Different Forces and Applications of Newton’s Laws

Different Forces and Applications of Newton’s Laws Types of Forces

Normal Force Weight vs. mass(gravitational mass = inertial mass) Apparent weight vs. true weight mg , g = 9.8 m/s2 Note: weight varies with location on earth, moon,… gmoon=1.6 m/s2

The Tension Force Massless rope: -T1=T2=mGg Massive rope: -T1=T2+mRg=(mG+mR)g>T2 Acceleration with massive rope and idealization of massless rope Pulley (massless & frictionless) mB mR X TR= (mB+ mR)a > TB= mB a = TR – mR a

Static and Kinetic Frictional Forces

Fluid Resistance and Terminal Speed Linear resistance at low speed f = k v Drag at high speed f= D v2 due to turbulence Newton’s second law: ma = mg – kv Terminal speed (a→0): vt = mg / k (for f=kv) , vt = (mg/D)1/2 (for f=Dv2) Baseball trajectory is greatly affected by air drag ! v0=50m/s

Applying Newton’s Laws for Equilibrium: Nonequilibrium:

Replacing an Engine (Equilibrium) Find: Tension forces T1 and T2 Another solution: Choose

Plane in Equilibrium

Example 5.9: Passenger in an elevator y Data: FN= 620 N, w = 650 N Find: (a) reaction forces to Fn and w; (b) passenger mass m; (c) acceleration ay . Solution: 0 • 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) Newton’s 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 Center of the Earth

Exam Example 9: How to measure friction by meter and clock? d) Find also the works done on the block by friction and by gravity as 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 = -μk FN L = -L μk mg cosθmax = -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) Data: m1, m2, μk, α1, α2, vx<0 m1 Find: (a) fk1x and fk2x ; (b) T1x and T2x ; (c) acceleration ax . X m2 Solution: Newton’s second law for block 1: FN1 = m1g cosα1 , m1ax= T1x+fk1x-m1g sinα1(1) block 2: FN2 = m2g cosα2 , m2ax= T2 x+fk2x+ m2g sinα2(2) α1 α2 X (a) fk1x= sμkFN1= sμkm1g cosα1 ; fk2x= sμkFN2= sμkm2g cosα2; s = -vx/v (c) T1x=-T2x, Eqs.(1)&(2)→ (b)

Exam Example 11: Hoisting a Scaffold Y Data: 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 m 0 Solution Newton’s second law: m • Newton’s 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 = 5·10 m = 50 m (pulley’s geometry)

Dynamics of Circular Motion Uniform circular motion: θ R Dimensionless unit for an angle: Period T=2πR/v , ac = v2/R = 4π2R/T2 Cyclic frequency f=1/T , units: [f] = Hz = 1/s Angular frequency ω = 2πf = 2π/T, units: [ω]=rad·Hz=rad/s 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 Force Sources of the centripetal force Rounding a flat curve (problem 5.44)

Exam Example 12: • Data: L, β, m • Find: • tension force F; • speed v; • period T. The conical pendulum(example 5.20) or a bead sliding on a vertical hoop (problem 5.119) Solution: Newton’s second law R Two equations with two unknowns: F,v Centripetal force along x: Equilibrium along y:

A pilot banks or tilts the plane at an angle θ to create the centripetal force Fc = L·sinθ Lifting force

Rounding a Banked Curve Example 5.22 (car racing): r = 316 m , θ = 31o

Uniform circular motion in a vertical circle Find: Normal force nT Newton’s second law Top: nT – mg = -mac Note: If v2 >gR , the passenger will be catapulted ! Bottom: nB – mg = +mac

Different Forces and Applications of Newton’s Laws