Work by a 2D Push/Pull

by Patrick Ford
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Alright guys, so let's check this one out here. We've got this 10 kg suitcase, that's the mass of the box. And we're also pulling with some applied force of 50 newtons and we're pulling it across the floor. And it's gonna be pulled some distance of 20 m. But which way it's gonna be pulled even though we're pulling it upwards like this, what happens is it's just gonna move along the floor, right? It actually wouldn't fly up like this. So what's gonna happen is it's gonna travel some displacement here. Let me draw this a little bit bigger. And we know that this delta X. Or D. Doesn't matter which letter we use um is gonna be 20 m. So we want to do this first part here is we want to figure out the work that's done by the pool. Right? So your pole here. So the work that's done by F. Well remember the work that's done by any force is always gonna be F. Always that force times deke data. So the work of my F. Is gonna be F. D. Co sign of data. Remember that? This angle is the angle between F. And D. So let's check this out. I have the magnitude of the force. That's 50. I have my ex. Or my D. Right? It's just equal to 20 m. So I can just plug this into my work equation. This is gonna be 50 times one. Sorry, 20 times the cosine of the angle that's between these two vectors. We can take a look here, they're actually not parallel. Right? The force actually points this way but your displacement points along the floor like this. So this is the angle that we use between these vectors. And we know that this angle here from the problem is equal to the 37°. So we're just gonna plug this in 50 times 20 times the cosine of 37. You go ahead and work this out. What you're gonna get is you're gonna get 800 jewels. That's the answer to party. Alright so now in part B what we have to do is we're gonna have to decompose are forced into its components FX and Fy and then calculate the work done by each one of those components. So for part B we have the work is done by F. X. Remember what we said? The work done by any forces that force. So F. X. Times D. Cosine of theta. Now we actually don't have F. X. In our diagram just yet. So we're gonna do is we're gonna take this force and we're gonna break it down into its components like this. So this is gonna be my F. X. And I know to figure that out. I'm just gonna use F Times cosign of data here. So this is just gonna be 50 times the cosine of actually let me go ahead and write it somewhere else. So I've got that F. X. Is just equal F. Cosine Theta. So it's just equal to 50 times the cosine of 37. So if you go ahead and plug this in, you're gonna get 40. And the same thing is gonna happen with F. Y. Except now we're just gonna use sign, right? So F. X. F. Y. Is F. F. Sine theta. That's basically this vector like this, this has been my F. Y. And this is gonna be 50 times the sine of 37. So you're gonna get 30 here. So those are the components, it's just basically a 30 40 50 triangle. Which makes sense. So now we have to do is we have to calculate our work. So we're gonna take the magnitude of fx. Which we know this fx is really just equal to 40. Remember this is gonna be my 40. Now the distance that it travels is still going to be the 20. That doesn't change. But now, what's the cosine between the angle between? What's the angle between these two vectors? Well, now what happens is your FX. Remember actually points along the horizontal in the same direction as your displacement. So what's gonna happen here is you're gonna have 40 times 20. And the co signed between these two angles is actually gonna be between these two vectors is actually zero. Right? Because basically your fx points to the right and your distance also points to the right here. So what happens is you get a cosine of zero that we know that equals one if you plug this in, you're gonna get 800 jewels. So you get the same exact number. But and that's basically because the only four that's actually doing work on this object is this Fx here. That's what gives you the 800 jewels. One way you can also think about this is that we use F. D. Cosine theta When we calculated part A. And part B. What happens is you already kind of have a cosigner of data that is absorbed in this term here. Right? It's kind of wrapped up inside this Fx because really this is already F. Times the cosine of theta. So you're doing FD cosine theta here. But you're doing F. Times Cosine Theta times D. When you're doing it this way. So you're just gonna get the same exact number here. Alright, so to finish off the problem, we're gonna take a look at part C. This is gonna be the work that's done by F. Y. So this is gonna be my F. Y. Times D. Cosine of Theta. So we take a look we got F. Y. Remember this, that's just the 30 that we calculated over here times the displacement. Which is 20. But now what's the cosigner, what's the angle between these two vectors? Well basically with this F. Y. Is vertical. But we know our displacement actually points to the right like this. So the angle between these two is actually 90. And so therefore the cosine of 90 is gonna be zero and the work gets done is just basically gonna cancel out to zero jewels. That makes sense because there's a component of your force that's pulling vertically, but if you're traveling horizontally then there's no work that that force does. Alright, so that's it for this one. Guys, let me know if you have any questions.