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forces,resolution of force,equilibrium of forces,types of forces

forces,resolution of force,equilibrium of forces,types of forces

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  • 1) You are asked for the weight. 2) You are given the weight on Earth and the strength of gravity on Jupiter. 3) Fw = mg. 4) First, find the person’s mass from the weight and strength of gravity on Earth: m = (490 N) ÷ (9.8 N/kg) = 50 kg Next, find the weight on Jupiter: F w = (50 kg) × (23 N/kg) = 1,150 N (259 lbs)
  • 1) You are asked to find force. 2) You are given a mass of 10 kilograms. 3) The force of the weight is: F w = mg and g = 9.8 N/kg. 4) The word “supported” means the ball is hanging motionless at the end of the rope. That means the tension force in the rope is equal and opposite to the weight of the ball. F w = (10 kg) × (9.8 N/kg) = 98 N. The tension force in the rope is 98 newtons.
  • 1) You are asked for the force of friction (F f ). 2) You are given the weight (F w ), the applied force (F) and the coefficient of sliding friction (μ) 3) The normal force is the sum of forces pushing down on thefloor (F f = μFn). 4) First, find the normal force: F n = 100 N + 10 N = 110 N Use F f = μFn to find the force of friction: F f = (0.25) x (110 N) = 27.5 N
  • 1) You are asked for the force to overcome static friction (F f ). 2) You are given the weight (F w ) and both surfaces are steel. 3) F f = μsF n . 4) F f = (0.74) x (50 N) = 37 N
  • 1) You are asked for the acceleration (a). 2) You are given the applied force (F), the mass (m), and the coefficient of rolling friction (μ). 3) Relationships that apply: a = F ÷ m, F f = μF n , (F w = mg and g = 9.8 N/kg). 4) The normal force equals the weight of the car: Fn = mg = (500 kg)(9.8 N/kg) = 4,900 N. The friction force is: F f = (0.07)(4,900 N) = 343 N. The acceleration is the net force divided by the mass: a = (1,000 N - 343 N) ÷ (500 kg) = (657 N) ÷ (500 kg) = 1.31 m/sec 2
  • 1) You are asked for acceleration. 2) You are given mass and force. 3) a = F ÷ m. 4) F = -75N - 25N +45N +55N = 0 N, so a = 0.
  • 1) You are asked for the force. 2) You are given one of two forces and the mass. 3) Relationships that apply: net force = zero, F w = mg and g = 9.8 N/kg. 4) The weight of the boat is F w = mg = (150 kg) (9.8 N/kg) = 1,470 N. Let F be the force in the other chain, The condition of equilibrium requires that: F + (600 N) = 1,470 N, therefore F = 870 N.

Transcript

  • 1. CPO ScienceFoundations of Physics Chapter 9 Unit 2, Chapter 6
  • 2. Unit 2: Motion and Force in One Dimension Chapter 6: Forces and Equilibrium 6.1 Mass, Weight and Gravity 6.2 Friction 6.3 Equilibrium of Forces and Hooke’s Law
  • 3. Chapter 6 Objectives Calculate the weight of an object using the strength of gravity ( g) and mass. Describe the difference between mass and weight. Describe at least three processes that cause friction. Calculate the force of friction on an object when given the coefficient of friction and normal force. Calculate the acceleration of an object including the effect of friction. Draw a free-body diagram and solve one-dimensional equilibrium force problems. Calculate the force or deformation of a spring when given the spring constant and either of the other two variables.
  • 4. Chapter 6 Vocabulary Terms mas s  n o rmal  c o m p r e s s io n w e ig h t force  s p r in g w e ig h t l e s s  e x t e n s io n c o n s tan t g -f o r c e  net force  d e f o r m a t io n f r ic t io n  f r e e -b o d y  r e s t o r in g d ia g r a m force s t a t ic f r ic t io n  l u b r ic a n t  c o e f f ic ie n t s l id in g  e q u il ib r iu m o f f r ic t io n f r ic t io n  b a l l b e a r in g  e n g in e e r in g r o l l in g  d im e n s io n  d e s ig n c y c l e f r ic t io n  s p r in g  s u b s c r ip t v is c o u s  prototype
  • 5. 6.1 Mass, Weight, and Gravity Mass is a measure of matter. Mass is constant. Weight is a force. Weight is not constant.
  • 6. 6.1 Mass, Weight, and Gravity The weight of an object depends on the strength of gravity wherever the object is. The mass always stays the same.
  • 7. 6.1 Weight Gravity (9.8 m/sec2)Weight force (N) Fw = mg Mass (kg)
  • 8. 6.1 Free fall and weightlessness An e l e v a t o r is a c c e l e r a t in g d o w n w a r d a t 9 .8 m /s e c 2 . T h e s c a l e f e e l s n o f o r c e b e c a u s e it is f a l l in g a w a y f r o m y o u r f e e t a t t h e s a m e r a t e y o u a r e f a l l in g . As a r e s u l t , y o u a r e w e ig h t l e s s .
  • 9. 6.1 Calculate weight  How much would a person who weighs 490 N ( 110 lbs) on Earth weigh on Jupiter?  The value of g at the top of J upiter’s atmosphere is 23 N kg. /  (Since J upiter may not actually have a surface, “on” means at the top of the atmosphere.)
  • 10. 6.1 Calculate force  A 10-kilogram ball is supported at the end of a rope. H much force ow (tension) is in the rope?
  • 11. 6.1 Mass, Weight, and Gravity Key Question: W h a t is s p e e d a n d h o w is it meas ured ?*Students read Section 6.1 BEFORE Investigation 6.1
  • 12. 6.2 Friction Friction results from relative motion between objects. Frictional forces are forces that resist or oppose motion.
  • 13. 6.2 Types of Friction Static friction Sliding friction Rolling friction
  • 14. 6.2 Types of Friction Air friction Viscous friction
  • 15. 6.2 Friction Normal force (N)Friction force (N) Ff = µ Fn Coefficient of friction
  • 16. 6.2 Calculate force of friction A 10 N force pushes down on a box that weighs 100 N. As the box is pushed horizontally, the coefficient of sliding friction is 0.25. Determine the force of friction resisting the motion.
  • 17. 6.2 Sliding Friction Normal force (N)Friction force (N) Ff = µ sFn Coefficient of sliding friction
  • 18. Table of friction coefficients
  • 19. 6.2 Calculate using friction A s t e e l p o t w it h a w e ig h t o f 5 0 N s it s on a s teel countertop. H o w m u c h f o r c e d o e s it t a k e t o s t a r t t h e p o t s l id in g ?
  • 20. 6.2 Calculate using friction  The engine applies a forward force of 1,000 newtons to a 500-kg car.  Find the acceleration of the car if the coefficient of rolling friction is 0.07.
  • 21. 6.2 FrictionKey Question:H o w c a n w e d e s c r ib e a n d m o d e l f r ic t io n ?*Students read Section 6.2 AFTER Investigation 6.2
  • 22. 6.3 Equilibrium and Hookes Law When the net force acting on an object is zero, the forces on the object are balanced. We call this condition equilibrium.
  • 23. 6.3 Equilibrium and Hookes LawNewton’s second law simply requires that for an object to be in equilibrium, the net force, or the sum of the forces, has to be zero.
  • 24. 6.3 Equilibrium and Hookes LawMany problems have more than one force applied to an object in more than one place.
  • 25. 6.3 Calculate net force Four people are pulling on the same 200 kg box with the forces shown. Calculate the acceleration of the box.
  • 26. 6.3 Calculate force using equilibrium  Two chains are used to lift a small boat. One of the chains has a force of 600 newtons.  Find the force in the other chain if the mass of the boat is 150 kilograms.
  • 27. 6.3 Equilibrium and Hookes Law The most common type of spring is a coil of metal or plastic that creates a force when it is extended ( stretched) or compressed ( squeezed) .
  • 28. 6.3 Equilibrium and Hookes Law  The force from a spring has two important characteristics: — The force always acts in a direction that tries to return the spring to its unstretched shape. — The strength of the force is proportional to the amount of extension or compression in the spring.
  • 29. 6.3 Hookes Law Deformation (m)Force (N) F=-kx Spring constant N/m
  • 30. 6.3 Calculate force A spring with k = 250 N m is extended by one / centimeter. H much force does the spring exert? ow
  • 31. 6.3 Equilibrium and Hookes Law T h e r e s t o r in g force from a w a l l is a l w a y s exactly eq ual a n d o p p o s it e t o th e fo rce yo u a p p l y , b e c a u s e it is c a u s e d b y t h e d e f o r m a t io n r e s u l t in g f r o m
  • 32. 6.3 Calculate using equilibrium The spring constant for a piece of solid wood is 1 × 10 8 N/m. Use Hooke’s law to calculate the deformation when a force of 500 N ( 112 lbs) is applied.
  • 33. 6.3 Equilibrium of Forces and Hookes LawKey Question:Ho w d o yo u p r e d ic t t h e force on a s p r in g ? *Students read Section 6.3 AFTER Investigation 6.3
  • 34. Application: The design of structures