Free-Body Diagrams

Free-Body DiagramsForceSymbol/FormulaDescriptionWeightFw = mgAlways directed downwardTension ForceTPulling forces directed away from the bodyNormal ForceFnForce exerted perpendicularly outward by a flat surface on an object pressed against itFrictional ForceFf = m FnForce which acts opposite the motion or an impending motion and parallel to the surface of contactAngle of Reposethe (minimum) angle a plane makes with the horizontal at which a body on the plane will begin to slide downdepends on the roughness of the surfacesthe rougher the surfaces, the greater the angle of reposeq = tan-1 mA crate weighing 562 N rests on an inclined plane 30o above the ground. Find the components of the force parallel and perpendicular to the plane.Chris is going down a water slide sloped at 37o. If he weighs 62kg, how fast is Chris going 5.0s after starting from rest? mk = 0.15Things to rememberFinding the Resultant Forcedraw a free body diagramuse the length of the arrow to indicate the vector's magnitude and the direction of the arrow to show its directionsimply find the sum of the vectorsa negative answer means that the force acts in the opposite direction to the one that you chose to be positive0 net force = balanced forces = 0 acceleration

EquilibriumAn object in equilibrium has both the sum of the forces acting on it and the sum of the moments of the forces equal to zero.If a resultant force acts on an object then that object can be brought into equilibrium by applying an additional force that exactly balances this resultant.EquilibrantThe equilibrant of any number of forces is the single force required to produce equilibrium, and is equal in magnitude but opposite in direction to the resultant force.

Friction forcesWhen the object is not movingstatic friction: Ff s FN

When the object is movingkinetic friction: Ff = k FN

Newton's Law of Universal GravitationEvery point mass attracts every other point mass by a force directed along the line connecting the two. This force is proportional to the product of the masses and inversely proportional to the square of the distance between them.

The magnitude of the attractive gravitational force between the two point masses,Fis given by:F = Gm1m2 r2where:Gis the gravitational constant,m1is the mass of the first point mass,m2is the mass of the second point mass andris the distance between the two point masses.

For example, consider a man of mass 80kg standing 10 m from a woman with a mass of 65kg. The attractive gravitational force between them would be:F = Gm1m2 r2 = (6.671011 Nm2/kg2)(80kg)(65kg)(10m)2 = 3.47 109 N

G = 6.67 x 10-11 N m2/kg2If the man and woman are 1 m apart, then the force is:F = Gm1m2r2 = (6,671011 Nm2/kg2)(80kg)(65kg)(1m)2 = 3.47107 N

Gravitational force between the Earth and the MoonThe mass of the Earth is5.981024kg, the mass of the Moon is7.351022kg and the Earth and Moon are3.8108m apart. The gravitational force between the Earth and Moon is: F = Gm1m2 r2 = (6.671011Nm2/kg2)(5.981024kg)(7.351022kg)(0.38109m)2 = 2.031020NComparative problems involve calculation of something in terms of something else that we know. For example, if you weigh 490 N on Earth and the gravitational acceleration on Venus is 0.903 that of the gravitational acceleration on the Earth, then you would weigh 0.903x490N=442.5N on Venus.Principles for answering comparative problems

Write out equations and calculate all quantities for the given situationWrite out all relationships between variable from first and second caseWrite out second caseSubstitute all first case variables into second caseWrite second case in terms of first case

A man has a mass of 70kg. The planet Zirgon is the same size as the Earth but has twice the mass of the Earth. What would the man weigh on Zirgon, if the gravitational acceleration on Earth is 9.8m/s2?

the mass of the man,mthe mass of the planet Zirgon (mZ) in terms of the mass of the Earth (mE),mZ=2mEthe radius of the planet Zirgon (rZ) in terms of the radius of the Earth (rE),rZ=rE

Situation on Earth wE = mgE = G mEm rE2 = (70kg)(9.8m/s2) = 686 N

Situation on Zirgon in terms of situation on EarthwZ = mgZ= GmZm rZ2 = G 2mEm rE2 = 2 GmEm rE2 = 2 wE = 2 (686N) = 1372N

Exercise

Two objects of mass 2m and 3m respectively exert a force F on each other when they are a certain distance apart. What will be the force between two objects situated the same distance apart but having a mass of 5m and 6m respectively?A:0.2 FB:1.2 FC:2.2 FD:5 F

As the distance of an object above the surface of the Earth is greatly increased, the weight of the object wouldA:increaseB:decreaseC:increase and then suddenly decreaseD:remain the sameA satellite circles around the Earth at a height where the gravitational force is 4 times less than at the surface of the Earth. If the Earth's radius is R, then the height of the satellite above the surface is:A:RB:2 RC:4 RD:16 RA satellite experiences a force F when at the surface of the Earth. What will be the force on the satellite if it orbits at a height equal to the diameter of the Earth:A:1FB:12FC:13FD:19F

The weight of a rock lying on surface of the Moon is W. The radius of the Moon is R. On planet Alpha, the same rock has weight 8W. If the radius of planet Alpha is half that of the Moon, and the mass of the Moon is M, then the mass, in kg, of planet Alpha is:A:M2B:M4C:2 MD:4 M

Impulse and MomentumMomentum the product of a moving bodys mass and its velocity the tendency of an object to continue to move in its direction of travelp= mvF= maF= m(vf -vi)tFt=mvf -mvi impulse-momentum eq. Ft = impulse

The change in momentum is equal to the impulse acting.Ft = m(vf vi)The change in momentum divided by time is equal to the force acting.F= m(vf -vi)t

A 1.0kg ball is kicked from rest, giving it a velocity of 12.0 m/s.Find the impulseIf the ball and foot were in contact for 0.22s, what was the average force exerted by the foot?Calculating the Total Momentum of a SystemTwo billiard balls roll towards each other. They each have a mass of 0.3kg. Ball 1 is moving atv1=1m/sto the right, while ball 2 is moving atv2=0.8m/s to the left. Calculate the total momentum of the system.

ptotal = p1 + p2 = m1v1 + m2v2 = 0.06 kg m/s to the rightChange in MomentumConsider a tennis ball (mass = 0.1 kg) that is dropped at an initial velocity of 5 m/s and bounces back at a final velocity of 3 m/s.

pi = m vipf = m vfDp = pf piDp = 0.8 kg m/s upwards Impulse and Change in momentumA 150 N resultant force acts on a 300 kg trailer. Calculate how long it takes this force to change the trailer's velocity from 2 m/s to 6 m/s in the same direction. Assume that the forces act to the right.

Fnet Dt = m (vf vi)Dt = 8 secConservation of MomentumThe total linear momentum of an isolated system is constant. An isolated system has no forces acting on it from the outside.

pi=pfm1vi1+m2vi2=m1vf1+m2vf2

A toy car of mass 2 kg moves westwards with a speed of 2 m/s. It collides head-on with a toy train. The train has a mass of 1.5 kg and is moving at a speed of 1.5 m/s eastwards. If the car rebounds at 2.05 m/s, calculate the velocity of the train.

vf1 = 3.9 m/s westwards

2 kgInitial momentumBefore collisionFinal momentumAfter collisionChange in momentum(final minus initial)2.0 kg toy car(2.0 kg) (-2 m/s) = -4.0 kg m/s(2 .0 kg) (2.05 m/s) = 4.1 kg m/s4.1 -4.0 = 8.1 kg m/s1.5 kg toy train(1.5 kg) (1.5 m/s) = 2.25 kg m/s(1.5 kg) (-3.9 m/s) = -5.85 kg m/s-5.85 2.25 = -8.1 kg m/sCombined-1.75 kg m/s-1.75 kg m/sRelated conceptsTransportation safetySeatbeltsAirbagsCrumple zonesPadding as protectionFollow through in sportsRecoil

Ohms LawV = IRV voltage, I current, R resistanceR = V/I1 ohm = 1 volt/1 ampere

Series circuitR = R1 + R2 + R3Is = IR1 = IR2 = IR3Vs = VR1 + VR2 + VR3

Parallel Circuit Vs = VR1 = VR2 = VR3Is = IR1 + IR2 +IR3

Power & EnergyP = IV = (V/R) VP = V2/RP power1 watt = 1 ampere x 1 voltE = P TE electric energy , P power rating, T time1 kilowatt hour = 1 kilowatt x 1 hour