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Now we are going to add things to exactly the same problem to show how things change slightly. We will now move to a lift moving at constant velocity. Remember if velocity is constant then acceleration is zero.

A man with a mass of 100 kg stands on a scale (measuring newtons) inside a lift that is moving downwards at a constant velocity of 2 m · s - 1 . What is the reading on the scale?

  1. We are given the mass of the man and the acceleration of the lift. We know the gravitational acceleration that acts on him.

  2. Once again we can use Newton's laws. We know that the sum of all the forces must equal the resultant force, F r .

    F r = F g + F N
  3. F g = m g = ( 100 kg ) ( 9 , 8 m · s - 2 ) = 980 kg · m · s - 2 = 980 N downwards
  4. The scale measures this normal force, so once we have determined it we will know the reading on the scale. Because the lift is moving at constant velocity the overall resultant acceleration of the man on the scale is 0. If we write out the equation:

    F r = F g + F N m a = F g + F N ( 100 ) ( 0 ) = - 980 N + F N F N = 980 N upwards
  5. The normal force is 980 N upwards. It exactly balances the gravitational force downwards so there is no net force and no acceleration on the man. The reading on the scale is 980 N.

In the previous two examples we got exactly the same result because the net acceleration on the man was zero! If the lift is accelerating downwards things are slightly different and now we will get a more interesting answer!

A man with a mass of 100 kg stands on a scale (measuring newtons) inside a lift that is accelerating downwards at 2 m · s - 2 . What is the reading on the scale?

  1. We are given the mass of the man and his resultant acceleration - this is just the acceleration of the lift. We know the gravitational acceleration also acts on him.

  2. Once again we can use Newton's laws. We know that the sum of all the forces must equal the resultant force, F r .

    F r = F g + F N
  3. F g = m g = ( 100 kg ) ( 9 , 8 m · s - 2 ) = 980 kg · m · s - 2 = 980 N downwards
  4. The resultant force can be calculated by applying Newton's Second Law:

    F r = m a F r = ( 100 kg ) ( - 2 m · s - 2 ) = - 200 N = 200 N down
  5. The sum of all the vertical forces is equal to the resultant force, therefore

    F r = F g + F N - 200 N = - 980 N + F N F N = 780 N upwards
  6. The normal force is 780 N upwards. It balances the gravitational force downwards just enough so that the man only accelerates downwards at 2 m · s - 2 . The reading on the scale is 780 N.

A man with a mass of 100 kg stands on a scale (measuring newtons) inside a lift that is accelerating upwards at 4 m · s - 2 . What is the reading on the scale?

  1. We are given the mass of the man and his resultant acceleration - this is just the acceleration of the lift. We know the gravitational acceleration also acts on him.

  2. Once again we can use Newton's laws. We know that the sum of all the forces must equal the resultant force, F r .

    F r = F g + F N
  3. F g = m g = ( 100 kg ) ( 9 , 8 m · s - 2 ) = 980 kg · m · s - 2 = 980 N downwards
  4. The resultant force can be calculated by applying Newton's Second Law:

    F r = m a F r = ( 100 kg ) ( 4 m · s - 2 ) = 400 N upwards
  5. The sum of all the vertical forces is equal to the resultant force, therefore

    F r = F g + F N 400 N = - 980 N + F N F N = 1380 N upwards
  6. The normal force is 1380 N upwards. It balances the gravitational force downwards and then in addition applies sufficient force to accelerate the man upwards at 4m · s - 2 . The reading on the scale is 1380 N.

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Source:  OpenStax, Maths test. OpenStax CNX. Feb 09, 2011 Download for free at http://cnx.org/content/col11236/1.2
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