Dimensional Analysis - Online Tutor, Practice Problems & Exam Prep

Dimensional analysis ensures that physics equations are dimensionally consistent, meaning the units on both sides must match. For example, in calculating distance using velocity and time, the equation must yield meters on both sides. Hooke's law illustrates how to derive units for unknown variables, such as the force constant (k), which is measured in newtons per meter. This method allows for understanding relationships without numerical calculations, emphasizing the importance of unit compatibility in physics.

Dimensional Analysis

Hey, guys. So one of the 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 what I'm going to 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 only 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 dimensional analysis is really just an easy way to check if your equations make sense without actually having to plug a single number into your calculator. So guys, I am going to show you a really simple way on how to do this. Let's just go ahead and check out this problem. We're going to walk at a constant speed, v equals 5 meters per second for a time of 2 seconds. We have these two equations down here and we're going to figure out which one of them would be appropriate for determining the distance. So you've got these 2 equations. They look kind of similar. So here's what we're going to do. To check if your equations are dimensionally consistent, first, we're just going to replace all the variables with units. So for example, we know that the unit for distance is going to 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 replace with just meters per second. And now finally, we know that t is just measured in seconds. So here, we're going to replace this with one second. And then over here, what happens is we have t raised to the second power. So we actually have to stick another exponent of 2 in front of that s. The second thing is we're just going to ignore all negative signs and all numbers. Basically, anything like 2 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, I’ve got meters per second times seconds. So what I can do is I can cancel out one factor of seconds in the bottom with 1 at the top. You can kind of cancel them out exactly how you would numbers or fractions. So we're going to cancel out seconds with seconds. And then over here on this side, we have seconds that will cancel out with only 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 and 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 does turn out to be the correct equation for distance: it’s just velocity 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 going to run into problems in which you're going to 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 jump into an example, and I am going to show you the steps. So we've got this equation, or this law. It’s called Hooke'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. And we're going to 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 given any numbers. There’s no plugging into any calculator. You don't actually even have to know how this force equation works or what it means. All we’re going to do is just look at the units and figure out the units for this constant here. So we’re going to use the same process. First, we're just going to 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. So that’s Newtons, 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 replacing variables with units. The next thing is I can just ignore any minus signs. So any minus signs or factors, don't have to worry about those. We’re just looking at the units. And now lastly, we just have to isolate the unknown variable and then just go ahead and solve. So we’ve got all these 3 steps here. So basically, I’m just going to 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 f$k$≈newtons/meter is the units for this force constant k here. Now, again, didn't know anything about how this equation works, but I can 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.

A box moving with an initial speed v is accelerated horizontally. If x is measured in [m],
v in [m/s], a in [m/s²], t in [s] which of the following equations is correct for solving the distance x?

• A. x = a/t²
• B. x = v + ¹/₂ at
• C. x = vt + ¹/₂ at²

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

where F is in [ kg·m / s²], m₁ and m₂ are masses in [ kg], and r is the distance in [ m] between them.

Determine the units of the Universal Constant G.

• A. kg·s² / m³
• B. m³ / (kg·s²)
• C. m / s²
• D. m³ / s²