Concept of Force G G Force is a vector quantity Fatotal ≡ m

The magnitude of the total force is defined to be Concept of Force and F = (mass) x (magnitude of the acceleration) Newton’s Laws of Motion The direction of the total force on a body is the 8.01 same as the direction of the acceleration. W03D1 The SI units for force are newtons (N): 1 N = 1kg . m/s2

Superposition Principle G G Forces Apply two forces and on a body, F1 F2 Gravitation the total force is the vector sum of the two forces: GGG total Electric and magnetic forces FFF=+12 Notation: The force acting on body 1 due to the G Elastic forces (Hooke’s Law) interaction between body 1 and body 2 is denoted by F12 Frictional forces: static and kinetic friction, fluid resistance Example: The total force exerted on Contact forces: normal forces and static m3 by m1 and m2 is: GGG friction total FFF33132=+ Tension and compression

1 Free Body Diagram Newton’s First Law

1. Represent each force that is acting on the object by an arrow on a free body force diagram that indicates the direction of GGG the force T FFF=++⋅⋅⋅12 Every body continues in its state of rest, or of 2. Choose set of independent unit vectors and draw them on uniform motion in a right line, unless it is free body diagram. compelled to change that state by forces G impressed upon it. 3. Decompose each force F in terms of vector components. G i Fijk=++FFFˆˆˆ G G G iixiyiz,,, F0=⇒= vconstant 4. Add vector components to find vector decomposition of the ∑ i total force TTT FFxxx=++⋅⋅⋅1, F 2, TTT FFyyy=1, + F 2, +⋅⋅⋅

TTT FFzzz=++⋅⋅⋅1, F 2,

Reference Systems Relatively Inertial Reference Frames Use coordinate system as a ‘reference frame’ to Two reference frames. describe the position, velocity, and acceleration of objects. Origins need not coincide.

One moving object has different position vectors in different frames G G G rRr12= + Relative velocity between the two reference frames G G VR= ddt is constant since the relative acceleration is zero G G G AV= ddt= 0

2 Law of Addition of Velocities Newton’s First Law Suppose the object is moving; then, observers in different reference frames will measure different Newton’s First Law in relatively inertial reference velocities frames: If there is no net force impressed on an object at rest in Frame 2, then there is also no G G net force impressed on the object in Frame 1. Velocity of the object in Frame 1: vr11= ddt G G vr= ddt Velocity of the object in Frame 2: 22 An object that is at rest in Frame 2 is moving at a constant velocity in reference Frame 1. Velocity of an object in two different reference GGG frames drrR 12=+ dt dt dt GGG vVv12=+

Table Problem Newton’s Second Law

Suppose two cars, Car A, and Car B, are traveling along roads that are perpendicular to The change of motion is proportional to the each other. One observer is at rest with respect motive force impresses, and is made in the direction to the ground. A second observer is in Car A. According to the observer on the ground, Car A of the right line in which that force is impressed, G ˆ G is moving with velocity vj AA = v , and Car B is G G ˆ Fa= m . moving with velocity vi BB = v . What is the velocity of Car B according to the observer in When multiple forces are acting, Car A? Express the velocity of Car B according N G to the observer in Car A as a vector (specify the G ∑Fai = m . components) and give the direction and i=1 magnitude. In Cartesian coordinates: NNN ∑∑∑Fxi,,,===ma x, F yi ma y , F zi ma z . iii===111

3 Newton’s Third Law Force Law: Newtonian Induction To every action there is always opposed an equal • Definition of force has no predictive content. reaction: or, the mutual action of two bodies upon • Need to measure the acceleration and the mass in order to define the each other are always equal, and directed to contrary force. parts. GG • Force Law: Discover experimental relation between force exerted on object and change in properties of object. FF1,2=− 2,1 • Induction: Extend force law from finite measurements to all cases within some range creating a model.

• Second Law can now be used to predict motion! Action-reaction pair of forces cannot act on same body; they act on different bodies. • If prediction disagrees with measurement adjust model.

Force Law: Gravitational Force Empirical Force Law: Hooke’s near the Surface of the Earth Law Consider a mass m attached to a spring

Near the surface of the earth, the gravitational Stretch or compress spring by interaction between a body and the earth is mutually different amounts produces attractive and has a magnitude of G different accelerations Fgrav= mg grav G Hooke’s law: ||F = klΔ m where grav is the gravitational mass of the body and g is a positive constant.

−2 Direction: restoring spring to equilibrium g = 9.80665 m ⋅s Hooke’s law holds within some reasonable range of extension or compression

4 Concept Question: Hooke’s Law Two identical springs with spring constant k are attached to each other. A block of mass m is suspended from the lower spring. What is the equivalent spring constant keq of the system of two springs? Note the block displaces Lecture Demo: the same amount in the two Spring Constant figures. 1) keq = k

2) keq = k/2

3) keq = 2k

4) an unknown relation; there is not enough information.

Group Problem: Hooke’s Law Force Laws: Contact Forces and Free Body Diagram Between Surfaces The spring in configuration (a) is stretched a The contact force between two surfaces is denoted by the vector distance Δx . The same spring in G G FCtotal ≡ configuration (b) will stretch a distance surface, hand Normal Force: Component of the contact force perpendicular to 1) 2 Δx surface and is denoted by G G 2) Δx normal FN≡ 3) Δx/2 surface, hand Friction Force: Component of the 4) 0 contact force tangent to the surface and is denoted by 5) Not sure G G GGG Fftangent ≡ surface, hand CNf≡+ Therefore the contact force can be modeled as a vector sum

5 Concept Question: Car-Earth Concept Question: Normal Force Interaction

Consider a person standing in an elevator that Consider a car at rest. We can conclude that the is accelerating upward. The upward normal downward gravitational pull of Earth on the force N exerted by the elevator floor on the car and the upward contact force of Earth on person is it are equal and opposite because

1) larger than 1) the two forces form an interaction pair. the net force on the car is zero. 2) identical to 2) 3) neither of the above. 3) smaller than 4) unsure

the downward force of gravity on the person.

Kinetic Friction Static Friction

The kinetic frictional force fk is proportional to the Varies in direction and magnitude depending on normal force, but independent of surface area of applied forces: contact and the velocity. 0 ≤ fs ≤ fs,max = μs N

The magnitude of fk is Static friction is equal to it’s maximum value f = μ N f = μ N s,max s k k

where µk is the coefficients of friction.

Direction of fk: opposes motion

6 Friction on Motorcycle Wheel Tension in a Rope A motorcycle has mass m and is accelerating 1. W μ forwards forwards with 2. Wμ/2 forwards The tension in a rope at a distance x from one end acceleration a. The 3. Wμ backwards of the rope is the magnitude of the action-reaction coefficient of friction of 4. Wμ/2 backwards the rear wheel with the pair of forces acting at that point , 5. Ma forwards road is μ. The weight W G G Ma backwards is equally divided 6. Tx()==FFleft,right () x right,left () x between front and rear 7. Ma/2 forwards wheels. The force of 8. Ma/2 backwards friction on the rear wheel 9. None of above is:

Concept Question: Tension in Rope You are trying to pull a rock resting on the ground with a heavy rope (the rope has non-zero mass). Just before the rock slips and starts to move, the magnitude of the tension in the rope is

1) greater than the magnitude of the pulling force?

2) equal to the magnitude of the pulling force?

3) less than the magnitude of the pulling force?

4) Not enough information is given to answer.

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