Lecture 03: 1-D Statics and DynamicsNewtons LawsForcesFree Body DiagramsSpringsTension2-D Statics ExamplesDynamicsApparent Weight

Newtons 1st Law of MotionIf the sum of all external forces on an object is zero, then its speed and direction will not change. (Know as the Law of Inertia.)

Example: air hockey tableWhen the puck travels with no net force on it, it moves in a straight line with constant speed.The only time the speed or direction changes is when a net force (from the walls or mallets) is exerted on the puck!

Newtons 2nd Law of MotionIf a nonzero net force is applied to an object its motion will change. (The change is described by the equation F= ma.)

Example: dropping a bookWhen you drop a Calculus book from your 10th story dorm window, there is a net force acting on it (gravity) thus it begins to accelerate toward the street below!

Newtons 3rd Law of MotionIn an interaction between two objects, the forces that each exerts on the other are equal in magnitude and opposite in direction.

Example: rowing a boatWhen you row, you are pushing the water backwardsThus the water pushes your paddle (and therefore you) forward!

Forces Quantifies interactions between objects.

4 Fundamental ForcesGravityElectromagneticStrong NuclearWeak Nuclear

Forces are VectorsHave magnitude and directionSI units are Newtons (N)

GravityAny two objects with mass are attracted.

On the surface of the earth, the force of attraction is referred to the force of gravity or weight. rearth = 6.4x106 m, mearth = 6x1024 kg, G = 6.67x10-11 N-m2/kg2Thus near the earth: Weight = Fg = m*g (mass in kg, g = 9.8 m/s2).

Example: a persons weightConsider a person with a typical mass of 80 kg.Near the earth, his weight (or force of gravity pulling him down toward the center of the earth) would be Fg = 784 N.

Contact ForcesAn object in contact with a surface may have a normal force and force of friction acting on it. The normal force is always perpendicular to the surface. There is no equation for the normal force its magnitude depends on the situation.The frictional force is always parallel to the surface (opposing motion). The frictional force depends on the coefficient of friction between the object and surface.Kinetic friction (object sliding on a surface): Ff = k*FNStatic friction (no sliding): there is no definite equation for static friction, but its maximum value is given by: Ff s*FN

Example: a sled sliding across a snowy fieldThe ground is pushing up on the sled (to keep it from falling into the earth), and the ground is pushing against the sleds motion (slowing it down a little).

Free Body DiagramsThe key to success in Physics 121!Simple picture of just the object of interest.Choose coordinate system (x,y).Identify all forces acting on object and draw arrows showing the directions.

Example: a dresser being pushed down the hall of your apartment as you move in.There are four forces: gravity, the normal force, friction, and you pushing on the dresser.FfFNFpFg

SummaryNewtons Laws of MotionInertiaF=maEqual and Opposite Force PairsForces:Non-Contact: GravityContact: Friction and NormalFree Body DiagramsIsolate One ObjectEach Direction is Independent

Example: The Dresser

Lets return to the dresser FBD and do some calculations(I will solve the horizontal direction and you will solve the vertical direction)

FfFNFpFgGiven: the dresser has a mass of 120 kg and you are pushing on it with a force of 345 N to make it slide down the hall at a constant velocity.

Example: The Dresserm = 120 kg and Fp = 345 N

Once the FBD is drawn and we have picked a coordinate system we must apply Newtons Second Law:

FfFNFpFgF = ma

We know the horizontal acceleration is zero (the velocity is constant).

F = 0

Example: The Dresserm = 120 kg and Fp = 345 N

Now start to substitute in what is known:

FfFNFpFgF = 0

We know there are two forces in the x-direction (the push and friction).

Fp Ff = 0

Example: The Dresserm = 120 kg and Fp = 345 N

Solve for the unknown:

FfFNFpFgFp Ff = 0

The force of the push is given as 345 N.

(345 N) Ff = 0

Ff = 345 N

Example: The Dresser

Follow-up: now we can find the coefficient of kinetic friction between the dresser and the floor

FfFNFpFgFf = k*FN

(345 N) = k*(1176 N)

k = 0.29

Contact Force: SpringsForce exerted by a spring: Fspring = k*xThe greater the compression or extension of the spring, the greater the force.The greater the spring constant (a characteristic of the spring), the greater the force.

Contact Force: SpringsForce exerted by a spring: Fspring = k*xThe greater the compression or extension of the spring, the greater the force.The greater the spring constant (a characteristic of the spring), the greater the force. Example: When a 5 kg mass is suspended from a spring, the spring stretches 8 cm. Determine the spring constant.Fs- Fg = 0Fs = Fgk x = m gk = 612 N/mFBD:Newtons Second Law:F = ma

Contact Force: TensionTension in an Ideal String: Magnitude of tension is equal everywhere.Direction is parallel to string (only pulls).

Contact Force: TensionTension in an Ideal String: Magnitude of tension is equal everywhere.Direction is parallel to string (only pulls).

Example: Determine force applied to string to hold a 45 kg mass hanging over pulleyFT - Fg = 0FT = FgFT = m gFT= 440 NFBD:Newtons Second Law:F = ma

SummaryContact Force: SpringCan push or pull, force proportional to displacementF = k xContact Force: TensionAlways Pulls, tension equal everywhereForce parallel to stringNext, A Two Dimensional Example:Choose coordinate systemAnalysis of each direction is independent

Example: Force at AngleA person is pulling a 150 kg refrigerator across a floor with mk= 0.4 at a constant speed. If she is pulling down at an angle of = 25 with respect to the horizontal, what is the magnitude of the force she exerts on the refrigerator (the tension)?

We will choose a standard x,y coordinate system.There are 4 forces in the FBD (Gravity, Normal, Friction, Tension).FfFTFgFN

Example: Force at AngleA person is pulling a 150 kg refrigerator across a floor with mk= 0.4 at a constant speed. If she is pulling down at an angle of = 25 with respect to the horizontal, what is the magnitude of the force she exerts on the refrigerator (the tension)?

Tension is in both the x and y directions, so we must break it down into its x and y components.

FfFTcosFgFNFTsin

Example: Force at AngleA person is pulling a 150 kg refrigerator across a floor with mk= 0.4 at a constant speed. If she is pulling down at an angle of = 25 with respect to the horizontal, what is the magnitude of the force she exerts on the refrigerator (the tension)?

Now we will write Newtons Second Law for each direction.In each direction the acceleration is zero.

Ffx-direction:FgFNFTsinF = maFTcos - Ff = 0

F = ma FN - FTsin - Fg = 0

y-direction:FTcos

Example: Force at AngleA person is pulling a 150 kg refrigerator across a floor with mk= 0.4 at a constant speed. If she is pulling down at an angle of = 25 with respect to the horizontal, what is the magnitude of the force she exerts on the refrigerator (the tension)?

In the first equation, replace Ff with *FN.Solve the second equation for FN, and replace Fg with m*g.Now substitute the expression for FN into the first equation and solve.

FTcos - Ff = 0

FTcos = FN

FN - FTsin - Fg = 0

FN = FTsin + mgFTcos = (FTsin + mg)

FT = mg / (cos - sin)

x-directiony-directioncombine

Example: Force at AngleA person is pulling a 150 kg refrigerator across a floor with mk= 0.4 at a constant speed. If she is pulling down at an angle of = 25 with respect to the horizontal, what is the magnitude of the force she exerts on the refrigerator (the tension)?

Finally, substitute in the given numbers! (note: it is usually easier to wait until the very end to do this)

FT = mg / (cos - sin) FT = 798 N

Dynamics: F = maWe have already dealt with situations where a = 0.

But when the net force is not zero, there IS an acceleration!

Dynamics ExampleA tractor is pulling a trailer with a constant acceleration. If the forward acceleration is 1.5 m/s2, Calculate the force on the trailer (m = 400 kg) due to the tractor. (Consider just the trailer.)

x-direction:F = maFT = maFT = 600 N

F = ma FN - Fg = 0FN = 3920 N

y-direction:FNFBD:FT = 600 N

Apparent WeightRecall: SF = m aConsider person accelerating in an elevator.Draw FBDApply SF = m aFN mg = maFN = m ( g + a )

33Apparent weight is normal force from scale or floor.Note: in free fall a = -g so N = 0.

Apparent Weight ExampleYou are standing on a scale inside an elevator. Normally you weigh 600 N, but the scale reads 720 N. What is your acceleration?

First, note that your mass is (600 N)/(9.8 m/s2) = 61.2 kg.Second, use: FN = m ( g + a ).

(720 N) = (61.2 kg) ( 9.8 m/s2 + a )a = +1.96 m/s2Your acceleration is 1.96 m/s2 UP.

Summary of Concepts F = m a Draw Free Body Diagram Write down equations Solve

Apparent WeightIf object is accelerating in vertical direction weight appears differentAccelerating up increases apparent weightAccelerating down decreases apparent weight

Blow up hydrogen balloon here?Maybe simple demo, add twice as much mass, get twice the stretching before showing the slide. Also make note that this is NOT as big a deal as Newtons laws. Just an example of a contact force.Maybe simple demo, add twice as much mass, get twice the stretching before showing the slide. Also make note that this is NOT as big a deal as Newtons laws. Just an example of a contact force.