθ = 60o , sin 60° = 0.87, cos 60° = 0.50μk = 0.15, μs = 0.30g = 10 m/s2a. What is the tension FT in the string.The acceleration is 0 because the system is not moving.ΣF = max-direction for m1: FT - f - m1gsinθ = 0 FT = f + m1gsinθy-direction for m1: FN - m1gcosθ = 0 FN = m1gcosθf = μsFNf = μsm1gcosθFT = f + m1gsinθFT = μsm1gcosθ + m1gsinθFT = (0.30)(10kg)(10 m/s2)(0.50) + (10kg)(10 m/s2)(0.87)FT = 102 N
138y-direction for m1: FN - m1gcosθ = 0 FN = m1gcosθ f = μsFN m1 = 10 kgm2 = 10 kgx = 2 mθ = 60o , sin 60° = 0.87, cos 60° = 0.50μk = 0.15, μs = 0.30g = 10 m/s2b. On the diagram above, draw and label all the forces acting on the box that is on the plane.c. Determine the magnitude of the frictional force acting on the box on the plane.y-direction for m1: FN - m1gcosθ = 0 FN = m1gcosθf = μsFNf = μsm1gcosθf = 0.30(10kg)(10 m/s2)(0.50)f = 15 Nm1FNfm1gFT60o
139x-direction of m1: f - m1gsinθ = -m1a m1 = 10 kgm2 = 10 kgx = 2 mθ = 60o , sin 60° = 0.87, cos 60° = 0.50μk = 0.15, μs = 0.30g = 10 m/s2d. The string is then cut and the left-hand box slides down the inclined plane. What is the magnitude of its acceleration?ΣF = max-direction of m1: f - m1gsinθ = -m1ay-direction for m1: FN - m1gcosθ = 0 FN = m1gcosθf = μkFN = μkm1gcosθm1a = m1gsinθ - fa = (m1gsinθ - μkm1gcosθ) / m1a = ((10kg)(10 m/s2)(0.87) - (0.15)(10kg)(10 m/s2)(0.50)) / 10kga = 7.95 m/s2m1m260oafFNm1g
140m1m2MvθMasses ml and m2, are connected by a light string. They are further connected to a block of mass M by another light string that passes over a pulley. Blocks l and 2 move with a constant velocity v down the inclined plane, which makes an angle θ with the horizontal. The kinetic frictional force on block 1 is f and that on block 2 is 2f.a. On the figure above, draw and label all the forces on block m1, m2, and M.Express your answers to each of the following in terms of ml, m2, g, θ, 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.d. The string between blocks 1 and 2 is now cut. Determine the acceleration of block 1 while it is on the inclined plane.
141m1m2Mvθa. On the figure above, draw and label all the forces on block m1, m2, and M.m1vm1gFT1fFNa = 0m2m2gFT2MMgθ
142Express your answers to each of the following in terms of ml, m2, g, θ, and f. b. Determine the coefficient of kinetic friction between the inclined plane and block 1.f = μFNΣF = maFN - m1gcosθ = 0FN = m1gcosθf = μm1gcosθμ = f /(m1gcosθ)c. Determine the value of the suspended mass M that allows blocks 1 and 2 to move with constant velocity down the plane.m1: ΣF = ma M: FT2 - Mg = 0 FT1 - m1gsinθ - f = 0 FT2 = Mg FT1 = m1gsinθm2: ΣF = ma FT2 - FT1 - m2gsinθ - 2f = 0 Mg - m1gsinθ - f - m2gsinθ - 2f = 0 Mg = m1gsinθ + m2gsinθ + 3f M = (m1gsinθ + m2gsinθ + 3f) / g
143d. The string between blocks 1 and 2 is now cut d. The string between blocks 1 and 2 is now cut. Determine the acceleration of block 1 while it is on the inclined plane.x-direction: ΣF = ma f - m1gsinθ = -m1a m1gsinθ - f = m1a a = (m1gsinθ - f) / m1m1m1gfFNaθ
145Fluid ResistanceWhen first learning dynamics discussed cases in which an object moves with any friction. Then we began to take the frictional force into account between an object and its surface. Now we are going to begin taking into account air resistance.vFresistance
146Fluid Resistance No resistance v = vo + gt vo = 0 a is constant yNo resistancev = vo + gtvo = 0a is constantv increasingmg = maa = g
147Fluid ResistanceIf we do not take into account air resistance then the object will fall with a constant acceleration.If air resistance however is taken into account then the objects acceleration will eventually be cancelled out by the drag force.Since the acceleration will essentially become zero, at some point in time your velocity will be constant.
148Fluid Resistance With resistance FRa=gvo=0In this example the equation for the resistive force is given asF = kv, but in other problems the resisitive force equation can be different.When the falling object stops accelerating it reaches its terminal velocity.
149If the equation for the force due to air resistance 28If the equation for the force due to air resistanceis kv2. What is the terminal velocity reached by the object?ABCDE