The Physics PHYA2 thread! 5th June 2013
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Original post by PickwickianGeek
Can someone please help on question 2 b) on page 98 of the AQA textbook? The answer is 6.2 but I got 0.8...
It really doesn't make sense to me.
It really doesn't make sense to me.
Okay, I'm going to make the assumption that you've drawn the diagram correctly. If you've done that then the question is 90% solved. All that's left for you to do is take moments about a sensible point and then use the principle of moments to find the value of W.
I have to stress that the key to this question is marking on your values correctly. After that, you need to consider what values you know and what you want to find out as this will help you decide where to take moments about.
Consider this; you're given two weights with known masses that lie on one side of a pivot that is placed at the centre. You have another mass of unknown weight placed on the other side of the pivot. You're not told the mass/weight of the ruler itself and so you essentially have two unknowns.
Can you think of a place to take moments about that will cancel out the unwanted unknown?
Original post by jack93o
the pivot point is at the centre of mass of the ruler, which is the middle of the ruler - the 50cm mark
if you draw the diagram of the ruler with all the masses, you should have the 3n and 2n weights on one side of the pivot (the 0-50cm half of the metre rule), and 80n would be on its own in the other side of the pivot (the 50-100cm half of the metre rule). The 3n and 2n have the same direction of rotation (for example clock-wise), and the w weight would have the opposite direction of rotation to oppose them (i.e anti-clockwise)
you assume the whole thing is in equilibrium and the ruler is in balance, you take moments about the pivot point (the middle of the ruler):
3*0.45 + 2*0.25 = w*0.3
1.85 = w*0.3
1.85/0.3 = w
6.1666.... = w
rounded to 2sf would give w as 6.2n
if you draw the diagram of the ruler with all the masses, you should have the 3n and 2n weights on one side of the pivot (the 0-50cm half of the metre rule), and 80n would be on its own in the other side of the pivot (the 50-100cm half of the metre rule). The 3n and 2n have the same direction of rotation (for example clock-wise), and the w weight would have the opposite direction of rotation to oppose them (i.e anti-clockwise)
you assume the whole thing is in equilibrium and the ruler is in balance, you take moments about the pivot point (the middle of the ruler):
3*0.45 + 2*0.25 = w*0.3
1.85 = w*0.3
1.85/0.3 = w
6.1666.... = w
rounded to 2sf would give w as 6.2n
thank you thank you!
Anyone able to explain question 4 on page 100 of the AQA textbook? The book is terrible at explaining the mechanics stuff.
(edited 10 years ago)
When doing moment questions and they don't tell you where to take the moments from, how do you work out where to take the moments from??
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Original post by Jimmy20002012
When doing moment questions and they don't tell you where to take the moments from, how do you work out where to take the moments from??
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You can take them from wherever you like, as long as you're clear about what you're doing. Generally you do it so you have one unknown however sometimes this is impossible and you substitute things in (however I could be getting confused here with maths M2, and you may not need to know this for this)
It tends to be one of the ends or a pivot you take moments about, as long as you're clear and say 'taking moments about point A' or M(A) : etc you should be alright
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Original post by PickwickianGeek
Anyone able to explain question 4 on page 100 of the AQA textbook? The book is terrible at explaining the mechanics stuff.
Ah yes, a very simple question with a little twist to cause confusion.
They key in this question is recognising that when the plank lifts off the brick that is at the end of the plank, the reaction force at that end then becomes zero. The question is now much simpler as you only have three forces; the weight of the plank; the reaction force at the pivot that is 1m away from the end and the unknown weight W.
Because the reaction force at the brick which is at the end of the plank, say A, is zero, that means that the reaction at the second support must be 150+W. However, that's irrelevant as you need to take moments about the second pivot which cancels out the reaction force.
That leaves you with:
M(P2);
1(150) = 1(W)
Therefore the weight of W is also 150 Newtons. Again, an accurate diagram is basically 90% of the question, as having a visual makes it much easier to break down the question into steps.
Original post by lebron_23
ah yes, a very simple question with a little twist to cause confusion.
They key in this question is recognising that when the plank lifts off the brick that is at the end of the plank, the reaction force at that end then becomes zero. The question is now much simpler as you only have three forces; the weight of the plank; the reaction force at the pivot that is 1m away from the end and the unknown weight w.
Because the reaction force at the brick which is at the end of the plank, say a, is zero, that means that the reaction at the second support must be 150+w. However, that's irrelevant as you need to take moments about the second pivot which cancels out the reaction force.
That leaves you with:
M(p2);
1(150) = 1(w)
therefore the weight of w is also 150 newtons. Again, an accurate diagram is basically 90% of the question, as having a visual makes it much easier to break down the question into steps.
They key in this question is recognising that when the plank lifts off the brick that is at the end of the plank, the reaction force at that end then becomes zero. The question is now much simpler as you only have three forces; the weight of the plank; the reaction force at the pivot that is 1m away from the end and the unknown weight w.
Because the reaction force at the brick which is at the end of the plank, say a, is zero, that means that the reaction at the second support must be 150+w. However, that's irrelevant as you need to take moments about the second pivot which cancels out the reaction force.
That leaves you with:
M(p2);
1(150) = 1(w)
therefore the weight of w is also 150 newtons. Again, an accurate diagram is basically 90% of the question, as having a visual makes it much easier to break down the question into steps.
Question 3 on page 100 of the AQA textbook anyone?
Original post by PickwickianGeek
Question 3 on page 100 of the AQA textbook anyone?
Lol I guess it's my go again
Okay so this question instantly screams moments about two separate points, as you've got two forces you want to find. Firstly, as always, draw out a good neat diagram indicating all of the forces and distances.
Once you've got that that, you need to decide on a sensible place to take moments. In this case it's rather obvious, as you have to take moments about the first support and then the second support as you can't work with two unknowns.
So your first equation would look something like this;
M(R1):
4.5(60kN) + 7.5(1200kN) = 15(R2)
Where R1 and R2 are the two support forces respectively.
What you've got to be careful of here is getting your distances right. He midpoint of the bridge is 8.5 meters, but it's only 7.5 meters away from R1. Similarly, the truck is at 5.5 meters on the bridge, buts its distance of 4.5 meters away from R1.
So you go ahead with the equation and you get something like this;
270kN + 9000kN = 15R1
9270kN = 15R1
R1 = 618kN
Now I'm not sure here but I think the book have rounded to 2sf.
Okay so once you've got that you don't really have to make a second equation unless you want to have fun. All you need to recall is the definition of equilibrium. (Not really, just realise that the bridge isn't flying off anywhere so up must equal down)
You know that the total of the reaction forces must equal the total forces acting downwards and so you can basically do 1260kN - 618kN to give you the support force at R2.
Really hope that made sense
(edited 10 years ago)
Original post by lebron_23
Lol I guess it's my go again
Okay so this question instantly screams moments about two separate points, as you've got two forces you want to find. Firstly, as always, draw out a good neat diagram indicating all of the forces and distances.
Once you've got that that, you need to decide on a sensible place to take moments. In this case it's rather obvious, as you have to take moments about the first support and then the second support as you can't work with two unknowns.
So your first equation would look something like this;
M(R1):
4.5(60kN) + 7.5(1200kN) = 15(R2)
Where R1 and R2 are the two support forces respectively.
What you've got to be careful of here is getting your distances right. He midpoint of the bridge is 8.5 meters, but it's only 7.5 meters away from R1. Similarly, the truck is at 5.5 meters on the bridge, buts its distance of 4.5 meters away from R1.
So you go ahead with the equation and you get something like this;
270kN + 9000kN = 15R1
9270kN = 15R1
R1 = 618kN
Now I'm not sure here but I think the book have rounded to 2sf.
Okay so once you've got that you don't really have to make a second equation unless you want to have fun. All you need to recall is the definition of equilibrium. (Not really, just realise that the bridge isn't flying off anywhere so up must equal down)
You know that the total of the reaction forces must equal the total forces acting downwards and so you can basically do 1260kN - 618kN to give you the support force at R2.
Really hope that made sense
Okay so this question instantly screams moments about two separate points, as you've got two forces you want to find. Firstly, as always, draw out a good neat diagram indicating all of the forces and distances.
Once you've got that that, you need to decide on a sensible place to take moments. In this case it's rather obvious, as you have to take moments about the first support and then the second support as you can't work with two unknowns.
So your first equation would look something like this;
M(R1):
4.5(60kN) + 7.5(1200kN) = 15(R2)
Where R1 and R2 are the two support forces respectively.
What you've got to be careful of here is getting your distances right. He midpoint of the bridge is 8.5 meters, but it's only 7.5 meters away from R1. Similarly, the truck is at 5.5 meters on the bridge, buts its distance of 4.5 meters away from R1.
So you go ahead with the equation and you get something like this;
270kN + 9000kN = 15R1
9270kN = 15R1
R1 = 618kN
Now I'm not sure here but I think the book have rounded to 2sf.
Okay so once you've got that you don't really have to make a second equation unless you want to have fun. All you need to recall is the definition of equilibrium. (Not really, just realise that the bridge isn't flying off anywhere so up must equal down)
You know that the total of the reaction forces must equal the total forces acting downwards and so you can basically do 1260kN - 618kN to give you the support force at R2.
Really hope that made sense
Yeah that made sense! Thanks again XD You seem quite well prepared for this exam haha
Original post by PickwickianGeek
Yeah that made sense! Thanks again XD You seem quite well prepared for this exam haha
You're welcome. And I wouldn't say I'm really that prepared, moments is just my strong point, as its basically the same as M1 just much easier. My waves knowledge is also fairly weak so I'm only as prepared as the next guy!
Original post by lebron_23
You're welcome. And I wouldn't say I'm really that prepared, moments is just my strong point, as its basically the same as M1 just much easier. My waves knowledge is also fairly weak so I'm only as prepared as the next guy!
What is it that you struggle on waves?
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In the exam will they always tell you where to take moments from?
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Original post by Jimmy20002012
Not necessarily , but just take it from an end or a pivot and you should be alright as long as you're clear about it.
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Original post by SortYourLife
Not necessarily , but just take it from an end or a pivot and you should be alright as long as you're clear about it.
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Well they have in every past paper I have seen, they told us where to take the moments from,n hopefully in this one they will
otherwise I just get a bit confused 
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Original post by Jimmy20002012
Ah, it's mainly down to my inability to write things properly. I understand the majority of it, it's just trying to answer questions that gets me
.And in terms of pivots and moments, the best thing to do if you're not told where to take moments about is first to look at how many unknowns you have. The less unknowns the easier it is to decide where to take moments about (i.e question 4 on page 100). If there's more than one unknown then you immediately need to start thinking about finding moments from two different places or, as it was the case in question 3 on page 100, just apply the law of conservation of energy.
There are a few things to remember, however. If the plank/rod is about to tilt about a pivot then the reaction force at that pivot will be zero. That's key, because that tells you that the weight of the beams and whatever is on it is balanced only by the other pivot(s). It's also worth noting when taking moments, it's always to important to sort out which way is your clockwise and which is you anticlockwise. Simple, I know, but it's a very easy mistake to make. Don't always assume that the forces acting down are all acting in the same direction as if the force is on a different side of the pivot the chances are it's acting in the other direction. I find it helps to put your pen on he pivot and twist it in the direction of the forces, then you know what's going where
Original post by lebron_23
Ah, it's mainly down to my inability to write things properly. I understand the majority of it, it's just trying to answer questions that gets me .
And in terms of pivots and moments, the best thing to do if you're not told where to take moments about is first to look at how many unknowns you have. The less unknowns the easier it is to decide where to take moments about (i.e question 4 on page 100). If there's more than one unknown then you immediately need to start thinking about finding moments from two different places or, as it was the case in question 3 on page 100, just apply the law of conservation of energy.
There are a few things to remember, however. If the plank/rod is about to tilt about a pivot then the reaction force at that pivot will be zero. That's key, because that tells you that the weight of the beams and whatever is on it is balanced only by the other pivot(s). It's also worth noting when taking moments, it's always to important to sort out which way is your clockwise and which is you anticlockwise. Simple, I know, but it's a very easy mistake to make. Don't always assume that the forces acting down are all acting in the same direction as if the force is on a different side of the pivot the chances are it's acting in the other direction. I find it helps to put your pen on he pivot and twist it in the direction of the forces, then you know what's going where
.And in terms of pivots and moments, the best thing to do if you're not told where to take moments about is first to look at how many unknowns you have. The less unknowns the easier it is to decide where to take moments about (i.e question 4 on page 100). If there's more than one unknown then you immediately need to start thinking about finding moments from two different places or, as it was the case in question 3 on page 100, just apply the law of conservation of energy.
There are a few things to remember, however. If the plank/rod is about to tilt about a pivot then the reaction force at that pivot will be zero. That's key, because that tells you that the weight of the beams and whatever is on it is balanced only by the other pivot(s). It's also worth noting when taking moments, it's always to important to sort out which way is your clockwise and which is you anticlockwise. Simple, I know, but it's a very easy mistake to make. Don't always assume that the forces acting down are all acting in the same direction as if the force is on a different side of the pivot the chances are it's acting in the other direction. I find it helps to put your pen on he pivot and twist it in the direction of the forces, then you know what's going where
You seem to be pretty prepared for this exam, could you help me with this moment problem:
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Original post by Jimmy20002012
You seem to be pretty prepared for this exam, could you help me with this moment problem:
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Okay here goes again..
The first part of the question simply want you to multiply the weight of he sprinter by the distance of her weight from her fingertips. This giving;
0.26*520 = 135.2 N
The second part is just a matter of taking moments about F and then employing conservation of energy.
When you take moments about F, your clockwise moment is her weight (your answer to part a) and the anticlockwise moment is forces X and Y multiplied by their respective distances from F. This gives you
0.21(W) = 0.41(Y) + 0.63(X)
Which gives
0.26(520) = 0.41(180) + 0.63(X)
Rearranging to find X you get
(135.2-73.8)/0.63 = X
Therefore X = 97.5 N
Original post by lebron_23
Okay here goes again..
The first part of the question simply want you to multiply the weight of he sprinter by the distance of her weight from her fingertips. This giving;
0.26*520 = 135.2 N
The second part is just a matter of taking moments about F and then employing conservation of energy.
When you take moments about F, your clockwise moment is her weight (your answer to part a) and the anticlockwise moment is forces X and Y multiplied by their respective distances from F. This gives you
0.21(W) = 0.41(Y) + 0.63(X)
Which gives
0.26(520) = 0.41(180) + 0.63(X)
Rearranging to find X you get
(135.2-73.8)/0.63 = X
Therefore X = 97.5 N
The first part of the question simply want you to multiply the weight of he sprinter by the distance of her weight from her fingertips. This giving;
0.26*520 = 135.2 N
The second part is just a matter of taking moments about F and then employing conservation of energy.
When you take moments about F, your clockwise moment is her weight (your answer to part a) and the anticlockwise moment is forces X and Y multiplied by their respective distances from F. This gives you
0.21(W) = 0.41(Y) + 0.63(X)
Which gives
0.26(520) = 0.41(180) + 0.63(X)
Rearranging to find X you get
(135.2-73.8)/0.63 = X
Therefore X = 97.5 N
Thanks, that really helped? Sorry for all these questions, its just my teacher didn't really go over moments with me. So in every situation the arrows pointing in the same direction either have a clockwise or anticlockwise moment, and the other arrow(s) would have an anticlockwise moment. This may sound really sound but how do you know why one is clockwise and which one is anticlockwise?
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Original post by Jimmy20002012
Thanks, that really helped? Sorry for all these questions, its just my teacher didn't really go over moments with me. So in every situation the arrows pointing in the same direction either have a clockwise or anticlockwise moment, and the other arrow(s) would have an anticlockwise moment. This may sound really sound but how do you know why one is clockwise and which one is anticlockwise?
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Okay this is going to be a little difficult to put into context, but imagine a seesaw. Seesaws have a pivot at the centre, which is the point about which the seesaw rotates. When a child site at one end, he makes the seesaw move in a direction, either clockwise or anticlockwise. If another child were to sit on the other end then he would cause the seesaw to rotate in the other direction. Now, how much the seesaw rotates would depend on the masses of the children and where they say in relation to rate pivot, but the point is that a force applied to one side of a pivot will cause it to move in that direction.
Now imagine two pivots, one at the centre and one at one end of the seesaw. The pivots both exert reaction forces but when taking moments about the centre, the centre pivot's reaction force cancels out. So you now have one force acting down on one side of the pivot and a force acting up as well as a force acting down on the other side of the pivot. Now it's rather obvious that as the force that is on it's own moves down, the second force (which is the reaction force of the second pivot) moves upwards, hence they act in the same direction.
I appreciate that my explanation may have been terrible so I've attached two diagrams that might just help out. Sorry for the bad handwriting though.
Also, they may be upside down
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