Dimensional Analysis - Video Tutorials & Practice Problems

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concept

Dimensional Analysis

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Hey, guys. So one of things I mentioned earlier was that units have to speak the same language or be compatible with each other in order for your physics equations to make sense. There's a conceptual way to do this called dimensional analysis that's gonna show you how to do in this video again. It is kind of conceptual. So just make sure you really need to know this before you move on. So, guys remember, equations work on Lee. If they are dimensionally consistent, Really? That just means that the units on both sides of the equation are going to be equal to each other. If I get meters on the left side of an equation, I must get meters on the right side of an equation. The units have to balance out on the left and right. And dementia analysis is really just an easy way to check if your equations makes sense without actually having to plug a single number into your calculator. So, guys, I'm going to see. I want to show you a really simple way and how to do this. Let's just go ahead and check out this problem. So we're gonna walk a constant speed. V equals 5 m per second for a time of two seconds. We have these two equations down here, and we're gonna figure out which one of them would be appropriate for determining the distance. So you've got these two equations. They look kind of similar. So here's what we're gonna dio. So check of your equations. Air dimensionally consistent. First, we're just gonna replace all the variables with units. So, for example, we know that the unit for distance is gonna be meters and then from the problem, we're also told that the unit for velocity is meters per second. So this V here we replaced with just meters per second. And now, finally, we know that tea is just measured in seconds. So here we're gonna replace this with one second, and then over here, what happens is we have a T race to the second power, so we actually have to stick another exponents of of too in front of That s the second thing is, we're just gonna ignore all negative signs and all numbers, basically anything like two or a fraction, Anything like that. Anything that's not a variable kind of just gets ignored because we're only just looking at the variables here. So that's the second step. The third step is now we just have to multiply and divide to cancel out units. So, for example, I've got on the left. I got meters and on the right, have got meters per second time seconds. So what I can do is I can cancel out one factor of seconds in the bottom with one of the top. You can kind of cancel them out. Exactly how you would numbers or fractions. Um, so we're gonna cancel out seconds with seconds and then over here on this side, we have seconds that will cancel the with on Lee one of the factors of seconds in the second squared. So that's the third step. And now we just have to check if the units on the left on the units on the right actually match up. So for this equation here, we've got units of meters on the left, and then here what happens is the seconds cancel out. So we just have meters on the right, so that means this equation is dimensionally consistent. Whereas on the other hand, this equation here we get meters on the left and we get meters times seconds because one of the factors of seconds didn't cancel out. So notice how the units on the left don't equal the ones on the right. Therefore, this is a dimensionally inconsistent equation. So what that means is that this equation cannot possibly be correct for determining the distance. And this one, which is consistent, actually is does turn out to be the correct equation for distance. Times are for distance. It's just philosophy. Times time. Notice how I didn't plug in anything into my calculator, but just looking at the units, I was able to figure out which one of these equations made sense. Alright, guys, let's move on. So a lot of times you're gonna run into problems in which you're gonna have to figure out the units of unknown variables, and we can also use dimensional analysis to solve these problems. It's a very similar process to what we did before. Let's just just jump into let's just jump into an example and I will show you the steps. So we've got this equation or this law. It's called Hook's Law, and it states that a restoring force, which is measured in Newtons in a spring, is related to the distance. By this equation here, we're gonna figure out the units of this force Constant K. So we have to figure out this equation here. So notice how again we're not giving any numbers. There's no plugging into any calculator. You don't actually even have to know how this for this force equation works or what it means. All we're gonna do is just look at the units and figure out the units for this constant here. So we're gonna use the same process. First, we're just gonna set up the equation and replace the variables with units. So I've got f equals negative K Times X. I'm told that the force is measured in a unit called the Newton. Oh, that's Newton's. And then I have K negative k over here. And then I have X where X is related to the distance, and we know the distance from before is just measured in meters. So now I've got my, um, my replacing variables with units. The next thing is, I could just ignore any minus signs, so any minus signs or factors don't have to worry about. Those were just looking at the units. And now, lastly, we just have to annoy, isolate the unknown variable, and then just go ahead and solved. So we've got all these three steps here, So basically, I'm just gonna isolate this K by moving the meters to the other side. What I get is that Newtons per meter ends up being the force constant. And this is actually correct. This is the units for this force. Constant K here now again. Didn't know anything about how this equation works. But I can't figure out what the units of this force Constant are. Alright, guys, that's it for this one. Let me know if you have any questions.

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Problem

Problem

A box moving with an initial speed v is accelerated horizontally. If x is measured in [m], v in [m/s], a in [m/s^{2}], t in [s] which of the following equations is correct for solving the distance x?

A

x = a/t^{2}

B

x = v + ^{1}/_{2} at

C

x = vt + ^{1}/_{2} at^{2}

3

Problem

Problem

Newton's Law of Gravitation describes the attraction force between two masses. The equation is ,

where F is in [ kg·m / s^{2}], m_{1} and m_{2} are masses in [ kg], and r is the distance in [ m] between them.

Determine the units of the Universal Constant G.

A

kg·s^{2} / m^{3}

B

m^{3} /(kg·s^{2})

C

m / s^{2}

D

m^{3} /s^{2}

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