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#1

Hi, just come across this question, and I seem to be getting the answer wrong for the first part. It’s definitely got something to do with my diagram because in the markscheme they also have a ‘1Sin30’ acting in the anti-clockwise direction, but I don’t have a clue where this is. A similar question to think came up somewhere else and for that particular question, I showed there was a normal force acting on the wall, but someone mentioned to me there would be no normal reactions, as the the rope/string is Light, meaning it has no mass. So bearing that in mind, I’ve attempted this question, but got it wrong. If anybody knows where that other moment is acting from, and why, could you please let me know, thank you. It’s a shame, because I can do all these ladder questions In the book, but then in the exam, they always throw in a rope or something, which really confuses me because idk how the forces work on those.
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4 weeks ago
#2

Hi, just come across this question, and I seem to be getting the answer wrong for the first part. It’s definitely got something to do with my diagram because in the markscheme they also have a ‘1Sin30’ acting in the anti-clockwise direction, but I don’t have a clue where this is. A similar question to think came up somewhere else and for that particular question, I showed there was a normal force acting on the wall, but someone mentioned to me there would be no normal reactions, as the the rope/string is Light, meaning it has no mass. So bearing that in mind, I’ve attempted this question, but got it wrong. If anybody knows where that other moment is acting from, and why, could you please let me know, thank you. It’s a shame, because I can do all these ladder questions In the book, but then in the exam, they always throw in a rope or something, which really confuses me because idk how the forces work on those.
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#3
Ah nah it’s fine, just tried again and managed it. More of a case of me rushing the question I think. Thanks anyway.
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#4

Anyone know how to do part B ? I just wanna know if the force exerted by the wall on the beam at A is Normal force, acting in the A to D direction, that acts due to the beam being pressed against the wall ? I’m also stuck on the actual question, had a look at the markscheme and still don’t understand. When I resolve horizontally, I get x = Tsin30 (I’m calling x the apparent normal force acting at A), whereas in the markscheme, they do X=TSin30, and it basically looks like they ignore the X when doing whatever they do next.
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4 weeks ago
#5
Anyone know how to do part B ? I just wanna know if the force exerted by the wall on the beam at A is Normal force, acting in the A to D direction, that acts due to the beam being pressed against the wall ? I’m also stuck on the actual question, had a look at the markscheme and still don’t understand. When I resolve horizontally, I get x = Tsin30 (I’m calling x the apparent normal force acting at A), whereas in the markscheme, they do X=TSin30, and it basically looks like they ignore the X when doing whatever they do next.
From part (a) you should've found .

If you consider the horizontal forces, you should notice that is a force acting to the left ... but what is the force acting equally to the right? This is precisely the HORIZONTAL COMPONENT of the normal reaction at point A.

If you consider the fact that vertical forces must balance, you should notice that . So there must be another force acting directly down to help out the 20g force. This is precisely the VERTICAL COMPONENT of the normal reaction at point A.

Knowing the two components, you can determine the direction in which the normal reaction is acting.
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#6
I got 679 N for T part A. The markscheme did say that too, but idk it could be wrong maybe. For part B, this is what I got:

But I’m not sure at all how to determine the direction of force from that ?
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4 weeks ago
#7
I got 679 N for T part A. The markscheme did say that too, but idk it could be wrong maybe. For part B, this is what I got:

But I’m not sure at all how to determine the direction of force from that ?
Looks good.

x = Tsin(30) is your horizontal component.
y = Tcos(30) - 20g is your vertical component.

Just interpret this normal reaction as a force vector with these components. State the angle that it makes with the beam in order to describe the direction, or give your answer as a bearing.
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#8
Ah but this is where I’m getting confused. Because y has just been added as a constant, but x actually has some value, right ? As in, isn’t x supposed to be the normal force acting at A ? Idk I’m just really confused by this.
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4 weeks ago
#9
Ah but this is where I’m getting confused. Because y has just been added as a constant, but x actually has some value, right ? As in, isn’t x supposed to be the normal force acting at A ? Idk I’m just really confused by this.
The normal force is acting INTO the beam at A. This is your x force.

But the reaction force in this case is not the same as the normal force, because the reaction force is acting in a certain direction with the normal force as its horizontal component.
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#10

But is this force that I’ve circled, not also a force acting ? Isn’t that the force it’s talking about in part B ?
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4 weeks ago
#11
But is this force that I’ve circled, not also a force acting ? Isn’t that the force it’s talking about in part B ?
That is your x force. It is one part of the overall reaction force which is asked for in part B.
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#12
What’s the other part ? The upwards force ?
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4 weeks ago
#13
What’s the other part ? The upwards force ?
No, downwards. As I said, the vertical force from the rope is greater than the weight force, so in order for the beam to balance there must be another force which is also acting downwards; this is the second piece of the reaction force.
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#14
Oh wow thank you very much, I’ve managed to get the answer. Just a couple of quick questions -
1)Whenever it talks about Direction of Force, is that pretty much a giveaway, that we need to consider both x and y components ?
2) I’ve attempted pretty much all the questions in the Edexcel Textbook, yet have never seen a question where the vertical component isn’t equated, normally you just assume it is. So I was wondering, is it rare to do this, as in for instance with what we’ve done here : Resolve Vertically, and then use whatever’s left over as the downward force. So weird because it’s like the textbook purposefully miss out on these sorta stuff to mess you up for the exam lol.
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4 weeks ago
#15
Oh wow thank you very much, I’ve managed to get the answer. Just a couple of quick questions -
1)Whenever it talks about Direction of Force, is that pretty much a giveaway, that we need to consider both x and y components ?
2) I’ve attempted pretty much all the questions in the Edexcel Textbook, yet have never seen a question where the vertical component isn’t equated, normally you just assume it is. So I was wondering, is it rare to do this, as in for instance with what we’ve done here : Resolve Vertically, and then use whatever’s left over as the downward force. So weird because it’s like the textbook purposefully miss out on these sorta stuff to mess you up for the exam lol.
1) More or less. You should always consider that vertical and horizontal forces must balance. If they do not, then you either did something wrong in the calculations, or there is an extra force which is not obvious that is lurking about.

2) It is not something that often occurs, which means there is less exposure to students hence many of them struggle with this concept often. Q9 in the June 2018 Edexcel Mech & Stats paper has a part which revolves around this concept; and a lot of TSR threads were made asking for help about this in the past year. Have a look at it if you want more practice.
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#16
Yeah the 2018 paper 3 was horrible in general. Did decent in Pure, and then absolutely bottled paper 3 and got like 55/100, and I think I fully messed up that Question. I’m decent with the Ladder problems in the book, they just add so many more elements in the real thing, which makes it so much harder. I’ll give that question a go probably tomorrow and see how it goes, need to finish off this current question tonight ahah.
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