# Clarification needed regarding acceleration of a lift

Consider a lift (or an elevator, as Americans call it) with a block inside it. The lift is accelerating upwards, and we are looking at this system from the ground frame.

Wouldn't the net acceleration of the block be $$(g-a)$$? This is because gravity is pulling the block downwards, but the lift's acceleration is moving it upwards.

Similarly, when the lift accelerates downwards, the block should experience an acceleration of $$(g+a)$$, but it is exactly the opposite, in both cases. Why is this so?

• Do you want an intuitive explanation or a mathematical proof? Jan 21 at 12:25
• An intuitive explanation. Jan 21 at 13:48
• It's the same as feeling "heavy" when an elevator goes up and "light" when it goes down. What makes you think the accelerations of the block in each should be different from what it is really? Jan 21 at 15:56
• Your "heavy" and "light" example really does the job for me. I now understand the idea of what the block would be feeling. Thanks! Jan 21 at 17:42

In standard Newtonian physics the acceleration of the lift is just $$\vec a$$. There are two forces acting on the block. One is gravity $$m\vec g$$ and the other is the normal force $$\vec N$$.

Newton’s 2nd law says that the sum of the forces is the mass times the acceleration so $$\vec N + m\vec g=m\vec a$$

This is the vector expression regardless of if the lift is accelerating up or down. You can then look at the vertical component of this vector equation to solve it. Only then does the sign of the vertical component of $$\vec g$$ and $$\vec a$$ come in.

• Yes, what you said is true, but this wasn't what I am confused about. I am confused about the acceleration of the block. Obviously, it would be (g+a) if both (lift and block) are moving downward, and (g-a) if moving upwards. But its the complete opposite. Why is that so? Jan 21 at 13:54
• the acceleration is always a, given the way the problem is structured. It is never (g+a) or (g-a). Were it to be (g+a), that would mean the elevator is accelerating down (not just going down), and within the elevator, something is accelerating the block towards the bottom of the elevator (in the elevator's frame) Jan 21 at 16:27
• @Haider the acceleration is just $\vec a$ in standard Newtonian mechanics
– Dale
Jan 21 at 18:24
• @Dale Oh, okay. So gravity only plays a part in mg (which is always downwards) and nowhere else. I now understand. Thank you. Jan 22 at 11:20

When you are standing on a scale in a stationary elevator, gravity is pulling you down, compressing the spring in the scale. That spring compresses until its upward force equals your weight, and you read your weight on the scale. The upward force provided by the spring is actually the normal force that prevents you from sinking further down on the scale.

When the elevator accelerates up, there is an additional force causing that acceleration, and that force is added to the normal force on the spring, increasing the spring compression and the reading on the scale. Due to this, the scale reading increases when the elevator accelerates you upward. Obviously, the opposite happens when you are accelerating downward.

This is because gravity is pulling the block downwards, but the lift's acceleration is moving it upwards.

Your confusion arises because you are forgetting the gravity is a fictitious force like centrifugal force.

A fictitious force is a force that is only apparent in an accelerating reference frame, but not in an inertial reference frame.

For example, imagine we have a mass of 100 Kgs sitting on a weighing scale on the floor of the elevator, that is initially stationary with respect to the surface of the Earth. The weighing scales would indicate a mass of 100 Kgs.

Now to a free falling observer, who considers himself to be stationary, there appears to be a mysterious force accelerating the mass upwards at 9.8 $$m/s^2$$ but there is NO force of gravity acting downwards on the mass. If there was a force acting downwards on the mass the forces would be balanced out and the mass would not be accelerating relative to the observer, which is not the case from the inertial point of view.

Now if we accelerate the lift itself upwards at say 2g, there would be total upward force equal to 100 Kgs $$\times$$ 9.8 $$\times$$ 3 and the weighing scales would indicate a mass of 300 Kgs. From the inertial observer's point of view all the forces are acting in the same direction, so they are additive.

An observer inside the lift is an accelerating observer and they measure an upward acceleration using an accelerometer (a small mass on a spring) and since the 100 Kg mass is stationary relative to them, they assume there must be a force acting downwards to balance the forces and we call this imaginary force the force of gravity.

Since in everyday life we live in the gravitational field of the Earth, we forget we are not inertial observer's and we think of gravitational force as a real force (which it is not).