# M1- Resolving Forces Question

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A glider of weight 2500N is being towed with constant speed by a four-wheel drive vehicle. The tow-rope is inclined at 25 degrees to the horizontal and the glider is inclined at 30 degrees to the horizontal as shown in the diagram. The air resistance has magnitude R N and the lift has magnitude L N; the directions in which the act are shown in the diagram. Calculate the values of R and L, given that the tension in the rope is 3000N.

I have been stuck on this for awhile, I tried resolving vertically and horizontally but I got 2 simultaneous equations that were very long and wehn I eventually solved them I got the wrong answer. My next thought was maybe I should resolve paralel/perpendicular to the plane but there are 2 planes by the look of it. So I am quite confused about how to approach this

Thanks for help in advance!

I have been stuck on this for awhile, I tried resolving vertically and horizontally but I got 2 simultaneous equations that were very long and wehn I eventually solved them I got the wrong answer. My next thought was maybe I should resolve paralel/perpendicular to the plane but there are 2 planes by the look of it. So I am quite confused about how to approach this

Thanks for help in advance!

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#2

Try resolving parallel and perpendicular to the plane of the glider's flight; it should make the calculations much easier. It should have been possible resolving vertical and horizontal too, but it will have been much more complicated so you may well have made a mistake.

Do you have access to the answers?

Do you have access to the answers?

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(Original post by

Try resolving parallel and perpendicular to the plane of the glider's flight; it should make the calculations much easier. It should have been possible resolving vertical and horizontal too, but it will have been much more complicated so you may well have made a mistake.

Do you have access to the answers?

**Implication**)Try resolving parallel and perpendicular to the plane of the glider's flight; it should make the calculations much easier. It should have been possible resolving vertical and horizontal too, but it will have been much more complicated so you may well have made a mistake.

Do you have access to the answers?

I will attempt try parallel and perpendicular now :P

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#4

Okay good; those are the answers I got resolving parallel and perpendicular haha! Let us know if you get stuck resolving this way

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(Original post by

Okay good; those are the answers I got resolving parallel and perpendicular haha! Let us know if you get stuck resolving this way

**Implication**)Okay good; those are the answers I got resolving parallel and perpendicular haha! Let us know if you get stuck resolving this way

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#6

Bear with me and I'll edit this post and draw you a diagram I think the bend you see is just an issue with the diagram; ignore it. There is a force of 3000N acting on the centre of mass of the glider (classic M1 assumption), which is at 25 degrees to the horizontal. That's all you need!

From this diagram, you should be able to work out the angle on the right - the one between 3000N and its component parallel to the plane. From this you can then see that the parallel component is the cosine of that angle and the perpendicular component is the sine of that angle.

From this diagram, you should be able to work out the angle on the right - the one between 3000N and its component parallel to the plane. From this you can then see that the parallel component is the cosine of that angle and the perpendicular component is the sine of that angle.

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(Original post by

Bear with me and I'll edit this post and draw you a diagram I think the bend you see is just an issue with the diagram; ignore it. There is a force of 3000N acting on the centre of mass of the glider (classic M1 assumption), which is at 25 degrees to the horizontal. That's all you need!

From this diagram, you should be able to work out the angle on the right - the one between 3000N and its component parallel to the plane. From this you can then see that the parallel component is the cosine of that angle and the perpendicular component is the sine of that angle.

**Implication**)Bear with me and I'll edit this post and draw you a diagram I think the bend you see is just an issue with the diagram; ignore it. There is a force of 3000N acting on the centre of mass of the glider (classic M1 assumption), which is at 25 degrees to the horizontal. That's all you need!

From this diagram, you should be able to work out the angle on the right - the one between 3000N and its component parallel to the plane. From this you can then see that the parallel component is the cosine of that angle and the perpendicular component is the sine of that angle.

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Ive tried it but Im not getting something because this doesnt seem right to me...

R(Parallel): 3000sin(25) - R - 250sin(30) = 0

R(Perpen): L - 250cos(30) - (perpen component of the 3000N which I cant figure out) = 0

But neither of these will give the right answer

R(Parallel): 3000sin(25) - R - 250sin(30) = 0

R(Perpen): L - 250cos(30) - (perpen component of the 3000N which I cant figure out) = 0

But neither of these will give the right answer

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#9

(Original post by

I'll try to use this, but surely with the diagram like that the vertical component=the parallel one as the glider is at the tip of the triangle. Ive never used a triangle like this in a calculation, I'll try to understand what I need for each component :P

**Genesis2703**)I'll try to use this, but surely with the diagram like that the vertical component=the parallel one as the glider is at the tip of the triangle. Ive never used a triangle like this in a calculation, I'll try to understand what I need for each component :P

When you resolve a vector (such as a force) into parallel and perpendicular components, that is the sort of diagram that is always used. For most people a diagram is unnecessary most of the time, as the angle and trigonometric functions to be used can be seen easily, but the diagram is still

*implied*. I suspect vector notation or maybe matrices could be used, but the equation for calcuating components of forces id derived from that kind of triangle...

It comes from the "triangle law" of vectors - if you have vectors

**a**,

**b**and

**c**that form a closed triangle as in the diagram below, then

**a + b**=

**c**.

So, if we have a force represented by the vector

**c**, then we can stick in two other forces represented by

**b**and

**c**, forming a closed triangle, and know that the vector addition of

**a**and

**b**gives

**c**. If we want, we can draw the triangle such that the angle between

**a**and

**b**is 90 degrees, and we can see that these are the components of

**c**parallel and perpendicular to a given plane. Then, given

**c**and the angle between

**a**and

**c**or

**b**and

**c**, we can use trigonometry to calculate

**a**and

**b**- the parallel and perpendicular components. This is what we do when we resolve forces.

Wrt your issue with the positioning of the glider; this does

**not**matter for the triangle. The triangle I showed you is a vector triangle or triangle of forces; it is

**not**a representation of 3-D (or even 2-D) space. The location of the glider is completely irrelevant to the horizontal and vertical components of the force.

If we were getting more complicated than M1, the place at which the force acts

*would*affect the overall force but, at this stage, one of the assumptions you are expected to make is that all the forces act on the glider's centre of mass. Because of this, "where" the force acts is irrelevant to the question

Sorry if I've confused you even more; this sort of thing is

*very*hard to explain over the internet!

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#10

(Original post by

Ive tried it but Im not getting something because this doesnt seem right to me...

R(Parallel): 3000sin(25) - R - 250sin(30) = 0

R(Perpen): L - 250cos(30) - (perpen component of the 3000N which I cant figure out) = 0

But neither of these will give the right answer

**Genesis2703**)Ive tried it but Im not getting something because this doesnt seem right to me...

R(Parallel): 3000sin(25) - R - 250sin(30) = 0

R(Perpen): L - 250cos(30) - (perpen component of the 3000N which I cant figure out) = 0

But neither of these will give the right answer

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(Original post by

You've never used that diagram? Gahh, how did they teach you to resolve then? Just say "if it's opposite use sine, if it's adjacent use cosine"? How confusing |:

Sorry if I've confused you even more; this sort of thing is

**Implication**)You've never used that diagram? Gahh, how did they teach you to resolve then? Just say "if it's opposite use sine, if it's adjacent use cosine"? How confusing |:

Sorry if I've confused you even more; this sort of thing is

*very*hard to explain over the internet!Ive barely come across vectors in my life, so your explanation is quite confusing, but Im trying to understand it. However I am still confused by this

(Original post by

I can't see what you've done exactly so I'm not too sure, but I think you've used the wrong angles to resolve. Can you post your full working so we can see where you've gone wrong? Or would you like me to post a full solution?

**Implication**)I can't see what you've done exactly so I'm not too sure, but I think you've used the wrong angles to resolve. Can you post your full working so we can see where you've gone wrong? Or would you like me to post a full solution?

I do not want the full answers, I feel bad enough asking for help as I really ought to be able to do this myself, I get quite paranoid about this sort of thing, please assist me though

All the teachers said to do FP1 in Jan as it is the hardest unit this year. I cannot believe that, Mecanincs is exponentially harder than FP1 in my opinion. Maths involving images and interpretations just make no sense to me...

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#12

M1 is definitely harder than FP1 if you don't study Physics or if you do Physics with an exam board that isn't mechanics-heavy. I've heard a few people say that FP1 is harder, but I think that's because mechanics just comes to them - they instinctively understand things like free-body diagrams and resolving forces. I guess you're like me and prefer the straight mathematics

Anyway, how about I just post the steps you should follow when questions like this?

1. Draw a diagram with all the forces on it (a free-body diagram).

2. Decide whether to resolve horizontal-vertical or parallell-perpendicular.

3. Write down in a list all the forces that are already entirely perpendicular or parallel to the plane you are considering, recording their direction.

4. Look at one of the forces that is not entirely perpendicular/parallel. Using the triangle law or vector addition, resolve this force into two components - one parallel to the plane and one perpendicular to it. Now add these two components to your list, recording their direction. Repeat this with all forces that that are not entirely perpendicular or parallel.

5. Draw a new free-body diagram with all the parallel and perpendicular components rather than the original forces. It should look something like this, where the letters represent the sum of all the forces in their respective directions:

6. As the system is at equilibrium (constant velocity), all the components opposite each other must be equal. So a = d and b = c.

7. Use these equations to calculate any unknown values.

I'm not sure it's possible to just "see what lines affect the paralel/perpendicular components and just put them together to make 2 equations", especially in an example as complicated as this. You should be resolving each force

Anyway, how about I just post the steps you should follow when questions like this?

1. Draw a diagram with all the forces on it (a free-body diagram).

2. Decide whether to resolve horizontal-vertical or parallell-perpendicular.

3. Write down in a list all the forces that are already entirely perpendicular or parallel to the plane you are considering, recording their direction.

4. Look at one of the forces that is not entirely perpendicular/parallel. Using the triangle law or vector addition, resolve this force into two components - one parallel to the plane and one perpendicular to it. Now add these two components to your list, recording their direction. Repeat this with all forces that that are not entirely perpendicular or parallel.

5. Draw a new free-body diagram with all the parallel and perpendicular components rather than the original forces. It should look something like this, where the letters represent the sum of all the forces in their respective directions:

6. As the system is at equilibrium (constant velocity), all the components opposite each other must be equal. So a = d and b = c.

7. Use these equations to calculate any unknown values.

I'm not sure it's possible to just "see what lines affect the paralel/perpendicular components and just put them together to make 2 equations", especially in an example as complicated as this. You should be resolving each force

*independently*i.e. on its own, and then putting them all together once you have done this.
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This is question 16/16 on an exercise im doing, so I know the method on simpler scenarios

if I do parallel the forces affecting it are: r, weight, and tension in rope

Perpendicularly: L, weight, tension in rope

R(Parallel): 3000sin(25) - R - 2500Sin(30) = 0 therefore R=3000sin(25) - 2500sin(30)

This gives an incorrect value...

R(Perpendicular): L - 2500cos(30) - 3000 = 0

This also gives L as an incorrect value, i did -3000 as it lies on the same line as L.

I honestly cant see the problem :/

if I do parallel the forces affecting it are: r, weight, and tension in rope

Perpendicularly: L, weight, tension in rope

R(Parallel): 3000sin(25) - R - 2500Sin(30) = 0 therefore R=3000sin(25) - 2500sin(30)

This gives an incorrect value...

R(Perpendicular): L - 2500cos(30) - 3000 = 0

This also gives L as an incorrect value, i did -3000 as it lies on the same line as L.

I honestly cant see the problem :/

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#14

(Original post by

This is question 16/16 on an exercise im doing, so I know the method on simpler scenarios

if I do parallel the forces affecting it are: r, weight, and tension in rope

Perpendicularly: L, weight, tension in rope

R(Parallel): 3000sin(25) - R - 2500Sin(30) = 0 therefore R=3000sin(25) - 2500sin(30)

This gives an incorrect value...

R(Perpendicular): L - 2500cos(30) - 3000 = 0

This also gives L as an incorrect value, i did -3000 as it lies on the same line as L.

I honestly cant see the problem :/

**Genesis2703**)This is question 16/16 on an exercise im doing, so I know the method on simpler scenarios

if I do parallel the forces affecting it are: r, weight, and tension in rope

Perpendicularly: L, weight, tension in rope

R(Parallel): 3000sin(25) - R - 2500Sin(30) = 0 therefore R=3000sin(25) - 2500sin(30)

This gives an incorrect value...

R(Perpendicular): L - 2500cos(30) - 3000 = 0

This also gives L as an incorrect value, i did -3000 as it lies on the same line as L.

I honestly cant see the problem :/

Considering those with a parallel component:

- R - 2500cos60 + 3000cos55 = 0

Where 2500cos60 is the parallel component of the weight and 3000cos55 is the parallel component of the tension.

Considering those with a perpendicular component:

L - 2500sin60 - 3000sin55 = 0

Where 2500cos60 is the perpendicular component of the weight and 3000sin55 is the perpendicular component of the tension.

You seem to be using either the incorrect angles or the incorrect trig functions or both. I can't think how to show you what you're doing wrong without going back to the basic vector triangle, though. I think you

*may*be resolving parallel and perpendicular while still using the angles given to the

*horizontal*plane. What you want to use, when resolving parallel-perpendicular to a given plane, are the angles to

*that*plane.

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(Original post by

Okay I get you. The trouble, I think, is that you aren't looking at the correct components.

Considering those with a parallel component:

- R - 2500cos60 + 3000cos55 = 0

Where 2500cos60 is the parallel component of the weight and 3000cos55 is the parallel component of the tension.

Considering those with a perpendicular component:

L - 2500sin60 - 3000sin55 = 0

Where 2500cos60 is the perpendicular component of the weight and 3000sin55 is the perpendicular component of the tension.

You seem to be using either the incorrect angles or the incorrect trig functions or both. I can't think how to show you what you're doing wrong without going back to the basic vector triangle, though. I think you

**Implication**)Okay I get you. The trouble, I think, is that you aren't looking at the correct components.

Considering those with a parallel component:

- R - 2500cos60 + 3000cos55 = 0

Where 2500cos60 is the parallel component of the weight and 3000cos55 is the parallel component of the tension.

Considering those with a perpendicular component:

L - 2500sin60 - 3000sin55 = 0

Where 2500cos60 is the perpendicular component of the weight and 3000sin55 is the perpendicular component of the tension.

You seem to be using either the incorrect angles or the incorrect trig functions or both. I can't think how to show you what you're doing wrong without going back to the basic vector triangle, though. I think you

*may*be resolving parallel and perpendicular while still using the angles given to the*horizontal*plane. What you want to use, when resolving parallel-perpendicular to a given plane, are the angles to the*parallel*plane.And also where on earth did you get the 55 from, there is no angle 55 or 35 anywhere!?!?!!

EDIT: Is this diagram correct???

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#16

(Original post by

Oh wow.... I assume cos30=sin60 right?

And also where on earth did you get the 55 from, there is no angle 55 or 35 anywhere!?!?!!

**Genesis2703**)Oh wow.... I assume cos30=sin60 right?

And also where on earth did you get the 55 from, there is no angle 55 or 35 anywhere!?!?!!

55' is the angle between the tension (3000N) and the plane. This can be deduced from the diagram I drew earlier:

90-25=65 therefore the angle between the tension and the vertical is 65'

90-30=60 therefore the angle between the plane (parallel component) and the vertical is 60'

180-60-65=55 therefore the angle between the tension (3000N) and the plane is 55'

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(Original post by

Yeah, cos30=sin60. cosx = sin(90-x) and sinx = cos(90-x)

55' is the angle between the tension (3000N) and the plane. This can be deduced from the diagram I drew earlier:

90-25=65 therefore the angle between the tension and the vertical is 65'

90-30=60 therefore the angle between the plane (parallel component) and the vertical is 60'

180-60-65=55 therefore the angle between the tension (3000N) and the plane is 55'

**Implication**)Yeah, cos30=sin60. cosx = sin(90-x) and sinx = cos(90-x)

55' is the angle between the tension (3000N) and the plane. This can be deduced from the diagram I drew earlier:

90-25=65 therefore the angle between the tension and the vertical is 65'

90-30=60 therefore the angle between the plane (parallel component) and the vertical is 60'

180-60-65=55 therefore the angle between the tension (3000N) and the plane is 55'

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Well sadly I have to go to my mums now... so I'll just copy your notes and try to comprehend them until tommorrow

Thanks for your help though

Thanks for your help though

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#19

(Original post by

But the line for tension is in line with the L N line.... so there is no angle between them and L N is the perpendicular plane... so why is any trig function at all needed with the 3000, and surely in terms on the paralell plane it is 90 degrees from it :/

**Genesis2703**)But the line for tension is in line with the L N line.... so there is no angle between them and L N is the perpendicular plane... so why is any trig function at all needed with the 3000, and surely in terms on the paralell plane it is 90 degrees from it :/

*L*N is

**not**the same as that of the tension. The question doesn't say that the forces are in the same plane - so you can't assume that they are - and the diagram given by the question shows that they are not. A few diagrams and a little geometric manipulation should actually show you that they

*can't*be in the same plane! Try again with this knowledge and hopefully you'll get the right answer hehe

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(Original post by

Noo! The plane of

**Implication**)Noo! The plane of

*L*N is**not**the same as that of the tension. The question doesn't say that the forces are in the same plane - so you can't assume that they are - and the diagram given by the question shows that they are not. A few diagrams and a little geometric manipulation should actually show you that they*can't*be in the same plane! Try again with this knowledge and hopefully you'll get the right answer heheI sort of understand what to do, its a bottomless trianglem with line L sticking out of the left. its just angles.... line L is still parallel looking to the tension. the parallel equation makes sense but not the perpendicular one due to how line l looks parallel

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