Lifts and Apparent Weight

Lifts

So far we have looked at objects being pulled or pushed across a surface, in other words motion parallel to the surface the object rests on. Here we only considered forces parallel to the surface, but we can also lift objects up or let them fall. This is vertical motion where only vertical forces are being considered.

Image credit: H. Breakell & Co. (Blackburn) Ltd

Let us consider a \(\text{500}\) \(\text{kg}\) lift, with no passengers, hanging on a cable. The purpose of the cable is to pull the lift upwards so that it can reach the next floor or lower the lift so that it can move downwards to the floor below. We will look at five possible stages during the motion of the lift and apply our knowledge of Newton's second law of motion to the situation. The 5 stages are:

1. A stationary lift suspended above the ground.
2. A lift accelerating upwards.
3. A lift moving at a constant velocity.
4. A lift decelerating (slowing down).
5. A lift accelerating downwards (the cable snaps!).

We choose the upwards direction to be the positive direction for this discussion.

Stage 1:

The \(\text{500}\) \(\text{kg}\) lift is stationary at the second floor of a tall building.

The lift is not accelerating. There must be a tension \(\vec{T}\) from the cable acting on the lift and there must be a force due to gravity, \(\vec{F}_g\). There are no other forces present and we can draw the free body diagram:

We apply Newton's second law to the vertical direction: \begin{align*} \vec{F}_R & = m_{\text{lift}} \vec{a}\ \text{(we use signs to indicate direction)} \\ T - F_g & = m_{\text{lift}} (0) \\ T & = F_g \end{align*}

The forces are equal in magnitude and opposite in direction.

Stage 2:

The lift moves upwards at an acceleration of \(\text{1}\) \(\text{m·s$^{-2}$}\).

If the lift is accelerating, it means that there is a resultant force in the direction of the motion. This means that the force acting upwards is now greater than the force due to gravity \(\vec{F}_g\) (down). To find the magnitude of the \(\vec{T}\) applied by the cable we can do the following calculation: (Remember we have chosen upwards as positive.)

We apply Newton's second law to the vertical direction: \begin{align*} \vec{F}_R & = m_{\text{lift}} \vec{a}\ \text{(we use signs to indicate direction)} \\ T - F_g & = m_{\text{lift}} (\text{1}) \\ T & = F_g + m_{\text{lift}} (\text{1}) \end{align*}

The answer makes sense as we need a bigger force upwards to cancel the effect of gravity as well as have a positive resultant force.

Stage 3:

The lift moves at a constant velocity.

When the lift moves at a constant velocity, the acceleration is zero, \begin{align*} \vec{F}_R & = m_{\text{lift}} \vec{a}\ \text{(we use signs to indicate direction)} \\ T - F_g & = m_{\text{lift}} (0) \\ T & = F_g \end{align*}

The forces are equal in magnitude and opposite in direction. It is common mistake to think that because the lift is moving there is a net force acting on it. It is only if it is accelerating that there is a net force acting.

Stage 4:

The lift slows down at a rate of \(\text{2}\) \(\text{m·s$^{-2}$}\). The lift was moving upwards so this means that it is decelerating or accelerating in the direction opposite to the direction of motion. This means that the acceleration is in the negative direction. \begin{align*} \vec{F}_R & = m_{\text{lift}} \vec{a}\ \text{(we use signs to indicate direction)} \\ T - F_g & = m_{\text{lift}} (-\text{2}) \\ T & = F_g -\text{2}m_{\text{lift}} \end{align*}

As the lift is now slowing down there is a resultant force downwards. This means that the force acting downwards is greater than the force acting upwards.

This makes sense as we need a smaller force upwards to ensure that the resultant force is downward. The force of gravity is now greater than the upward pull of the cable and the lift will slow down.

Stage 5:

The cable snaps.

When the cable snaps, the force that used to be acting upwards is no longer present. The only force that is present would be the force of gravity. The lift will fall freely and its acceleration.

Apparent Weight

Your weight is the magnitude of the gravitational force acting on your body. When you stand in a lift that is stationery and then starts to accelerate upwards you feel you are pressed into the floor while the lift accelerates. You feel like you are heavier and your weight is more. When you are in a stationery lift that starts to accelerate downwards you feel lighter on your feet. You feel like your weight is less.

Weight is measured through normal forces. When the lift accelerates upwards you feel a greater normal force acting on you as the force required to accelerate you upwards in addition to balancing out the gravitational force.

When the lift accelerates downwards you feel a smaller normal force acting on you. This is because a net force downwards is required to accelerate you downwards. This phenomenon is called apparent weight because your weight didn't actually change.