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Force: is a push or a pull.

Force: is a push or a pull. There are contact forces such as pushing, pulling, or friction. There are field forces such as gravitation and electrostatics. Force is measured in Newtons, N. Force, Mass, and Acceleration.

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Force: is a push or a pull.

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  1. Force: is a push or a pull. • There are contact forces such as pushing, pulling, or friction. • There are field forces such as gravitation and electrostatics. • Force is measured in Newtons, N.

  2. Force, Mass, and Acceleration • Isaac Newton’s 2nd law of motion states: the acceleration of an object is directly proportional to the net force on an object and indirectly proportional to the acceleration of an object. That is, • a = Fnet / m • The net force, Fnet, is defined as the sum of all forces acting along a given line of action. • Example: 7N + -4N = 3N In this example, 3N is the net force.

  3. Weight: is the attractive force between any mass and the mass of the earth. • Weight = mass x gravitational acceleration • FW = m g where g = -10 m/s/s So, the weight of 1 kilogram is -10 Newtons (N). • FW = m g • FW = (1 kg) (-10 m/s/s) = -10 N

  4. Find:A) the weight of the block.B) the force applied to the block, if the acceleration of the block is 0 m/s/s.C) the net force on the block.D) the force applied to the block, if the acceleration of the block is 1.2 m/s/s. to the right. F = ? Friction = 5N M = 10 kg

  5. Find:A) the weight of the block. (W = 98 N)B) the force applied to the block, if the acceleration of the block is 0 m/s/s. (F = 5N)C) the net force on the block. (Fnet = 0 N)D) the force applied to the block, if the acceleration of the block is 1.2 m/s/s. to the right. (F = 17 N) F = ? Friction = 5N M = 10 kg

  6. FindA) the weight of the block.B) the tension in the rope tied to the block, if the acceleration of the block is 1.2 m/s/s upwards. T = M = 10 kg Weight =

  7. FindA) the weight of the block. (W = 98 N) B) the tension in the rope tied to the block, if the acceleration of the block is 1.2 m/s/s upwards. ( T = 110 N) T = M = 10 kg Weight =

  8. Newtons Laws Practice Probs.(accelerating objects)1. Calculate the force that must be applied to produce an acceleration of 1.8 g’s (18 m/s/s) on a 1.2 kg puck sliding free of friction on an air table.

  9. Newtons Laws Practice Probs.1. Calculate the force that must be applied to produce an acceleration of 1.8 g’s (18 m/s/s) on a 1.2 kg puck sliding free of friction on an air table. F = 21.6 N

  10. 2What will be the acceleration of a sky diver when air resistance is half the weight of the sky diver?

  11. 2What will be the acceleration of a sky diver when air resistance is half the weight of the sky diver?0.5g

  12. Find: The weight of the 1200 kg elevator. The upward force on the elevator when it is pulled upward at 2 m/s/s.

  13. Find: The weight of the 1200 kg elevator. F = W = (1200 kg)(-10 m/s/s) = -12,000 N The upward force on the elevator when it is pulled upward at 2 m/s/s. Fnet = m a F - 12,000 N = (1200 kg)(2 m/s/s) F = 2,400 N +12, 000 N F = 14, 400 N

  14. A 2 kg hanging block accelerates the 8 kg block on the frictionless table. What is the acceleration of the two block system?

  15. A 2 kg hanging block accelerates the 8 kg block on the frictionless table. What is the acceleration of the two block system? a = 20 N/10 kg = 2 m/s/s

  16. A 2 kg hanging block accelerates the 8 kg block on the frictionless table. What is the tension in the string supporting the 2 kg mass?

  17. A 2 kg hanging block accelerates the 8 kg block on the frictionless table. The tension, T, in the string on the 2kg mass F net = ma T - (2 kg)(10 m/s/s) = (2 kg)(-2 m/s/s) T = 20 N - 4 N T = 16 N

  18. Find the rate of acceleration of the 8 kg and 2 kg blocks. What is the tension in the string?

  19. The acceleration of mg is down (negative), while the acceleration of Mg is up (negative) as both are connected via the string. T T mg Mg Mg – T = Ma T – mg = ma Adding these yields: Mg – mg = (M + m) a a = g(M –m) / (M +m) so a =- 10 (6/10) = -6 m/s/s

  20. The acceleration of mg is down (negative), while the acceleration of Mg is up (negative) as both are connected via the string. T T mg Mg To find the tension in the string ( which is the same throughout the string), we may use: T – mg = ma T = ma + mg T = m(a + g) = 2(-6 + 10) = 8 N Note: g is NOT -10 as the sign is part of the eqtn.

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