2 is the most volcanically active body in the solar system. Think about the following questions: What is this object? Where is it? Why does it look like that?Erupting Volcano!!IO is a moon of JupiterCompeting forces between Jupiter and the other Galilean moons cause the center of Io compress and melt. Consequently Iois the most volcanically active body in the solar system.
4 What is a force?IPC definition: A push or a pull exerted on some objectBetter definition: Force represents the interaction of an object with its environmentThe Unit for Force is a Newton
5 Two major types of forces Contact Forces: Result from physical contact between two objectsExamples: Pushing a cart, Pulling suitcaseField Forces: Forces that do not involve physical contactExamples: Gravity, Electric/Magnetic Force
6 Force is a vector! (yay more vectors ) The effect of a force depends on magnitude and direction
7 Force Diagrams (p. 126)Force Diagram: A diagram that shows all the forces acting in a situation
8 Free Body Diagrams p.127Free Body Diagrams (FBDs) isolate an object and show only the forces acting on itFBDs are essential! They are not optional! You need to draw them to get most problems correct!
9 How to draw a free body diagram Situation: A tow truck is pulling a car(p. 127)We want to draw a FBD for the car only.
10 Steps for drawing your FBD Step 1: Draw a shape representing the car (keep it simple)Step 2: Starting at the center of the object, Draw and label all the external forces acting on the objectForce of Tow Truck on Car=5800 N
11 Add force of gravity Force of Tow Truck on Car= 5800 N Gravitational force (Weight of car)=14700 NForce of Tow Truck on Car=5800 N
12 Add force of the road on the car (Called the Normal Force) Gravitational force (Weight of car)=14700 NForce of Tow Truck on Car=5800 NNormal Force =13690 N
13 Finally add the force of friction acting on the car Gravitational force (Weight of car)=14700 NForce of Tow Truck on Car=5800 NNormal Force =13690 NForce of Friction= 775 N
14 A Free Body Diagram of a Football Being Kicked FgFkick
15 A person is pushed forward with a force of 185 N A person is pushed forward with a force of 185 N. The weight of the person is 500 N, the floor exerts a force of 500 N up. The friction force is 30 N.FN= 500 NFg= 500 NFapp= 185 NFf= 30 N
16 Forces you will need Symbol of Force Description Fg Gravitational Force is the Weight of the Object (equal to mass x g= mg)FNNormal Force= Force acting perpendicular to surface of contactFfFrictional Force- Opposes applied force; acts in direction opposite of motionFappApplied Force
17 Sample Problem p. 128 #3Draw a free body diagram of a football being kicked. Assume that the only forces acting on the ball are the force of gravity and the force exerted by the kicker.
18 Newton’s 1st Law of Motion The Law of InertiaAn object at rest remains at rest, and an object in motion continues in motion with constant velocity (constant speed in straight line) unless the object experiences a net external forceThe tendency of an object not to accelerate is called inertia
19 AccelerationThe net external force (Fnet) is the vector sum of all the forces acting on an objectIf an object accelerates (changes speed or direction) then a net external force must be acting upon it
20 EquilibriumIf an object is at rest (v=0) or moving at constant velocity, then according to Newton’s First Law, Fnet =0When Fnet =0, the object is said to be in equilibrium
21 How do we use this information? Sample Problem p. 133 #2 A crate is pulled to the right with a force of 82.0 N, to the left with a force of 115 N, upward with a force of 565 N and downward with a force of 236 N.A. Find the net external force in the x directionB. Find the net external force in the y directionC. Find the magnitude and direction of the net external force on the crate.
22 Step 1: Draw a FBD Fup = 565 N Fright = 82 N Fleft = 115 N Fdown = 236 N
23 Find the vector sum of forces A. 82 N + (-115 N )= -33 NB. 565 N + (-236 N) = 329 NC. Find the resultant of the two vectors from part a and b.33 N329 NR = 331 N at 84.3 North of West
24 Newton’s 1st Law Review Newton’s 1st Law: When Fnet=0, an object is in equilibrium and will stay at rest or stay in motionIn other words, if the net external force acting on an object is zero, then the acceleration of that object is zero
25 Newton’s 2nd Law (p.137)The acceleration of an object is directly proportional to the net external force acting on the object and inversely proportional to the object’s mass
26 Example p. 138 # 4A 2.0 kg otter starts from rest at the top of a muddy incline 85 cm long and slides down to the bottom in 0.50 s. What net external force acts on the otter along the incline?
27 Solving the problem To calculate Fnet, we need m and a M=2.0 kg What is a?Vi= 0 m/s, t=0.50 s,displacement=85 cm=.85 mWelcome back kinematic equations!
29 Newtons’ 3rd Law Forces always exist in pairs For every action there is an equal and opposite reaction
30 Action- Reaction Pairs Some action-reaction pairs:
31 Although the forces are the same, the accelerations will not be unless the objects have the same mass.
32 Everyday Forces Weight= Fg = mg Normal Force= FN= Is always perpendicular to the surface.Friction Force= FfOpposes applied forceThere are two types of friction: static and kinetic
33 Static FrictionForce of Static Friction (Fs) is a resistive force that keeps objects stationaryAs long as an object is at rest:Fs = -Fapp
34 Kinetic FrictionKinetic Friction (Fk) is the frictional force on an object in motion
35 Coefficients of Friction The coefficient of friction (μ) is the ratio of the frictional force to the normal forceCoefficient of kinetic FrictionCoefficient of Static Friction
36 Sample Problem p. 145 #2A 25 kg chair initially at rest on a horizontal floor requires a 365 N horizontal force to set it in motion. Once the char is in motion, a 327 N horizontal force keeps it moving at a constant velocity.A. Find coefficient of static frictionB. Find coefficient of kinetic friction
37 Coefficient of Static Friction In order to get the chair moving, it was necessary to apply 365 N of force to overcome static friction. Therefore Fs = 365 N.The normal force is equal to the weight of the chair (9.81 x 25= 245 N)
38 Coefficient of Kinetic Friction The problem states that the chair is moving with constant velocity, which means Fnet=0. Therefore, Fapp must equal -Fk.Fapplied= 327 NFk= 327 N
40 Forces at an angleA woman is pulling a box to the right at an angle of 30 above the horizontal. The box is moving at a constant velocity. Draw a free body diagram for the situation.
41 FBD F app,y F app,x FN= Normal Force Fapp= Applied Force Ff= Friction ForceFg=Weight
42 What is Fnet?Since the suitcase is moving with constant velocity, Fnet=0.That means the forces in the x direction have to cancel out and the forces in y direction have to cancel outFk = Fapp,xFN + Fapp,y = FgNOTICE THAT NORMAL FORCE DOES NOT EQUAL WEIGHT IN THIS SITUATION
43 Let’s do an example. P. 154 #42A 925 N crate is being pushed across a level floor by a force F of 325 N at an angle of 25 above the horizontal. The coefficient of kinetic friction is Find the magnitude of the acceleration of the crate.
44 What do we need to know? So we need mass and Fnet. We have weight (925 N). So what is mass?How to find Fnet?Find vector sum of forces acting on crate.
45 FBD F app,y F app,x FN= Normal Force Fapp= 325 N Ff= Friction Force Fg=Weight=925 N
46 Finding Fnet,y Is box accelerating in y direction? No. Therefore Fnet in y direction is 0So FN + Fapp,y = FgSo FN = Fg- Fapp,y= 925 N- 325sin(25)FN= N
47 Finding Fnet,x Is box accelerating in x direction? Yes. Therefore Fnet,x is not 0Fnet,x= Fapp,x – FfFapp,x = Fappcos(25)=294.6 NUse coefficient of friction to find FfFf=μFN=(0.25)(787N)=197 N
48 Finish the Problem Fnet,x = 294 N – 197 N= 97 N So now we know that the Fnet on the box is 97 N since Fnet,y is 0
49 Another example. P. 154 #54 part a A box of books weighing 319 N is shoved across the floor by a force of 485 N exerted downward at an angle of 35° below the horizontal.If μk between the floor and the box is 0.57, how long does it take to move the box 4.00 m starting from rest?
57 Closer look at gravity triangle. Solve for Fg,y and Fg,xFgFg,xFg,yθ
58 Coordinate system for inclined planes Y axisX axis
59 Fnet in the y directionWhen a mass is sliding down an inclined plane, it is not moving in the y direction.Therefore Fnet,y =0 and all the forces in the y direction cancel out.
60 Forces In the y-direction So what are the forces acting in the y direction?Look at your FBDWe have normal force and Fg,ySince they have to cancel out…FN= mgcos(θ)
61 Forces in the x direction What is the force that makes the object slide down the inclined plane?Gravity…but only in the x direction
62 Remember that Vectors can be moved parallel to themselves!! FFFNFgθFg,yθFg,x
63 Forces in the x direction So what are the forces acting in the x direction?Friction Force (Ff) and Gravitational Force (Fg,x)If the box is in equlibriumFg,x = FfIf the box is acceleratingFnet= Fg,x - Ff
64 What if there is an additional applied force? Example: a box is being pushed up an inclined plane…FappFNFg,xFfFg,yθ
65 In that case… FN= mgcosθ Fnet = Fapp- Fg,x – Ff If the object is in equilibrium thenFapp= Fg,x + Ff
66 An Example p. 153 #40A 5.4 kg bag of groceries is in equilibrium on an incline of angle. Find the magnitude of the normal force on the bag.