CAPE Physics - May 2026 Unit 1 Question 3 (d) (Terry David)

Added: 2026-05-13

[00:00:16]Hey guys. Um, so I hope you guys can hear me. All right. So again, this is Terry David here and I got some feedback regarding the Cape Physics Unit One. I heard a lot of students had trouble with um that third question in the paper involving Young's modulus. So a lot of students weren't sure how to work the problem. So um I'm going to basically work that that only that part of the problem for you guys. Um just to show you or give you an idea as to how we're going to solve that problem. Okay. So first of all uh just as a reminder right I am the sole author of the official Cape Physics unit one and unit 2 study guides. All right you may see other names on the book but I am the person who wrote the entire text for these two books. All right. So let's have a look at this uh question here. So you have a bench. All right. You have a bench and it has two legs which differ in length by d is equal to 050 mm causing the bench to wobble slightly.

[00:01:26]The shorter leg is 1 m in length. The bench is loaded with a mass of 290 kg in such a way that both legs are compressed and the table no longer wobbles. The legs are wooden cylinders with cross-sectional area a equal to 1.0 cm squared and the young's modulus for the wood is 1.3x 10 the 1.3 by 10 the 10 newton per meter squared. The shorter leg is compressed by an amount delta L short and the longer leg is compressed by an amount delta L long. All right.

[00:02:07]Now the question is a bit wordy. Okay.

[00:02:10]So you have to kind of try to understand what is happening here. So basically we have um we have two legs, right? We have two legs. So basically you have a bench, right? So let's say this is the top of the bench, right? Okay. Now trying to visualize what's happening here will help you work out the problem. So in the question they said the bench will wobble. So the shorter leg has a length of 1 m. Okay.

[00:02:36]So we have one leg here. All right. And I'm just going to assign that a length of 1 m. Okay. So this distance here from here to here. All right. Let's call that 1 meter.

[00:02:52]So that's 1.0 m. All right. And then we have um Right. So that's a shorter length. We have a longer length now.

[00:03:02]Right now I'm just going to exaggerate my diagram a little bit. Okay.

[00:03:09]Because in the question here, we have some um some things in millimeter. So I'm just exaggerating it here to make it make sense. Okay. All right. Um so the shorter length is 1.0. The bench has two legs which differ. So again I'm saying that I'm exaggerating the length, right?

[00:03:25]I'm exaggerating the length. So the distance between the two lengths. All right, this is just way to see what's happening here.

[00:03:32]The distance between these two lengths here is given by uh 0.5 um millimeters. So what I'm going to do since the question has varying units here because we have the length in meters, we have the um the difference in millimeters and we even have the area in cm squ. So we need to be consistent when it comes to the units. So in this case here um this here let's convert this to meters one time.

[00:03:59]So that's going to be 0 0 right um 0 five m. So that's the difference in the length between the two legs of this um bench. Now what else do we have here? Um all right so the bench is loaded with a mass of 290 kg. So basically we are placing a mass on top of this here.

[00:04:23]Right? You have a mass on top of this and this mass here is 290 kg. So before I even work the problem I need to visualize what is actually taking place here. So that's 290 kg on top of this thing. All right. Um yeah so what I'm going to do now let's see so we have the area in so one sec I just want to read the question properly okay so I will assume that they gave us g as 9.81 right we'll take g as 9.81 81 because I'll have to calculate the weight of this thing here in all right so they give us the cross-sectional area right they give us the cross-sectional area now the first thing is that what they said is that when you load the bench what is happening here is that the legs right um they compressed in such a way so that the table no longer wobbles. So what's happening here is that the long side is going to be compressed the short side is also going to be compressed and then eventually they're going to reach the same length. That's that's basically what's going to happen. Now, if you read the question here, they said the shorter leg is compressed by an amount delta L short, right? So, the short leg, which is the one on the right hand side here, this one is going to compress. So, let's let's draw a line here to represent where the two legs end up being the same length. So, what's going to happen here is that we're going to get some compression on the shorter side, right?

[00:05:59]And that compression here, they're calling it delta L short. All right. So, this here is delta L. I'm just going to call it S, right?

[00:06:08]Instead of writing out the word short.

[00:06:11]In the case of uh the longer side, the longer side is also going to be compressed. And that compressed amount here is going to be called delta L.

[00:06:21]Let's call it long, right? Let's call it delta L L, right? Um, what else do we know here? And oh by the way in this case here if we know that the difference between the lengths is 0.005 0005.

[00:06:38]We also know, right? We also know that um yeah, we also know that the length of the bigger one, right? And the length of the bigger one here, this five, right?

[00:06:58]That's extra zero inside. Yeah, this should be five. Right now, the length of the long side here, right?

[00:07:08]This um trying to figure out where to write this. This length here, right, of the longer um leg of the table, right, is actually going to be 1.005 m. Okay? Because they give us the difference. So, this here represents the length of the long leg and 1 meter represents the leg length of the short leg. Now the first part of the question says to write an expression uh to show the relationship between the short and the long. So the long right delta L right L right so that's the compression for the um the longer leg that should be equal to your compression for the short leg plus the difference which is 0 05 right so this is our answer here for the first part right is the answer for the first part okay now the next thing. Now that's part one. For part two, now assume that the force acting on the short leg is f short. Deduce the equation, right? Deduce the equation to show the relationship between Fshort and this. So first of all in the question they gave us a Young's modulus for the um the material. And let's remember Young's modulus E is given by stress over strain. So that is actually F / A / E on L. All right, that's the formula that we use for Young's modulus. So we can say that E is equal to F on A multipli by L on E. Right? So in this case here, you want to work out the force here. So if I rearrange this formula, I'm going to get that force is equal to the Young's modulus multiplied by the cross-sectional area multiplied by the extension all over the length.

[00:09:08]Okay. Now in the question they gave us the Young's modulus. The Young's modulus was 1.3x 10 the 10. So E was equal to 1.3x 10 the 10.

[00:09:21]Um that's NMUS 2, right? That's Young's modulus. But they also gave us the area.

[00:09:26]They said the area the cross-sectional area is 1 cm squ. So the cross-sectional area A is 1 cm squared. But if I convert that into me squar that's going to be 1 by 10 the minus4 m squ. Okay. So the first part of the question they want an expression for F um F short right. So the force that is acting on the short. So remember this is our rearrange formula here for f. So this is going to be e a by the extension divided by l. So in this case here the young's modulus is going to be 1.3 by 10.

[00:10:09]All right multiply by the cross-sectional area which is 1x 10us 4 multiply by e. Now in this case here it is not an it's not an it's not an extension but it's a compression. So in this case here the compression right the compression is going to be um this is for the short side right so the compression is going to be delta ls right so that's the compression there and you're going to divide that by the original length of the leg so this is for the short leg so the short leg has a length of one so when I work this out here let's see what we get this will be 1.3 1.3 3 by 10 the 10 * 1 by 10 the - 4. So this is going to give us 1.3 by 10 to the 6 delta ls. Right? So that's the answer for part two. That's the formula here or the equation that shows the relationship between FS, right? That's the force on the short leg in relation to the compression on the short leg. So that's part two.

[00:11:26]Um for this one here, now we want to deduce the equation to show the relationship between F long and delta long. Okay, so we're going to use the same the same formula again, right? So the same formula here would have been um F. So F long, right, is equal to E A E over L. So that's e a e over l. So in this case here our young's modulus would have been 1.3x 10 multiply by the area which is 1x 10us4 multipli by the compression right in this case here it'll be delta l long right all over the length. Now remember the length here we said previously that the total length would have been 1.005 005 right so 1 05 all right so therefore f L is equal to so that's going to be 1.3x 10 to the 10 * 1 by 10 theus 4 right divided that now by 1.12 35 right and I'm getting 1 299 by 10 to the 6 delta L l okay so that's what we have here right um in this case here now that's one mark because they expect us to use the same expression we had from this one here now the next part here we want to calculate the values right you want to calculate the values of deltal L short and delta L long. So first of all, if we go back to the question here, right, we have this this bench that is being loaded with a mass of 290 kg. So the first thing I would do in a question like this here, I need to work out what is the weight, right? What's the force that is acting on this thing here? So weight is equal to m * g. So the mass is uh 290.

[00:13:37]So we're going to say 290 * 9.81. That's what I'm using as G. All right. I'm not sure what they gave them in the exam, but 2190 multiply by 9.81 and you're going to get 2845.

[00:13:54]All right. So 2845.

[00:13:57]All right. So now what do we know about these two things here? So the thing is this force here that is acting downwards which is the weight of this mass here.

[00:14:09]it is being applied onto the two legs.

[00:14:13]So what we can say is that the force on each leg that is compressing each leg when you add up those two forces it should equal to the weight of the um the block that is on top of it. Okay. So in this case here what we saying that's our expressions were L right? So f the force on the short leg plus the force on the long leg should equal to 2845.

[00:14:45]That is what's supposed to happen.

[00:14:46]Right? Now this is plenty marks here.

[00:14:48]There's seven marks here. So the force on the short leg we had an expression for that. Uh where was that? 1.36 by 10 the 6. So we have 1.36 by 10 6. All right. And that is delta S right multiply by delta S. So this is delta LS. Okay. Good.

[00:15:16]Right. Plus we're going to add the force um on the long leg which would have been 1.299 by 10 to the 6. So this is 1.299 by 1026.

[00:15:31]Delta L L right one second let me see something here I wrote something wrong this is 1.3 by 10 to the6 all right so when I add this up I'm going to get 2845 okay but if we go back to the beginning of the question right in the beginning of the question there was a relationship between the long side and the short side. So that relationship was delta LL.

[00:16:06]So this is what we had from before.

[00:16:08]Delta L is equal to delta L short plus that was 05.

[00:16:21]Right? So this here is from part one. So what I'm going to do, I'm going to use that expression and I'm going to use what I have now. That's this. And we're going to have to make a substitution, right? We're going to have to make a substitution. So what I'm going to do, I'm going to replace I'm going to say 1.3 by 10 6 delta LS, right? Plus 1.299 by 10 6, right? Now delta L, we're going to put this as delta LS plus 0.005.

[00:16:57]All right. and that here is going to be equal to 2845.

[00:17:03]So basically what we going to do here we have an equation that we have to try to solve. So let me just make it clear that I have substituted. So this is what you had from part one and you're going to substitute that into this here. So then what's going to happen now is we're going to say 1.3 by 10 6 delta LS plus let's remove that bracket there. So that'll be 1.299 by 106 delta LS plus. So this is 1.299 299 by 1036 multiplied by.123 right and I'm getting 64 649.5 is equal to 2845 right so on the left hand side here when I work this out I'm going to get 1.3 by 10 the 6 + 1 299 by 1036, right? And we're going to get So, I'm just kind of speeding up here a little bit, right? 2.599 uh delta LS is equal to 2845 minus 649.5.

[00:18:34]So therefore, delta LS is equal to So I'm just going to do a kind of quick thing for you here, right? So 2845 2845 - 649.5 all over.

[00:18:58]So this is 2.599 by 106.

[00:19:08]So this will be 2.599 by 106.

[00:19:13]Okay. So when I work out this here, I'm going to get 2845 - 649.5 / 2.599 by 1036 right I am getting 8 1 second 28 5US 649.5 / 2.599.

[00:20:10]All right. So, this is what I have here.

[00:20:11]I have 8 point One, two, one, two, one.

[00:20:18]So that's 8.45, right? By 10us 4, right? And this is a um change in length here, right? So that's meters. Okay. So that's your short. So that's L short I just worked out there to work out L um the long one.

[00:20:38]Now that's delta L L.

[00:20:40]So all we need to do is to take that answer we just got which is 8.45 by 10 theus 4 right and you're going to add it to this. So that's plus 0.005 0 05 in which case we're going to get plus right and I'm getting 1 3 4 by 10 the minus 3 all right so the two values that they wanted for seven marks here they want this the change in length of short side and the change in length sorry change in length of the long side and the change in length of the short side. So that's your answer there for that seven marks that they're talking about there. Okay.

[00:21:31]And then the final part of the question they want you to actually work out what is f short and fong. So fs we have to go back to our expression for fs. Uh that would have been here. So that's 1.3 by 10 the 6. So there's a 1.3 by 10 6 multiplied by um delta ls which is 8.45 by 10us 4 8.45 by 10us 4 right and let's work that out. So that's 1.3 10 6 * 8.45 45 by 10 theus 4 right and I'm getting about 1099 around there newtons so that's the force on the short side the force on the longer side will be that's 1.299 so this is 1.299 299 by 10 6 multiplied by your long side which is 1.3 by 10us 4 right hold on right so this is 1.3 by 10 - All right, hold on. Let me just check something here. This is 8.45 by 10 to the - 4, right? This should have been minus 3.

[00:23:32]Right? This is minus three.

[00:23:35]Right? So this would have been right. So 1.299 2996.

[00:24:16]Right? And when I work this out here, I'm getting 1741 Newtons. Right? So for those of you who are doing uh Cape Physics, right? And you all had this exam today. All right. Um right. Um I hope this solution helps you guys.

[00:24:37]Right. Uh like I said, a lot of students said it was uh tricky, right? Um so that's it for me here, guys. All right.