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A man of mass 70 kg is standing on a weighing scale in a lift. Calculate the reading on the scale when the lift is moving upward with an acceleration of 2 m/s², downward with an acceleration of 2 m/s²

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4 parts
a

826 N826 \text{ N}

b

546 N546 \text{ N}

c

686 N686 \text{ N}

d

0 N0 \text{ N}

Step-by-step solution

8 steps

The reading on the scale is the normal force (NN) exerted by the scale on the man. We use Newton's second law, ∑F=ma\sum F = ma. Taking the upward direction as positive, the forces acting on the man are the normal force (NN) upwards and his weight (mgmg) downwards.
So, N−mg=maN - mg = ma, which means N=m(g+a)N = m(g+a).
Given:
Mass of the man (mm) = 70 kg70 \text{ kg}
Acceleration due to gravity (gg) = 9.8 m/s29.8 \text{ m/s}^2

Part a· 2 steps

The lift is moving upward with an acceleration of 2 m/s22 \text{ m/s}^2.

  1. Identify the acceleration.

    The acceleration (aa) is +2 m/s2+2 \text{ m/s}^2 (upward is positive).

  2. Calculate the normal force.

    N=m(g+a)N = m(g+a) N=70 kg(9.8 m/s2+2 m/s2)N = 70 \text{ kg} (9.8 \text{ m/s}^2 + 2 \text{ m/s}^2) N=70 kg(11.8 m/s2)N = 70 \text{ kg} (11.8 \text{ m/s}^2) N=826 NN = 826 \text{ N}

    The reading on the scale is 826 N\boxed{826 \text{ N}}.

Part b

546 N546 \text{ N}

The lift is moving downward with an acceleration of 2 m/s22 \text{ m/s}^2.

  1. Identify the acceleration.

    The acceleration (aa) is −2 m/s2-2 \text{ m/s}^2 (downward is negative).

  2. Calculate the normal force.

    N=m(g+a)N = m(g+a) N=70 kg(9.8 m/s2−2 m/s2)N = 70 \text{ kg} (9.8 \text{ m/s}^2 - 2 \text{ m/s}^2) N=70 kg(7.8 m/s2)N = 70 \text{ kg} (7.8 \text{ m/s}^2) N=546 NN = 546 \text{ N}

    The reading on the scale is 546 N\boxed{546 \text{ N}}.

Part c

686 N686 \text{ N}

The lift is moving with constant velocity.

  1. Identify the acceleration.

    Constant velocity means the acceleration (aa) is 0 m/s20 \text{ m/s}^2.

  2. Calculate the normal force.

    N=m(g+a)N = m(g+a) N=70 kg(9.8 m/s2+0 m/s2)N = 70 \text{ kg} (9.8 \text{ m/s}^2 + 0 \text{ m/s}^2) N=70 kg(9.8 m/s2)N = 70 \text{ kg} (9.8 \text{ m/s}^2) N=686 NN = 686 \text{ N}

    The reading on the scale is 686 N\boxed{686 \text{ N}}.

Part d

0 N0 \text{ N}

The lift is in free fall.

  1. Identify the acceleration.

    In free fall, the acceleration (aa) is equal to the acceleration due to gravity, but in the downward direction, so a=−g=−9.8 m/s2a = -g = -9.8 \text{ m/s}^2.

  2. Calculate the normal force.

    N=m(g+a)N = m(g+a) N=70 kg(9.8 m/s2−9.8 m/s2)N = 70 \text{ kg} (9.8 \text{ m/s}^2 - 9.8 \text{ m/s}^2) N=70 kg(0 m/s2)N = 70 \text{ kg} (0 \text{ m/s}^2) N=0 NN = 0 \text{ N}

    The reading on the scale is 0 N\boxed{0 \text{ N}}.

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