# Normal force and the assumption of motion of an object in an inclined plane

I was watching a video of object's motion on an inclined plane on Khan Academy. https://youtu.be/Mz2nDXElcoM

The video describes the motion of an ice block on an inclined ice plane (meaning that the plane is frictionless). Keep in mind that I assume that I won't know what the motion of the ice block will be.

I know that the weight force of the ice block can be broken down into the x and y component. Here's the problem:

In the video, Sal Khan, the teacher, said that the block will not move along the y direction. Therefore, the normal force will be equal in magnitude to the y component of weight force, $$N=mg\cos\theta$$

However, assuming that I do not know anything about the motion of the ice block on the plane, how can we prove that the perpendicular component of the weight force is equal the normal force? Can't the normal force be bigger than it so that the ice block will accelerate upwards? Is it an assumption or it's a natural phenomenon and property of normal force? Can it even be proved?

Whatever we read in physics are completely based on observations. First we observe something happening and then formulate laws which suits the observation. It is not the other way that we first make laws and then try to observe the law in action.

So coming back to your question, I think you should try repeating that experiment. Check what you observe. Of course , you will notice that the block doesn't fly off the wedge of its own. Now since we noticed this , we should try to formulate laws.

It is a common observation also that a block moves only when there is a net push or pull on it. So we can conclude that we need something to move a block at rest. This is what physicist call a force. So if something is at rest, then we can use our formulated law and conclude that the net force on that block is zero.

In the same way we can say that the ice in the video has no net force in the perpendicular direction on the basis of observation and thus normal force equals the component of the weight.

Edit : it is almost impossible to independently calculate the magnitude of the normal force. Since it not only depends on Coulomb's law. Also we need many factors like the number of foreign atoms between the two surfaces in contact, the type of foreign atoms, their charge and many other things. So we generally ignore them and on the basis of observation, we conclude that it's total magnitude is equal to weight.

Hope it helps 🙂.

• @Womas was that helpful ? Nov 27 '20 at 15:55
• It did help me a little bit. I'm still quite frustrated that this isn't as rigorous as I thought it would be. Nov 27 '20 at 22:53
• What is 'not rigorous' about a reactive force can never exceed the active force in the action/reaction pair? You suffer from unrealistic expectations.
– Gert
Nov 28 '20 at 14:55
• It is because the normal reaction is not the reactive force of the force of gravity. It is the reactive force of the force that the object applies on a plane. And the force that the block applies on a plane is different from the one that the earth applies on the object Nov 28 '20 at 18:36
• @Womas you are absolutely correct. If gravity from the earth disappears then we all will be thrown up due to normal forces .. but as I said it is more like observation and on the basis of that we define both to be equal in magnitude.. Nov 28 '20 at 18:39

The Normal force $$N$$ is constraint force (contact force) not applied force. To obtain the Normal forces you "need" the constraint equation or you can use the free body diagram, in your case is the deflection toward the normal zero, this is how you obtain $$N=m\,g\,\cos(\theta)$$, If the deflection toward the normal zero not zero , you don't get contact thus the normal force is zero.

Can't the normal force be bigger than it so that the ice block will accelerate upwards? Is it an assumption or it's a natural phenomenon and property of normal force? Can it even be proved?

It's like having an object of mass $$m$$ rest on a table. The table, assuming it is strong enough, will provide enough normal force, i.e. $$N=mg$$, to prevent the object from 'falling through' the table. So there's no motion in the vertical, downward direction.

However, $$N=mg$$ is the maximum force the table can provide.

It cannot provide a normal force:

$$N>mg$$

which would cause upward motion of the object. $$N$$ is a reactive force which at best equals $$mg$$, it cannot exceed it.

It's basically the same for a mass $$m$$ on an incline (with angle $$\alpha$$), regardless even of friction:

The incline's surface needs to provide a reactive, Normal force $$N$$ to prevent the mass for crashing through the surface:

$$N=mg\cos\alpha$$

This reactive force however cannot exceed $$mg\cos\alpha$$.

• I know that normal force is part of a action reaction pair, but it's the force that the ice block applies on the plane which is in the pair, not the weight of the ice block. So how do we guarantee that the weight force is exactly equal to the force that the ice block, or book in your case, applies on the table? Nov 27 '20 at 22:52
• I also found this question which is basicly the same as mine. physics.stackexchange.com/questions/546228/… The answer is that we can theoretically prove that the normal force will never be strong enough to lift the object above the surface(Assuming the support plane is strong enough of course)? Nov 28 '20 at 3:36

However, assuming that I do not know anything about the motion of the ice block on the plane, how can we prove that the perpendicular component of the weight force is equal to the normal force? Can't the normal force be bigger than it so that the ice block will accelerate upwards? Is it an assumption or it's a natural phenomenon and property of normal force? Can it even be proved?

The force analysis doesn't require anything from the motion, we take to analyze the forces(causes of motion) at a point and predict future motion. The component of forces in the direction normal to the ramp must be zero, otherwise, it'd mean 'in the future', the block will 'jump' for the incline or 'depress' into the incline. Since the first possibility is not something physical and we assumed that the incline doesn't change shape, the total normal is zero. (How does it move into the incline without changing its shape??)

Here we are doing the force components in directions parallel and perpendicular to the plane.

• So if I'm getting it right, we analyse the informations of an object's motion at a certain time frame, and then base on the informations, we predict the future motions of the object? Nov 29 '20 at 3:56
• Yep that's the big picture of the free body analysis.. sometimes the forces (causes) will change mid way into the motion and you'll need to reanalyze the whole motion Nov 29 '20 at 6:35
• **not whole motion from that point. I can show you an example of such a case, imagine a ball rolling on the ground, first on floor without friction and then it encounters floor with friction. The forces on second part is different so the motion itself changes Nov 29 '20 at 6:36