Download presentation

Presentation is loading. Please wait.

Published byShane Lyle Modified over 2 years ago

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 ~ 10 -13 cm) Nuclear forces (residual, “Van der Waals”) Weak: lepton-quark ( n → p + e - + )L~10 -16 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 = - 0.45 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 = - 400 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

Similar presentations

OK

Force Chapter 6. Force Any push or pull exerted on an object.

Force Chapter 6. Force Any push or pull exerted on an object.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on importance of sports and games in students life Moving message display ppt on ipad Ppt on central limit theorem definition Ppt on needle stick injury product Ppt on area of equilateral triangle Ppt on marie curie nobel Ppt on personality development for teachers Ppt on uses of metals and nonmetals Ppt on advertisement in hindi language Ppt on earth and space