In a book I saw that the time period of a pendulum inside a elevator moving up is, $$T=2\pi\sqrt{\frac{L}{g+a}}$$ I was curious as to why we use $(g+a)$ as we know inside an elevator,




So my question is shouldn't we use $a_{net}$ instead of the acceleration due to the normal force?


3 Answers 3


The net force on the pendulum is:

$${\vec F}_{net}=-m{\vec g} -{\vec F}_n$$


$$m{\vec a}_{net}=-m{\vec g}-m{\vec a}_n$$

dividing both sides by $m$:

$${\vec a}_{net}= -{\vec g}-\vec a_n$$

where both $\vec g$ and $\vec a_n$ point downwards. You state instead that (and I think that this is the source of confusion) $\vec F_n$, and thus $\vec a_n$ points upwards, which is the case for the lift, but the normal reaction force (which the pendulum experiences), which is equal in magnitude to the acceleration of the lift $\vec a$ is pointing downwards. So the magnitude of the net acceleration is $g+a_n$, pointing downwards, which we obviously can't use as $a_n$.

  • $\begingroup$ The pendulum does not experience any normal force. $\endgroup$
    – Yashas
    Commented Apr 23, 2017 at 13:24
  • $\begingroup$ In reaction to the upward acceleration, the pendulum experiences a normal acceleration downwards (although a fictitious one). This is why $g$ is replaced by $g+a$, where $a$ is the acceleration upwards. $\endgroup$ Commented Apr 23, 2017 at 13:54
  • $\begingroup$ It is not called normal acceleration; it is pseudo acceleration. Normal forces are the forces which act perpendicular to a surface. There is nothing wrong in using $a_{net}$. You can approach the problem from two reference frames: ground frame and the lift frame. In the ground frame, there $a_{net} = a$ and in the lift frame, the pseudo force acts to compensate for the lack of acceleration (it is at rest relative to the lift). $\endgroup$
    – Yashas
    Commented Apr 23, 2017 at 13:54
  • $\begingroup$ In the ground frame, the net acceleration "experienced" by the pendulum is g+a, directed downwards (which doesn't imply that the pendulum should accelerate towards the earth), just as in the lift frame the acceleration in g+a, directed downwards. $\endgroup$ Commented Apr 23, 2017 at 14:13

What is normal force?

The normal force does not appear in general. The normal force by definition is a contact force which surfaces experience when they come in contact with each other.

When you stand on the ground, you experience a normal force. The gravity pulls you downwards with $mg$ and the normal force $N$ pushes you upwards with an equal force and this is why you stay in equilibrium.

If you jumped, your feet would leave the surface and hence would no longer experience a normal force. As the only force acting on you while you jump is the gravitation force, it pulls you down (otherwise, you would float away to space).

The normal force acts to prevent deformation of the surface. For example, if you had to accelerate through the ground due to gravity, you would have to make your way through the ground (you would have to tear apart the ground and dig a tunnel). The elastic forces of the ground do not let you do that. It provides a normal force such that you don't significantly pierce through it.

When you fall back to the ground after jumping, the normal force $N$ is greater than $mg$ because of which you decelerate.

If you had fallen on a thermoform sheet, you would have broken it as the sheet doesn't have enough strength to provide a normal force to counteract your weight as well as the momentum. However, if a fly were to sit on the sheet, the fly wouldn't break it as the sheet can sustain such a small weight.

Inertial and non-inertial frames of reference

An accelerating frame of reference is known as a non-inertial frame of reference. Non-accelerating frames are known as inertial frames of reference.

Newton's laws, as it is, do not work in a non-inertial frame of reference unless you tweak it. The laws are valid for inertial frames of reference frames only.

To use Newton's laws in a non-inertial reference frame, we introduce a concept known as pseudo-acceleration or pseudo-force.

If the non-inertial frame is accelerating with an acceleration $\vec{a}$, all objects in that frame experience a force in the opposite direction, $-m\vec{a}$.

Once you consider the pseudo-force in your non-inertial frame of reference, Newton's laws hold good.

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What's happening in the lift in the frame of the lift?

The lift is accelerating upwards with an acceleration $a$. Therefore, all the objects in the lift experience a pseudo-force downwards whose magnitude is given by $ma$.

The total force an object of an isolated mass $m$ is

$$F_{net} = ma + mg = m(a + g)$$

As the pseudo-force acts in the downward direction, the pseudo-force term can be combined with the gravitational force term to give a force which is known as effective gravitational force (the pseudo force isn't related to gravity, however).

This is why you get $a + g$ in the formula which gives the time period of the pendulum.

When you stand in such a lift, in the lift's frame of reference (which is non-inertial as it is accelerating), you don't move. Therefore, the net force on you must be zero.

$$F_{net} = N - ma - mg = 0$$

The normal force you feel is given by $$N = m(a + g)$$

As you feel a larger normal force, you feel heavier.

What is happening in the lift from the ground frame?

The ground frame is inertial and has no mysterious forces such as pseudo force acting.

In this frame of reference, you are accelerating upwards with an acceleration $a$. Therefore, there must be a net force of $ma$ acting on you.

$$F_{net} = ma = N - mg$$

If you rearrange the equation, you get:

$$N = m(a + g)$$

The above equation is identical to the equation we got earlier when we calculated the normal force from the lift's frame of reference.

They must agree. If they didn't, then physics would be wrong. You can choose to solve the problem from any frame of reference. You will always get the same answer.

  • $\begingroup$ When you fall back to the ground after jumping, the normal force 𝑁 is not greater than 𝑚𝑔, the deceleration is caused because the only force acting was weight during free fall and when the object makes contact with the ground the net force becomes zero. $\endgroup$
    – ten1o
    Commented Jan 12, 2021 at 14:02
  • $\begingroup$ The "ma" in this case is not a pseudo force. Pseudo forces such as centrifugal forces exist exclusively in the non-inertial frame, from an inertial frame, they do not exist. $\endgroup$
    – ten1o
    Commented Jan 12, 2021 at 14:15

The $a$ is $a_{net}$! There is no such thing as an acceleration due to each force. There is no acceleration specifically from the normal force alone, unless it is alone.

Newton's 2nd law says that all forces combined give an acceleration (times the mass):

$$\sum F=ma$$

Not that each force individually gives one acceleration each which are then summed. The summation symbol $\sum$ must not be ignored. You can have many forces, but they together give just one acceleration. There is no such thing as accelerations due to each force, only one acceleration due to the net force at any point in time.


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