# Mechanics Work

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Heya!

I’m quite stuck on this mechanics question and was wondering if anyone could help me out. I have annotated the diagram too with info but can’t work it out :/ I would really appreciate some help.

I’m quite stuck on this mechanics question and was wondering if anyone could help me out. I have annotated the diagram too with info but can’t work it out :/ I would really appreciate some help.

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#2

(Original post by

Heya!

I’m quite stuck on this mechanics question and was wondering if anyone could help me out. I have annotated the diagram too with info but can’t work it out :/ I would really appreciate some help.

**Lucia20613978**)Heya!

I’m quite stuck on this mechanics question and was wondering if anyone could help me out. I have annotated the diagram too with info but can’t work it out :/ I would really appreciate some help.

Can you set up any equations relating the forces involved?

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(Original post by

What are your thoughts so far?

Can you set up any equations relating the forces involved?

**davros**)What are your thoughts so far?

Can you set up any equations relating the forces involved?

And then, 0.9(5g)+0.2(2g)=1.8TQ

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#4

(Original post by

So, I was thinking that using Force= moment/distance might be something used here where 0.9(5g)+ 1.6(2g)=1.8TP

And then, 0.9(5g)+0.2(2g)=1.8TQ

**Lucia20613978**)So, I was thinking that using Force= moment/distance might be something used here where 0.9(5g)+ 1.6(2g)=1.8TP

And then, 0.9(5g)+0.2(2g)=1.8TQ

You will find it helps to organize your thoughts (and people marking your work) if you write down the principles you're using when you set up the equations.

So in one case you might say "the vertical forces balance each other, so resolving forces vertically gives..."

And for another case you know there is no net turning effect on the rod, so taking moments about a point will give you another equation relating TP and TQ.

It looks like you're trying to use 2 moment equations, which is fine also.

Can you say about which point you're taking moments in each case, and check your distances carefully?

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I am not fully sure about the distances.

I said that the 5Kg was at the Centre of mass so thats how I got the 0.9m and then from that for both sides I did 0.9(5g). Is this the case or would you ignore the 5Kg because it is equal on both sides?

0.2m is the distance from B to C so that was the 0.2(2g)

If the vertical forces balance each other, the resolving forces vertically would give zero as they are equal.

If there’s no net turning effect the moments will be equal for P and Q?

I’m really sorry that I have made this confusing, I just started the topic today and in my notes I wasn’t given anything on Tensions.

Usually, when doing tensions, I would use F=ma where F is equal to the downwards mass- the tension in the string.

However, the topic was on moments so I assumed using moments equations was the way to go :/

Thanks,

Lucia

I said that the 5Kg was at the Centre of mass so thats how I got the 0.9m and then from that for both sides I did 0.9(5g). Is this the case or would you ignore the 5Kg because it is equal on both sides?

0.2m is the distance from B to C so that was the 0.2(2g)

If the vertical forces balance each other, the resolving forces vertically would give zero as they are equal.

If there’s no net turning effect the moments will be equal for P and Q?

I’m really sorry that I have made this confusing, I just started the topic today and in my notes I wasn’t given anything on Tensions.

Usually, when doing tensions, I would use F=ma where F is equal to the downwards mass- the tension in the string.

However, the topic was on moments so I assumed using moments equations was the way to go :/

Thanks,

Lucia

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#6

(Original post by

I am not fully sure about the distances.

I said that the 5Kg was at the Centre of mass so thats how I got the 0.9m and then from that for both sides I did 0.9(5g). Is this the case or would you ignore the 5Kg because it is equal on both sides?

0.2m is the distance from B to C so that was the 0.2(2g)

If the vertical forces balance each other, the resolving forces vertically would give zero as they are equal.

If there’s no net turning effect the moments will be equal for P and Q?

I’m really sorry that I have made this confusing, I just started the topic today and in my notes I wasn’t given anything on Tensions.

Usually, when doing tensions, I would use F=ma where F is equal to the downwards mass- the tension in the string.

However, the topic was on moments so I assumed using moments equations was the way to go :/

Thanks,

Lucia

**Lucia20613978**)I am not fully sure about the distances.

I said that the 5Kg was at the Centre of mass so thats how I got the 0.9m and then from that for both sides I did 0.9(5g). Is this the case or would you ignore the 5Kg because it is equal on both sides?

0.2m is the distance from B to C so that was the 0.2(2g)

If the vertical forces balance each other, the resolving forces vertically would give zero as they are equal.

If there’s no net turning effect the moments will be equal for P and Q?

I’m really sorry that I have made this confusing, I just started the topic today and in my notes I wasn’t given anything on Tensions.

Usually, when doing tensions, I would use F=ma where F is equal to the downwards mass- the tension in the string.

However, the topic was on moments so I assumed using moments equations was the way to go :/

Thanks,

Lucia

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(Original post by

I thought post 3 was basically right, but your P and Q tensions should be swapped?

**mqb2766**)I thought post 3 was basically right, but your P and Q tensions should be swapped?

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#8

(Original post by

Really? I’m not too sure. This question is really a pain in the butt for me lol.

**Lucia20613978**)Really? I’m not too sure. This question is really a pain in the butt for me lol.

Last edited by mqb2766; 2 weeks ago

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(Original post by

Moment about A has TQ on the right. And moment about B has TP on the right. Just work out their values.

**mqb2766**)Moment about A has TQ on the right. And moment about B has TP on the right. Just work out their values.

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#10

(Original post by

So, does that mean that they’re right, I just got them mixed up?

**Lucia20613978**)So, does that mean that they’re right, I just got them mixed up?

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#11

**Lucia20613978**)

I am not fully sure about the distances.

I said that the 5Kg was at the Centre of mass so thats how I got the 0.9m and then from that for both sides I did 0.9(5g). Is this the case or would you ignore the 5Kg because it is equal on both sides?

0.2m is the distance from B to C so that was the 0.2(2g)

If the vertical forces balance each other, the resolving forces vertically would give zero as they are equal.

If there’s no net turning effect the moments will be equal for P and Q?

I’m really sorry that I have made this confusing, I just started the topic today and in my notes I wasn’t given anything on Tensions.

Usually, when doing tensions, I would use F=ma where F is equal to the downwards mass- the tension in the string.

However, the topic was on moments so I assumed using moments equations was the way to go :/

Thanks,

Lucia

**Lucia20613978**)

Really? I’m not too sure. This question is really a pain in the butt for me lol.

Let's take an example if you're working with moments. Take moments about the point A. Since TP acts through A its moment is 0.

In an anticlockwise sense you have the tension TQ pulling the rod upwards acting at a perpendicular distance of 1.8m from A, so the anticlockwise moment about A is 1.8TQ.

In a clockwise sense you have 2 weights: 5g acting at 0.9m from A and 2g acting at 1.6m from A (because the particle is 0.2m from the other end), so the clockwise moment is (5g x 0.9) + (2g x 1.6).

So our moments equation taking moments about A becomes:

1.8TQ = (5g x 0.9) + (2g x 1.6)

You can do the same for TP with the appropriate distances.

As a check, there is no net vertical force, so when you have worked out TP and TQ you should find that TP + TQ = 5g + 2g because the tensions acting vertically upwards are balanced out by the weights acting vertically downwards.

Does this help to clarify things?

Last edited by davros; 2 weeks ago

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(Original post by

Yes I think you have the 2 mixed up - I only had chance to have a quick look earlier but it looked like your distance was wrong in each case with the notation you're using.

Let's take an example if you're working with moments. Take moments about the point A. Since TP acts through A its moment is 0.

In an anticlockwise sense you have the tension TQ pulling the rod upwards acting at a perpendicular distance of 1.8m from A, so the anticlockwise moment about A is 1.8TQ.

In a clockwise sense you have 2 weights: 5g acting at 0.9m from A and 2g acting at 1.6m from A (because the particle is 0.2m from the other end), so the clockwise moment is (5g x 0.9) + (2g x 1.6).

So our moments equation taking moments about A becomes:

1.8TQ = (5g x 0.9) + (2g x 1.6)

You can do the same for TP with the appropriate distances.

As a check, there is no net vertical force, so when you have worked out TP and TQ you should find that TP + TQ = 5g + 2g because the tensions acting vertically upwards are balanced out by the weights acting vertically downwards.

Does this help to clarify things?

**davros**)Yes I think you have the 2 mixed up - I only had chance to have a quick look earlier but it looked like your distance was wrong in each case with the notation you're using.

Let's take an example if you're working with moments. Take moments about the point A. Since TP acts through A its moment is 0.

In an anticlockwise sense you have the tension TQ pulling the rod upwards acting at a perpendicular distance of 1.8m from A, so the anticlockwise moment about A is 1.8TQ.

In a clockwise sense you have 2 weights: 5g acting at 0.9m from A and 2g acting at 1.6m from A (because the particle is 0.2m from the other end), so the clockwise moment is (5g x 0.9) + (2g x 1.6).

So our moments equation taking moments about A becomes:

1.8TQ = (5g x 0.9) + (2g x 1.6)

You can do the same for TP with the appropriate distances.

As a check, there is no net vertical force, so when you have worked out TP and TQ you should find that TP + TQ = 5g + 2g because the tensions acting vertically upwards are balanced out by the weights acting vertically downwards.

Does this help to clarify things?

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(Original post by

Yes, you simply misspelled them. You could have called then TA and TB, so the moment about A would involve TB and vice versa.

**mqb2766**)Yes, you simply misspelled them. You could have called then TA and TB, so the moment about A would involve TB and vice versa.

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#14

(Original post by

Ahhh yes this helps tons! Thank you so much. I’m really really sorry I made this such trouble hahah. I really appreciate the help so much

**Lucia20613978**)Ahhh yes this helps tons! Thank you so much. I’m really really sorry I made this such trouble hahah. I really appreciate the help so much

Are you reading ahead of this in preparation for starting mechanics in the autumn?

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#15

When deciding at which point to take moments about, consider what you are trying to find and what info you have.

In this example, if you want to find TQ, then you would need to take moments about point A, since that will eliminate TP.

Similarly, if you want to find TP, then take moments about point B to eliminate TQ.

If you took moments about C, for example, you will have 2 unknowns: TP and TQ.

If it is in a state equilibrium, the upward forces are balanced by the downward forces (equal to each other). So you could take moments about either A or B to find TQ or TP, then calculate the total upward and downward forces and work out the remaining tension from that.

As mqb2766 said, when writing your moments calculations, you swapped the tension labels.

In this example, if you want to find TQ, then you would need to take moments about point A, since that will eliminate TP.

Similarly, if you want to find TP, then take moments about point B to eliminate TQ.

If you took moments about C, for example, you will have 2 unknowns: TP and TQ.

If it is in a state equilibrium, the upward forces are balanced by the downward forces (equal to each other). So you could take moments about either A or B to find TQ or TP, then calculate the total upward and downward forces and work out the remaining tension from that.

As mqb2766 said, when writing your moments calculations, you swapped the tension labels.

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(Original post by

That's OK

Are you reading ahead of this in preparation for starting mechanics in the autumn?

**davros**)That's OK

Are you reading ahead of this in preparation for starting mechanics in the autumn?

I know some of the stuff about tensions was in AS but It was one of the last things we did and I missed the last week before schools closed due to my mum being at such high risk. Hopefully, with more practise I will be able to get used to these ones haha.

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(Original post by

When deciding at which point to take moments about, consider what you are trying to find and what info you have.

In this example, if you want to find TQ, then you would need to take moments about point A, since that will eliminate TP.

Similarly, if you want to find TP, then take moments about point B to eliminate TQ.

If you took moments about C, for example, you will have 2 unknowns: TP and TQ.

If it is in a state equilibrium, the upward forces are balanced by the downward forces (equal to each other). So you could take moments about either A or B to find TQ or TP, then calculate the total upward and downward forces and work out the remaining tension from that.

As mqb2766 said, when writing your moments calculations, you swapped the tension labels.

**mathstutor24**)When deciding at which point to take moments about, consider what you are trying to find and what info you have.

In this example, if you want to find TQ, then you would need to take moments about point A, since that will eliminate TP.

Similarly, if you want to find TP, then take moments about point B to eliminate TQ.

If you took moments about C, for example, you will have 2 unknowns: TP and TQ.

If it is in a state equilibrium, the upward forces are balanced by the downward forces (equal to each other). So you could take moments about either A or B to find TQ or TP, then calculate the total upward and downward forces and work out the remaining tension from that.

As mqb2766 said, when writing your moments calculations, you swapped the tension labels.

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#18

Lucia20613978 - you're very welcome! There are lots of us here who are more than happy to help.

Out of interest, what books/resources are you using to prep for A2? Which exam board and course are you studying?

Out of interest, what books/resources are you using to prep for A2? Which exam board and course are you studying?

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(Original post by

Lucia20613978 - you're very welcome! There are lots of us here who are more than happy to help.

Out of interest, what books/resources are you using to prep for A2? Which exam board and course are you studying?

**mathstutor24**)Lucia20613978 - you're very welcome! There are lots of us here who are more than happy to help.

Out of interest, what books/resources are you using to prep for A2? Which exam board and course are you studying?

I’m glad to hear that

I am studying maths A2 for WJEC and I have been using resources our teacher gave us with the WJEC mathematics A2 revision book by Sophie Goldie & Rose Jewell. I have the WJEC maths book for Pure maths for that section and some resources on Statistics too which are also in the book I mentioned . Would you happen to know any good resources for this?

Lucia

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#20

(Original post by

Heya,

I’m glad to hear that

I am studying maths A2 for WJEC and I have been using resources our teacher gave us with the WJEC mathematics A2 revision book by Sophie Goldie & Rose Jewell. I have the WJEC maths book for Pure maths for that section and some resources on Statistics too which are also in the book I mentioned . Would you happen to know any good resources for this?

Lucia

**Lucia20613978**)Heya,

I’m glad to hear that

I am studying maths A2 for WJEC and I have been using resources our teacher gave us with the WJEC mathematics A2 revision book by Sophie Goldie & Rose Jewell. I have the WJEC maths book for Pure maths for that section and some resources on Statistics too which are also in the book I mentioned . Would you happen to know any good resources for this?

Lucia

I don't know how comfortable you are with the AS applied unit, or how WJEC teach the first year of mechanics. If you wanted to use the CGP books, it may be that you can go straight to the Y2 Applied book.

As I said, always feel free to ask anything here, or to private message for help. Personally, I believe it is really important for students to understand everything correctly, so if you don'tunderstand something, or want to check your understanding - ask away!

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