# Pseudo Force and Inertial and Non-Inertial frames

In the figure given below is block placed on an incline $$\theta$$. Now the lift is accelerating upwards with an acceleration $$a_0$$. Now if we make our measurements from the lift frame we will have to apply a pseudo force $$-ma_0$$. Which will have two components one in the direction of $$Mg\cos\theta$$. And other in the direction of $$Mg\sin\theta$$. Now $$Mg\sin\theta+Ma_0\sin\theta=Ma_\text{net}$$. Where $$a_\text{net}$$ is the net acceleration in that direction.

Now let's observe it from the ground or an inertial frame here the object has a net upward acceleration which has a component opposite to $$Mg\sin\theta$$. Therefore $$Mg\sin\theta=- Ma_0\sin\theta$$. Now what I thought was that this is not possible and hence there is another force acting opposite to $$Mg\sin\theta$$, $$Ma_\text{net}$$. Now this doesn't make any sense to me if there is a force acting opposite to $$Mg\sin\theta$$, and the net is also in that direction, then won't the object move upwards on the incline. Now that doesn't make any sense. Can someone tell me what Is happening and from where is this $$a_\text{net}$$ coming from when observing in the inertial frame?

• "Now let's observe it from the ground or an inertial frame here the object has a net upward acceleration" Does the block stay at rest relative to the incline due to friction? This seems to contradict your earlier analysis that there is some acceleration along the incline. Jul 3 '20 at 10:44

Mathematically, moving between inertial and non-inertial frames correspond to moving terms from one side of Newton's second law to the other side.

So, in your non-inertial frame accelerating with the incline you have for Newton's second law along the incline (using your notation) $$Mg\sin\theta+Ma_0\sin\theta=Ma_\text{net}$$

Moving to the inertial frame we have $$Mg\sin\theta=Ma_\text{net}-Ma_0\sin\theta=M(a_\text{net}-a_0\sin\theta)=Ma'_\text{net}$$

So we see that in the inertial frame we have an acceleration of $$a_\text{net}-a_0\sin\theta$$ in the direction of the incline.

Note that this is also equal to $$g\sin\theta$$, which makes sense because the only force that has a component along the incline is the force of gravity. However, don't get this acceleration in the inertial frame mixed up with acceleration down the incline. The incline is accelerating in the inertial frame too, so it also has an acceleration of $$g\sin\theta$$ along the incline. Therefore, just saying the acceleration along the incline is $$g\sin\theta$$ isn't very interesting, in my opinion.

Another way to see this is to think of it as adding up the acceleration (along the direction of the incline) of the block relative to the incline $$a_\text{net}$$ and the acceleration of the incline relative to the inertial frame $$-a_0\sin\theta$$. This gives the acceleration of the block relative to the inertial frame $$a_\text{net}-a_0\sin\theta$$. This is just classical relative acceleration addition (which is the time derivative of classical relative velocity addition).

In either case, the block moves down the incline for $$a_0>0$$ as in your diagram. This is true even when $$a_\text{net}-a_0\sin\theta<0$$ because in the inertial frame the net acceleration is relative to our inertial frame, not the incline. $$a_\text{net}$$ is still positive, so using your sign convention the block still accelerates down relative to the incline.

To get a better picture of everything, I suggest doing this same analysis for Newton's second law perpendicular to the incline. I think this is a good exercise, so I'll leave it for you to do.

• Where did the Mao come in the inertial frame? And why is it direction that? Jul 3 '20 at 10:58
• @BlackSusanoo Because the net acceleration in the direction along the incline in the inertial frame is $a_0-a\sin\theta$. I mean it needs to be somewhere. The incline itself is accelerating in the inertial frame. Why the down vote? Jul 3 '20 at 11:01
• @BlackSusanoo Think of it as adding up the acceleration of the block relative to the incline and the acceleration of the incline relative to the inertial frame. This gives the acceleration of the block relative to the inertial frame. Jul 3 '20 at 11:03
• I kind of get your answer but I'm still waiting for more answers thanks a lot mate Jul 3 '20 at 11:46
• @BlackSusanoo Parallel to the incline, the acceleration of the block relative to the inertial frame is the acceleration of the block relative to the incline $a_\text{net}$ plus the acceleration of the incline relative to the inertial frame $-a_0\sin\theta$. It is addition to relative accelerations. Why do you think the acceleration of the block in the inertial frame shouldn't depend on the acceleration of the incline in the inertial frame? Jul 4 '20 at 3:28

To avoid confusion caused by gravity, we will assume the lab frame is in an inertial frame, floating in space far from Earth. $$F = ma$$ works in this frame. In these coordinates, there is no net force on an object that stays at $$x = 0$$.

Working in the lab frame, you see the lift accelerated upward. Given no friction, the inclined plane exerts a normal forces on the block. This has an upward component and a leftward component. The leftward component makes the block slide down the plane as the upward component lifts it. The block accelerates upward as it slides along the plane, but not as fast as the lift.

To repeat the exercise in the lift frame, you have to pretend the lift is not accelerating. You choose a frame of reference where $$x^{'} = 0$$ is attached to the lift. It stays still, while the point $$x = 0$$ accelerates downward.

But now you are working in a frame where $$F^{'} = ma^{'}$$ gives the wrong answer. When $$F^{'} = 0$$, you see the block accelerating downward to keep up with the lab frame's $$x = 0$$. To make $$F^{'} = ma^{'}$$ work, you need to pretend that there is a downward force to explain the pretended downward acceleration.

In the lift frame, the downward force presses the block into the frictionless inclined plane. The plane presses back with a normal reaction force that has an upward and leftward component. The leftward component makes the block slide down the plane as the sum of the upward component and pretend forces accelerates it downward. The block accelerates downward as it slides along the plane.

Perhaps you can see it better with this figure. to apply NEWTON second law , you have to calculate the components of the position vector to the mass in inertial system.

$$\vec{R}= \pm\begin{bmatrix} x_0 \\ y_0 \\ \end{bmatrix}=\pm \left[ \begin {array}{c} s\cos \left( \vartheta \right) \\ s\sin \left( \vartheta \right) + y \left( \tau \right) \end {array} \right] \tag 1$$

where "+" from inertial system and "-" from lab system .

with equation (1) you can obtain the kinetic energy and with the potential energy $$U=m\,g\,R_y$$ you get the equation of motion:

$$M\,{\ddot{s}}\pm{M}\,g\sin \left( \vartheta \right) +{M}\,\sin \left( \vartheta \right) \underbrace{{\frac {d^{2}}{d{\tau}^{2}}}y \left( \tau \right)}_{a_0} =0$$

thus for "+" sign you get $$M\,a_{\text{net+}}=M\,g\,\sin(\vartheta)+{M}\,\sin(\vartheta)\,a_0\quad \Rightarrow\quad g\mapsto g+a_0$$

and for "-" sign you get:

$$M\,a_{\text{net-}}=-M\,g\,\sin(\vartheta)+{M}\,\sin(\vartheta)\,a_0\quad \Rightarrow\quad g\mapsto a_0-g$$

The acceleration of the mass, a, in the inertial frame is the sum of the acceleration of the elevator, $$a_o$$ and the acceleration of the mass relative to the incline, a', in the elevator. To avoid using the normal force, I'll chose the +x axis parallel to and up the incline. Then for x components: -mg sin(θ) = m$$a_x$$ = m($$a_o$$ sin(θ) + a') giving a' = -(g + $$a_o$$) sin(θ).

In the first part of your question you are looking at the frame of the elevator. Since the frame is accelerating, you have an $$Ma_0$$ pseudo force in the downward direction. So the free body diagram looks like

You used $$a_{net}$$ for acceleration along plane but I'm using $$a_{x_{el}}$$ because its a component of acceleration not the net acceleration. So the force equation along the incline (along $$x_{el}$$) is $$MgSin\theta+Ma_0Sin\theta=Ma_{x_{el}}$$ canceling $$M$$ gives us $$a_{x_{el}}=gSin\theta+a_0Sin\theta$$ If you look perpendicular to the incline $$a_{y_{el}}=0$$ as the incline is stationary in the frame of the elevator and the block does not have any acceleration perpendicular to the incline when its stationary. Upto this part you got it right. I Don't know what you did in the second part. w.r.t inertial frame, $$MgSin\theta \neq-Ma_0Sin\theta$$.

Anyway, looking from the inertial frame, the free body diagram looks like this Notice there is no component of a pseudo force due to $$a_0$$ in this frame as it is inertial. The force equation along the incline is $$MgSin\theta=Ma_{x_{in}}$$ therefore $$a_{x_{in}}=gSin\theta$$ This is because only force acting on it along the incline (along $$x_{in}$$) is gravity. That is the acceleration of the block along the incline in the inertial frame. So you might ask what difference does this make from a system in which incline is stationary. The difference is in this system, the acceleration of the block perpendicular to the incline has to be equal to the component of the acceleration of the elevator(and therefore the wedge the block is on) perpendicular to the incline. That is, $$a_{y_{in}}=a_0Cos\theta$$ Otherwise if the block and the incline didn't have the same acceleration in the $$y_{in}$$ direction they would separate or the block would go into the incline.

• So if the elevator was in free fall the block would still slide down the incline? Just because $a_{x_{in}}=g\sin\theta$ doesn't mean the block of moving down the incline. You are forgetting that the incline is also accelerating in the inertial frame. In any case to make things less confusing, if you call your acceleration in the inertial frame $a'_{x_{in}}$ to compare between the two cases, it is easy to show that it must be the case that $a'_{x_{in}}=a_{x_{in}}-a_0\sin\theta$ like I show in my answer. I think you just didn't understand my words/notation. Jul 6 '20 at 12:06
• Yes, the acceleration along the incline in the inertial frame is always $g\sin\theta=a'_{x_{in}}=a_{x_{in}}-a_0\sin\theta$, but that doesn't mean the acceleration relative to the incline is $g\sin\theta$. You are mixing up the acceleration relative to the inertial frame and the acceleration relative to the incline. Jul 6 '20 at 12:14
• I never said acceleration w.r.t incline is $gSin\theta$ (except I wrote $g_{net}=gSin\theta$ which was me getting confused with the bad notation that wasn't mine). $a_{x_{in}}$ is the acceleration in the inertial frame which is $gSin\theta$. $a_{x_{el}}$ is the acceleration in the elevator frame. In the inertial frame, it doesn't matter if it slides down the incline or not, it will have $gSin\theta$ in the down the incline DIRECTION which doesn't mean its actually sliding down relative to the incline for a general case. Jul 7 '20 at 6:01
• In this problem, since the acceleration is upwards as given with the up arrow and $a_0$ the block will slide down the incline as seen from the frame of incline. I saw your edit, so $a_{net}-a_0Sin\theta=gSin\theta$, what extra information do you get from $a_{net}-a_0Sin\theta$ since $a_{net}$ is an unknown in the inertial frame. Taking $Ma_0Sin\theta$ from one side of the elevator frame equation to the other doesn't change it from elevator frame to inertial frame. Jul 7 '20 at 6:21
• You can however say $gSin\theta+a_0Sin\theta-a_0Sin\theta=a_{x_{in}}$ where you're changing from non inertial to inertial frame by subtracting relative accelerations, but you'll still get the same answer and its the same as writing force equations in the inertial frame Jul 7 '20 at 6:21