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## Class 11 Physics (India)

### Course: Class 11 Physics (India) > Unit 9

Lesson 5: Normal force and contact force- Normal force and contact force
- Normal force in an elevator
- Breaking down forces for free body diagrams
- Free body diagram with angled forces: worked example
- Normal force (basic)
- Forces at an angle
- Angled forces review
- More on Normal force (shoe on floor)
- More on Normal force (shoe on wall)
- What is normal force?

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# Free body diagram with angled forces: worked example

Sal draws a free body diagram for a box held stationary against a wall with a force at an angle theta.

## Want to join the conversation?

- I don’t get it. Friction is against the direction of motion. So why is the frictional force(Ff) pointing upwards in the direction of motion instead of downwards?(14 votes)
- So in this scenario, the vertical component of the angled force is less than gravity. Therefore, if there was no friction, then the block would accelerate downwards. Since friction always counteracts the motion, the friction force would point upwards.

In order for the friction force to point downwards, the block would have to be trying to accelerate upwards. That would happen if the horizontal component of the angled force were greater than gravity.

Hope this helps!(11 votes)

- What about the equal and opposite force exerted by block which is pulling the earth?(2 votes)
- You are right in saying that the block is pulling up the earth, but it is really small. THis is seen through f=ma. THe forces have to be the same according to newton's 3rd law, but the mass would be far far far far far larger - the earth is very heavy, thus the acceleration is really really really really small as mass is indirectly proportional to acceleration when the force is the same. Because it is too small, it is mostly disregarded.(4 votes)

- whats the purpose of doing math without numbers?(0 votes)
- Math doesn't have to be all about numbers. Mathematicians like to generalize things so it can be applicable to all cases, when there are numbers.(2 votes)

- Non of those forces have to equal our right or is it just this particular set of question?(2 votes)
- (I know this is a year late, but I’ll answer the question anyway just in case someone else has the same one.)

The forces don’t always have to equal (even) out or add up to a net force of zero when we do these kinds of problems, but they do cancel out here.

In this problem, we’re told that the box is “held stationary”, meaning it’s at rest. Staying stationary/at rest means that the box has 0 velocity and 0 acceleration. Newton’s 2nd law says that the sum of the forces equals mass times acceleration ΣF=ma

We know the acceleration is 0m/s^2, so we can substitute it to get ΣF=m(0), which multiplies to ΣF=0.

ΣF=0 means that the sum of the forces equals 0, or that all of the forces have to add up to zero (cancel out). Even though the forces won’t always cancel out, that’s why they do here.(2 votes)

- At2:51, Sal says that he doesn't draw the arrow atop F sub g because he's talking about the magnitude of the vector. Do you always need to draw the arrow? It seems like a bit of extra work for the same purpose. Thanks!(1 vote)
- F sub g refers to the magnitude of the vector, while F sub g with an arrow refers to the whole vector (both magnitude and direction). Here the downward arrow already tells us the direction so it is fine to write only F sub g, which tells us the magnitude of the vector Fsub g. We need to draw the arrow if we want to refer to the “whole” vector and not just its magnitude.(4 votes)

- Think about it. If a force is there FsinQ then the friction should be exact opposite, but Sal draws it in the same direction. Why?(2 votes)
- Why is the normal force going sideways?(2 votes)
- According to Newton's 3rd law of motion, every force action (the horizontal component Fcosθ) exerted on an object (the box on the wall), has an opposite force reaction (the wall on the box) equal in magnitude and opposite in direction. This force is called the normal Force.

Hope it helps!(1 vote)

- If Fsinθ + Ff= Fg then wouldn't there be no movement because the forces are equal?(1 vote)
- Yes, there won't be any movement, that is why in the beginning of the video it was mentioned that the block was stationary.(3 votes)

- so what do i do if force isn't given in newtons? one of my homework questions has the magnitude of an angled force given in mg (milligrams) and no matter where i look, i can't find any help on how to deal with that.

edit: turns out the force was 1/2 of the mass * gravity, which had been written as 0.500mg. :( lost ten marks because the textbook couldn't write out full words...(1 vote)- If you have mg (milligrams), then you know that is the mass. To find the downwards force, you need to find out weight, which is w=mg, weight = mass x gravitational field strength. So if the object is on earth, you would do mass x 9.81 or mass x 10. Remember weight is a force, and is measured in newtons.(2 votes)

- Wouldn't the Normal Force be counteracting the force of gravity and the force of friction be counteracting the horizontal component? I don't understand why they're switched in the video.

The friction of the wall is on the right side of the body, so it would be acting against the right-facing force. Whereas the Normal Force would be the one to act against gravity. Why is this not the case?(1 vote)

## Video transcript

- [Instructor] So what
we have depicted here. We have a block and let's say that this block is completely stationary and it's being pushed up against this non-frictionless wall. So this wall does have
friction with the block. It's being pushed by this force of magnitude F and its direction makes an angle of theta
with the horizontal. What I want you to do is pause this video and construct a free body diagram for this block and include the vertical and horizontal components of this force right over here, but also include other things. Include the force, include the force of gravity acting on this block. Include the force of friction acting on this block and include the normal force of the wall acting on the block as well. Pause the video and
try to have a go at it. So before I even start to draw the free body diagram, let's break down this force into its vertical
and horizontal components. So the first thing, let me do its vertical component. So it would look like that, and it's horizontal component would look like this. And so what's the magnitude
of the vertical component? Well it is opposite
this angle that we know. This is the angle that is theta and so this is going to be, the vertical component is going to be F sine of theta. We've seen that in previous videos and it comes straight out of right triangle trigonometry. Encourage you to review that
if this looks unfamiliar. And the magnitude of the
horizontal component, that is going to be F cosine theta. This side right over here is adjacent to the angle, SOH CAH TOA. And now with that out of the way we can draw our free body diagram. So let me draw that free body, let me draw that block and I'm really just gonna focus on the block only. And we know a couple of
things that are going on. We know that we have
this horizontal force, F cosine theta, so let me draw that. So the magnitude there is F cosine theta. We know we have this vertical force, F sine theta, so let me draw that. So this would be F and that one is actually a little bit shorter, it's obviously not drawn perfectly to scale, but this would be F sine theta. Now let's think about these characters right over here. We have the force of gravity and so that's going to be acting downward. So it would look something like this. So we have, and it would
have magnitude F sub g. I'm not drawing the arrow now because I'm just talking about
the magnitude of this vector. Here I'm referring to the entire vector, I'm referring to its
magnitude and direction. Now what about the force of friction? Let's assume that we're in a situation where the magnitude of
the vertical component of this applied force, F sine theta is less than the magnitude of the force of gravity. Well in that situation
if there was no friction, the block would start
accelerating downward, because you would have
a net force downward. We haven't talked about the forces to the left and right yet. But as we mentioned,
this thing is stationary and the force of friction is going to act against the direction of motion. And so in this situation,
the force of friction will act upwards, so we would have a force of friction just like that and its magnitude would be F sub lowercase f. And then last but not least what about the normal force? Well if this block is not accelerating in any direction that means that the normal force must perfectly counteract this force to the right which is the horizontal component of this applied force. And so our normal force is going to go to the left and
it would look like this. So it's magnitude is F sub capital N. So there you have it, we have drawn a free body diagram for this scenario right over here. If these two were equal, then you would have no force of friction or the force of friction would be zero. There would be nothing for
it to be counteracting. These two would perfectly cancel out. And if the magnitude of
the vertical component of the applied force were greater than the force of
gravity without friction, it would start to accelerate upwards and so the force of friction would act against that motion, but let's just go with this
scenario right over here and start to appreciate why free body diagrams are so useful. Because we can start to set up equations that relate these magnitudes. We can say look if this
box is not accelerating in any way, there's zero net forces in the horizontal and
the vertical directions, well then we could say that F sub N completely counteracts
F times cosine of theta. So we could say F sub N is equal to, is equal to F cosine theta, and we could also say that F sine theta, F sine theta plus the magnitude of the force of friction. Plus the magnitude of
the force of friction that these would completely counteract the magnitude of the force of gravity, because it's going in
the opposite direction. So these would be equal to F sub g. So once you set up equations like this, if you know all but
one of these variables, you can figure out the other ones which is very useful in physics.