# M1- Resolving Forces Question

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#1
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! 0
9 years ago
#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.

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#3
(Original post by 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.

Yes one of the answers is 470 and the other is 4620 (doesnt state which is which in the back though)

I will attempt try parallel and perpendicular now :P
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9 years ago
#4
Okay good; those are the answers I got resolving parallel and perpendicular haha! Let us know if you get stuck resolving this way 0
#5
(Original post by Implication)
Okay good; those are the answers I got resolving parallel and perpendicular haha! Let us know if you get stuck resolving this way Im slightly stuck already because I dont know how to deal with the 3000N, as it is not connecting directly to the glider in the diagram (by the looks of it, the line extends upwards slightly before bending down to the 3000N) so I am unsure what angle/trig function to use for it paralel or perpendicular 0
9 years ago
#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.
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#7
(Original post by 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.
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
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#8
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 0
9 years ago
#9
(Original post by 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
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 |:

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!
0
9 years ago
#10
(Original post by 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 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?
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#11
(Original post by 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!
Yes thats how we were taught it, along with the whole, if you rotate the line to the horizontal/verticle and the angle closes/goes to 0 it is cos, and if it opens up it is Sin.

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 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?
What working? I just went and said to myself "what lines affect the paralel/perpendicular components" and just put them together to make those 2 equations. I have probably misunderstood and used the wrong angles/trig functions though.

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...
0
9 years ago
#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 independently i.e. on its own, and then putting them all together once you have done this.
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#13
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 :/
0
9 years ago
#14
(Original post by 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 :/
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 that plane.
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#15
(Original post by 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.
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!?!?!!

EDIT: Is this diagram correct???

0
9 years ago
#16
(Original post by 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!?!?!!
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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#17
(Original post by 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'
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 :/
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#18
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 0
9 years ago
#19
(Original post by 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 :/
Noo! The plane of LN 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 0
#20
(Original post by Implication)
Noo! The plane of LN 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 but.... you said the drawing in the book was wrong and that the tension directly effected the glider!

I 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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