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Net Force (Free Body) Diagram. Newton’s Second Law

Net Force (Free Body) Diagram. Newton’s Second Law. Force. Force is any push or pull SI unit: Newtons (N) 1 N = 1 kg * Force needed to cause a mass of one kilogram to accelerate at a rate of one meter per second squared Vector quantity: magnitude and direction. Friction Forces.

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Net Force (Free Body) Diagram. Newton’s Second Law

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  1. Net Force (Free Body) Diagram.Newton’s Second Law

  2. Force • Force is any push or pull • SI unit: Newtons (N) • 1 N = 1 kg * • Force needed to cause a mass of one kilogram to accelerate at a rate of one meter per second squared • Vector quantity: magnitude and direction

  3. Friction Forces • Friction Forces are forces that act on materials in contact with one another • Friction acts in the direction opposite to motion. • Constant velocity = frictional force balances out the applied force, net force equals zero • Fluid friction applies to liquids and gases. Most common is air resistance.

  4. Friction Forces • Static friction occurs between non-moving objects, constant friction. • Kinetic friction occurs between objects in motion, sliding friction • Coefficient of friction (μ): ratio of frictionalforce to normalforce • μ = or Ffrict = μ•Fnorm

  5. Net Force • Net Force is combination of all forces acting on an object (normal force, applied force, frictional force, and force of weight). • When net force is equal to zero, either the system is at rest or there is constant velocity (no acceleration). No acceleration means there is no change in velocity. Fn F applied F friction Fw Wood block Text book

  6. Forces

  7. Equilibrium • Equilibrium is the state of an object (or system) in which there is no change in motions (at rest or constant velocity) Fn y Fapplied F friction F w x Free body diagram textbook

  8. Equilibrium • Equilibrium: equations which describe the relationship among opposing forces directed along the x and y axis. • ∑Fx – the sum of the forces in the x direction ∑ - sigma, “the sum of” ∑Fx = right – left • ∑Fy – the sum of the forces in the y direction ∑Fy = up – down

  9. Net Force (Free Body) Diagram • Problem. A jug of juice is resting on the table when a rightward force of 25 N is applied. The force of friction is 10 N. The weight of the juice is 7 N. Draw a free body diagram and write equilibrium statement. ∑Fx = right – left y ∑Fx = 25 N – 10 N ∑Fx = 15 N x ∑Fy = up – down ∑Fy = 7 N – 7 N Fapp = 25 N Fn = 7 N ∑Fy = 0 N Ff= 10 N table Fw=7Ns juice

  10. Acceleration and Force • Force causes acceleration. • Acceleration is change in velocity per change of time • a = = • Acceleration is directly proportional to net force. • a ∞ Fnet Example: • apply more force to puck; puck accelerates • double the force applied; acceleration doubles

  11. Mass and Acceleration • Mass resists acceleration. • Mass is inversely proportional to acceleration. a ∞ Shopping Cart Example y direct Full cart: more mass, a ∞ y inverse difficult to push, slower (less acceleration) Empty cart: less mass, a ∞ easier to push, x x faster (more acceleration) double mass = acceleration Mass is a measure of inertia. The ability of an object to resist a change in motion

  12. Newton’s Second Law of Motion • The acceleration produced by a net force on an object is directly proportional to the magnitude of the net force, is in the same direction as the net force, and inversely proportional to the mass of the object. • F = m*a a = m = • SI units of mass: kilograms (kg) • SI units of acceleration: • SI units of force: Newtons (N) or

  13. Newton’s Second Law of Motion Examples • Example # 1. Determine the acceleration that result when a 12 N force is applied to a 3 kg object and then to a 6 kg object. F=12N F=12N m1=3kg a1= a1= a1=4 F=12N m2=6kg a1=? a2=? a2= a2= a2=2 3 kg 6 kg

  14. Newton’s Second Law of Motion Examples • Example # 2. A net force of 15 N is exerted on an encyclopedia to cause it accelerate at a rate of 5 . What is the mass of the encyclopedia? F=15N a=5 m = m= m=3kg m=?

  15. Newton’s Second Law of Motion Examples • Example # 3. Suppose that a sled is accelerating at a rate of 2 . If the net force is tripled and the mass is doubled, then what is the new acceleration of the sled? Two ways to solve: Fn2=3Fn1 1) a1==2 a2=3 m2=2m1 a1= a2= = a2=? a2= a1=2 2) a2= a2=3 a2=2*1.5

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