Presentation is loading. Please wait.

Presentation is loading. Please wait.

Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

Similar presentations


Presentation on theme: "Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces."— Presentation transcript:

1 Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces Strong: quark-gluon ( L ~ cm) Nuclear forces (residual, “Van der Waals”) Weak: lepton-quark ( n → p + e - + )L~ cm Nuclear β-decay A Z → A Z+1 +e - + Electromagnetic: charged particles exchanging by photons (long-range) Normal force and pressure Tension force and shear force Frictional forces Propulsion force Buoyant force Electric and magnetic forces Chemical bonds...

2 Normal Force Weight vs. mass (gravitational mass = inertial mass) Apparent weight vs. true weight mg, g = 9.8 m/s 2 Note: weight varies with location on earth, moon,… g moon =1.6 m/s 2

3 Pulley (massless & frictionless) The Tension Force Massless rope: - T 1 =T 2 =m G g Massive rope: -T 1 =T 2 +m R g=(m G +m R )g>T 2 Acceleration with massive rope and idealization of massless rope mBmB mRmR X T R = (m B + m R )a > T B = m B a = T R – m R a

4 Static and Kinetic Frictional Forces

5 Fluid Resistance and Terminal Speed Linear resistance at low speed f = k v Drag at high speed f= D v 2 due to turbulence Newton’s second law: ma = mg – kv Terminal speed (a→0): v t = mg / k (for f=kv), v t = (mg/D) 1/2 (for f=Dv 2 ) Baseball trajectory is greatly affected by air drag ! v 0 =50m/s

6 Applying Newton’s Laws for Equilibrium: Nonequilibrium:

7 Replacing an Engine (Equilibrium) Another solution: Choose Find: Tension forces T 1 and T 2

8 Plane in Equilibrium

9 Example 5.9: Passenger in an elevator y 0 Center of the Earth Data: F N = 620 N, w = 650 N Find: (a) reaction forces to F n and w; (b) passenger mass m; (c) acceleration a y. Solution: (a)Normal force –F N exerted on the floor and gravitational force –w exerted on the earth. (b) m = w / g = 650 N / 9.8 m/s 2 = 64 kg (c) Newton’s second law: ma y = F N – w, a y = (F N – w) / m = g (F N – w) / w = = 9.8 m/s 2 (620 N – 650 N)/650 N = m/s 2

10 How to measure friction by meter and clock? Exam Example 9: 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).

11 d) Work done by friction: W f = -f k L = -μ k F N L = -L μ k mg cosθ max = -9 J ; work done by gravity: W g = mgH = 10 J ; total work: W = mv || 2 /2 = 2 kg (1m/s) 2 /2 = 1 J = W g + W f = 10 J + 9 J = 1 J

12 Hauling a Crate with Acceleration

13 Exam Example 10: Blocks on the Inclines (problem 5.92) m 1 m2m2 X X α1α1 α2 α2 Data: m 1, m 2, μ k, α 1, α 2, v x <0 Solution: Newton’s second law for block 1: F N1 = m 1 g cos α 1, m 1 a x = T 1x +f k1x -m 1 g sinα 1 (1) block 2: F N2 = m 2 g cosα 2, m 2 a x = T 2 x +f k2x + m 2 g sinα 2 (2) Find: (a) f k1x and f k2x ; (b) T 1x and T 2x ; (c) acceleration a x. (a) f k1x = sμ k F N1 = s μ k m 1 g cosα 1 ; f k2x = sμ k F N2 = sμ k m 2 g cosα 2 ; s = -v x /v (c) T 1x =-T 2x, Eqs.(1)&(2)→ (b)

14 Exam Example 11: Hoisting a Scaffold Y 0 m Data: m = 200 kg Find: (a) a force F to keep scaffold in rest; (b) an acceleration a y if F y = N; (c) a length of rope in a scaffold that would allow it to go downward by 10 m Solution Newton’s second law: (a) Newton’s third law: F y = - T y, in rest a y = 0→ F(a=0)= W/5= mg/5 =392 N (b) a y = (5T-mg)/m = 5 (-F y )/m – g = 0.2 m/s 2 (c) L = 5·10 m = 50 m (pulley’s geometry)

15 Dynamics of Circular Motion θ Uniform circular motion: Period T=2 πR/v, a c = v 2 /R = 4π 2 R/T 2 Cyclic frequency f=1/T, units: [f] = Hz = 1/s Angular frequency ω = 2πf = 2π/T, units: [ω]=rad·Hz=rad/s R Dimensionless 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

16 Centripetal Force Rounding a flat curve (problem 5.44) Sources of the centripetal force

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

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

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

20 Uniform circular motion in a vertical circle Newton’s second law Top: n T – mg = -ma c Bottom: n B – mg = +ma c Note: If v 2 >gR, the passenger will be catapulted ! Find: Normal force n T


Download ppt "Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces."

Similar presentations


Ads by Google