Different Forces and Applications of Newtonâ€™s Laws Types of Forces Fundamental...
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Transcript of Different Forces and Applications of Newtonâ€™s Laws Types of Forces Fundamental...
Different Forces and Applications of Newtons LawsTypes of Forces
Fundamental ForcesNon-fundamental ForcesGravitational (long-range)Weight, tidal forcesStrong: quark-gluon ( L ~ 10-13 cm)Nuclear forces (residual, Van der Waals)Weak: lepton-quark( n p + e-+ )L~10-16cmNuclear -decayAZ AZ+1 +e- + Electromagnetic: charged particles exchanging by photons(long-range)Normal force and pressureTension force and shear forceFrictional forcesPropulsion forceBuoyant forceElectric and magnetic forcesChemical bonds . . .
Normal ForceWeight vs. mass (gravitational mass = inertial mass)Apparent weight vs. true weight mg , g = 9.8 m/s2Note: weight varies with location on earth, moon,gmoon=1.6 m/s2
Pulley (massless & frictionless)The Tension ForceMassless rope:
-T1=T2=mGgMassive rope: -T1=T2+mRg=(mG+mR)g>T2Acceleration with massive rope and idealization of massless ropemBmRXTR= (mB+ mR)a > TB= mB a = TR mR a
Static and Kinetic Frictional Forces
Fluid Resistance and Terminal SpeedLinear resistance at low speed f = k vDrag at high speed f= D v2 due to turbulenceNewtons second law: ma = mg kv
Terminal speed (a0):
vt = mg / k (for f=kv) ,
vt = (mg/D)1/2 (for f=Dv2)Baseball trajectory is greatlyaffected by air drag !v0=50m/s
Applying Newtons Laws for Equilibrium: Nonequilibrium:
Replacing an Engine (Equilibrium) Another solution: ChooseFind: Tension forcesT1 and T2
Plane in Equilibrium
Example 5.9: Passenger in an elevatory0Center of the EarthData: FN= 620 N, w = 650 N Find: (a) reaction forces to Fn and w; (b) passenger mass m; (c) acceleration ay . Solution:Normal force FN exerted on the floor and gravitational force w exerted on the earth.
(b) m = w / g = 650 N / 9.8 m/s2 = 64 kg
(c) Newtons second law: may = FN w ,
ay = (FN w) / m = g (FN w) / w = = 9.8 m/s2 (620 N 650 N)/650 N = - 0.45 m/s2
How to measurefriction by meter and clock?Exam Example 9:d) Find also the works done on the block by friction and by gravityas well as the total work done on the block if its mass is m = 2 kg (problem 6.68).
d) Work done by friction: Wf = -fkL = -k FN L = -L k mg cosmax = -9 J ; work done by gravity: Wg = mgH = 10 J ; total work: W = mv||2 /2 = 2 kg (1m/s)2 /2 = 1 J = Wg + Wf = 10 J + 9 J = 1 J
Hauling a Crate with Acceleration
- Exam Example 10: Blocks on the Inclines (problem 5.92) m1 m2 XX1 2 Data: m1, m2, k, 1, 2, vx
Exam Example 11: Hoisting a ScaffoldY0mData: m = 200 kg Find: (a) a force F to keep scaffold in rest;(b) an acceleration ay if Fy = - 400 N;(c) a length of rope in a scaffold that would allow it to go downward by 10 mSolutionNewtons second law: Newtons third law: Fy = - Ty , in rest ay = 0 F(a=0)= W/5= mg/5 =392 N
(b) ay= (5T-mg)/m = 5 (-Fy)/m g = 0.2 m/s2
(c) L = 510 m = 50 m (pulleys geometry)
Dynamics of Circular MotionUniform circular motion:Period T=2R/v , ac = v2/R = 42R/T2 Cyclic frequency f=1/T , units: [f] = Hz = 1/sAngular frequency = 2f = 2/T, units: =radHz=rad/sRDimensionless unit for an angle:Example: 100 revolutions per second f=1/T=100 Hz or T=1s/100=0.01 s Non-uniform circular motion:equation for a duration of one revolution T
Centripetal ForceRounding a flat curve (problem 5.44) Sources of the centripetal force
Data: L, , m Find: tension force F; speed v; period T. Solution:Newtons second lawCentripetal force along x:Equilibrium along y:Two equations with two unknowns: F,vThe conical pendulum (example 5.20) or a bead sliding on a vertical hoop (problem 5.119)Exam Example 12:R
A pilot banks or tilts the plane at an angle to create the centripetal force Fc = Lsin Lifting force
Rounding a Banked CurveExample 5.22 (car racing):r = 316 m , = 31o
Uniform circular motion in a vertical circleNewtons second lawTop: nT mg = -mac Bottom: nB mg = +macNote: If v2 >gR , the passenger will be catapulted !Find: Normal force nT