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Newton’s laws

m. Newton’s laws. Imagine a mass of m kg hung from a spring balance in a lift. The tension in the spring is given by the measurement on the scale of the balance. Case 1 :. Lift at rest. The lift problem. T. F = ma :. T = mg. Tension = weight. 0. mg. m. The lift problem. T. F = ma :.

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Newton’s laws

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  1. m Newton’s laws Imagine a mass of m kg hung from a spring balance in a lift. The tension in the spring is given by the measurement on the scale of the balance. Case 1 : Lift at rest The lift problem T F = ma : T = mg Tension = weight 0 mg

  2. m The lift problem T F = ma : T – mg = ma  T = m( g + a ) a Case 2 :Lift accelerating upwards apparent weight = T = mg + ma  mass apparently weighs more mg

  3. m The lift problem T F = ma : mg - T = ma  T = mg - ma a Case 3 :Lift accelerating downwards apparent weight = T = mg - ma  mass apparently weighs less mg

  4. m The lift problem T F = ma : T -mg = 0  T = mg a Case 4 :Lift moving up or down with constant speed T = weight mg

  5. Example 1 • A lift and its passengers have a total mass of 500 kg. Find the tension in the cable supporting the lift if (i) the lift is at rest (ii) the lift is moving at constant speed (iii) the lift is accelerating upwards at 0.6 ms-2 (iv) the lift is accelerating downwards at 0.4 ms-2 Solution (i) At rest T – mg = 0 T – 500 x 9.8 = 0 T = 4900 N

  6. (ii) Ascending at constant speed Constant speed or at rest will have the same tension. i.e. T = 4900 N (iii) Accelerating upwards gives T = 5200 N T – mg = ma T – 500 x 9.8 = 500 x 0.6 (iv) Accelerating downwards gives T = 4700 N T – mg = m(-a) T – 500 x 9.8 = 500 x (-0.4)

  7. Example 2 A man of mass 80 kg stands o the floor of a lift that is accelerating at 2 ms-2. Find the normal reaction exerted on the man by the lift floor if the lift is (i) going up (ii) going down Solution (i) R – 80g = 80a : R – 784 = 160  R = 944 (ii) R – 80g = 80(-a) : R – 784 = -160  R = 624

  8. Example 3 A dynamo of mass 1200 kg is placed in a cage of mass 600 kg, which is raised vertically by a cable from a crane. The tension in the cable is 18 000 N. Find the acceleration of the cage, and the contact force between the cage and the dynamo. 18 000 N For the cage-and-dynamo a ms-2 T – mg = ma :18000 – 17640 = 1800a So a = 0.2 17640 N For the dynamo alone R – 11760 = 1200 a 0.2 ms-2 R – 11760 = 1200 x 0.2 So R = 12000 N R 11760 N

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