Casio Posted June 23 Share Posted June 23 I have a right angle, the 90 degrees angle is at the top right corner. A force is applied along AB. AB is the horizontal line from the 90 degree corner marked A to the left marked B, with a length 10.5 cm. The force along AB is 102.31 N. The force AC is the vertical line of action measured at a length of 13.5 cm. The hypotenuse final length BC is 17 cm. Hope your with me to this point! According to a engineering book I've read that includes the methods for calculating the Traingle of Forces, where two or more forces are known, the side lengths of the triangle are measured and then multiplied by the acting force to record the results, however, either I'm missing something in the writing of the book, or information is missing! By example... The force acting along AB is 102.31 N. The length of AB is 10.5 cm. Therefore 10.5 x 102.31 = 1074 N Now the line of action changes direction from AB to BC, and this is where I'm questioning the book maths. The length of side BC is 17 cm, therefore 17 x 102.31 = 1740 N I could carryout the same procedure with the remining vertical force, which records 1381 N. What I questioning is this. The book example is using the Triangle of Forces Rules, but nowhere does the written instructions in the book take into consideration that these forces have changed direction, at angles of 52.41 degrees being angle ABC, and 37.59 degrees at angle BCA. Am I missing something here? I'm sure the angles should be included in the math!! Link to comment Share on other sites More sharing options...

joigus Posted June 23 Share Posted June 23 How come you multiply force times distance to get a force? The result would have to be a work, energy, torque... (Never mind the mixed units, as you should convert cm to m.) This should be homework, right? Link to comment Share on other sites More sharing options...

Casio Posted June 23 Author Share Posted June 23 51 minutes ago, joigus said: How come you multiply force times distance to get a force? The result would have to be a work, energy, torque... (Never mind the mixed units, as you should convert cm to m.) This should be homework, right? First answer, multiply force x distance to get a force? Your not creating a force doing that, you've already got the force, its just increasing along the line of action. Strickly speaking yes in the SI Units the length should or could be metres m, but equally so the units cm are also used. Common sense must prevail, if then your measuring a football pitch, by all means use metres m, but if your measuring something approximately 50 mm long, then I'd of thought using either mm or cm would suffice. The engineering book is using cm so I kept with that unit of length. This is not homework. I'm simply reading at home in an interest I have based on my job I do. I'm nearly 60 years old and have no interest nor energy left for speed chalk board exercises that also envelope many mistakes. Link to comment Share on other sites More sharing options...

joigus Posted June 23 Share Posted June 23 1 hour ago, Casio said: First answer, multiply force x distance to get a force? Your not creating a force doing that, you've already got the force, its just increasing along the line of action. You said, 2 hours ago, Casio said: The force acting along AB is 102.31 N. The length of AB is 10.5 cm. Therefore 10.5 x 102.31 = 1074 N This is not dimensionally consistent. All physical equalities must be dimensionally consistent. It can happen though that when a dimensional quantity is a universal constant, you can choose it to be one, and there is a dimensional reduction. AFAIK, the centimeter is not a universal constant. Link to comment Share on other sites More sharing options...

Casio Posted June 23 Author Share Posted June 23 While that maybe true in what you say, it however doesn't help with the orginal question asked! Link to comment Share on other sites More sharing options...

studiot Posted June 23 Share Posted June 23 3 hours ago, Casio said: According to a engineering book I've read that includes the methods for calculating the Traingle of Forces, where two or more forces are known, the side lengths of the triangle are measured and then multiplied by the acting force to record the results, however, either I'm missing something in the writing of the book, or information is missing! Yes I think you may be missing something. Is your triangle a real object like a triangular shelf bracket, say a length of strip metal bent into a triangular shape ? Link to comment Share on other sites More sharing options...

swansont Posted June 23 Share Posted June 23 Summing forces is just vector addition. A 1 N force at 30 degrees to the x axis is 0.5 N in the y direction and 0.866 N in the x direction. You can work this in the other way; the sum of the x and y force gives you 1N (0.5^2 + .866^2 = 1) at 30 degrees Link to comment Share on other sites More sharing options...

joigus Posted June 24 Share Posted June 24 17 hours ago, Casio said: While that maybe true in what you say, it however doesn't help with the orginal question asked! Sorry you see it as though I wasn't trying to help. I was. I thought it was more urgent that you learn to handle dimensions than it was for you to handle vector addition. As an old professor of mine said, 'If you want to be able to write Chinese poetry, your first goal should be to be able to write Chinese'. I wish you best of luck with your Chinese poetry. Link to comment Share on other sites More sharing options...

Casio Posted June 24 Author Share Posted June 24 On 6/23/2024 at 4:35 PM, studiot said: Yes I think you may be missing something. Is your triangle a real object like a triangular shelf bracket, say a length of strip metal bent into a triangular shape ? No, the triangle is representing forces. In this right triangle, there is only one force N applied. This force then meets and equal and opposite force at point B in the right triangle, and then that force changes direction by 52.41 degrees and travels along the hypotenuse BC, before changing direction again by 37.59 degrees back to point A which is the right angle 90 degrees, where the applied force starts. Because only one applied force is known, in my understanding, until the angle 52.41 degrees is taken into consideration to change the direction of the force applied, then the force along BC cannot correctly be calculated. So the engineering book I'm reading advises to measure the length of the sides of the triangle and multiply that length by the force applied to justify the force acting in that direction, such as the force from AB = 102.31 N times 10.5 = 1074 N, and then the force is applied again along BC = 102.31 N times 17 = 1740 N So by just calculating two sides, the applied force is increasing after changing direction of 52.41 degrees. I just can't accept that the method used is correct? Forces changing direction like this to me I'd think should reduce in force, not increase? Link to comment Share on other sites More sharing options...

studiot Posted June 25 Share Posted June 25 11 hours ago, Casio said: No, the triangle is representing forces. In this right triangle, there is only one force N applied. This force then meets and equal and opposite force at point B in the right triangle, and then that force changes direction by 52.41 degrees and travels along the hypotenuse BC, before changing direction again by 37.59 degrees back to point A which is the right angle 90 degrees, where the applied force starts. Because only one applied force is known, in my understanding, until the angle 52.41 degrees is taken into consideration to change the direction of the force applied, then the force along BC cannot correctly be calculated. So the engineering book I'm reading advises to measure the length of the sides of the triangle and multiply that length by the force applied to justify the force acting in that direction, such as the force from AB = 102.31 N times 10.5 = 1074 N, and then the force is applied again along BC = 102.31 N times 17 = 1740 N So by just calculating two sides, the applied force is increasing after changing direction of 52.41 degrees. I just can't accept that the method used is correct? Forces changing direction like this to me I'd think should reduce in force, not increase? Thank you for your reply. I am still not clear what is going on here and therefore what your actual question is. Can you (scan/photograph) and post the passage from the book as printed please. You need to be aware that there are (always) two triangles involved. There is the real world-object triangle which defines the geometry of the situation, both in terms of distances / positions and angles. And there is a purely theoretical abstact triangle others have referred to as a vector triangle. This second triangle is similar to, but not congruent with, the real world triangle. Two triangles are said to be 'similar' if they have the same angles, but not generally the same side lengths. Triangles which have both the same angles and the side lengths are said to be congruent. I am guessing that mixing up these two different triangles is the source of your difficulty, but I can't be sure without sight of the original. Link to comment Share on other sites More sharing options...

Casio Posted June 25 Author Share Posted June 25 8 hours ago, studiot said: Thank you for your reply. I am still not clear what is going on here and therefore what your actual question is. Can you (scan/photograph) and post the passage from the book as printed please. You need to be aware that there are (always) two triangles involved. There is the real world-object triangle which defines the geometry of the situation, both in terms of distances / positions and angles. And there is a purely theoretical abstact triangle others have referred to as a vector triangle. This second triangle is similar to, but not congruent with, the real world triangle. Two triangles are said to be 'similar' if they have the same angles, but not generally the same side lengths. Triangles which have both the same angles and the side lengths are said to be congruent. I am guessing that mixing up these two different triangles is the source of your difficulty, but I can't be sure without sight of the original. Thanks for your reply. The triangle is imaginary and is not similar to another triangle. I'm no professional in this subject, but another book I've researched suggests I should have posted this problem in the Statics forum and not the maths department. If I can find away of posting my drawing I'll do it as soon as I can, but I think you could also get a good general idea of my request by drawing a right triangle on paper yourself to give you the picture I'm referring to. The specifications... Right triangle. The 90 degrees angle MUST BE at the right corner of the horizontal line, i.e. __________________Here. Then follows the vertical line downwards, i.e. | This line touches the horizontal line marked "Here". That is the 90 degree angle point. The horizontal line length is 10.5 cm. The vertical line length is 13.5 cm. The line that represents the hypotenuse is 17 cm. The triangle has acute angles, the top left is 52.41 degrees and the bottom angle (sine) is 37.59 degrees. My problem... An applied force is present along the horizontal line from A to B. A is the 90 degree angle. B is the 52.41 degree angle to the left. When the force of 102.31 N is passed along the horizontal line from A to B, the line changes direction along BC, which is the hypotenuse, which then changes direction again by 37.59 degrees along CA to the 90 degree start point. The force must change its magnitude when it changes direction. What I'm trying to find out and learn is how to use the acute angles in the math to calculate the new forces along BC and CA. I hope you understand what I'm trying to do!! Link to comment Share on other sites More sharing options...

studiot Posted June 25 Share Posted June 25 1 hour ago, Casio said: The triangle is imaginary and is not similar to another triangle. Forgive me for knowing what i am talking about. 1 hour ago, Casio said: I'm no professional in this subject, but another book I've researched suggests I should have posted this problem in the Statics forum and not the maths department. You posted Applied Mathematics in the Analysis and Calculus subforum where it is certainly misplaced, but i thought it a small point, easily rectifyable. Perhaps @swansont would be kind enough to move it to Applied Maths where it truly belongs. 1 hour ago, Casio said: Right triangle. The 90 degrees angle MUST BE at the right corner of the horizontal line, i.e. __________________Here. Then follows the vertical line downwards, i.e. | This line touches the horizontal line marked "Here". That is the 90 degree angle point. The horizontal line length is 10.5 cm. The vertical line length is 13.5 cm. The line that represents the hypotenuse is 17 cm. The triangle has acute angles, the top left is 52.41 degrees and the bottom angle (sine) is 37.59 degrees. So what is the triangle if it is not part of the physical world? 1 hour ago, Casio said: An applied force is present along the horizontal line from A to B. A is the 90 degree angle. B is the 52.41 degree angle to the left. When the force of 102.31 N is passed along the horizontal line from A to B, the line changes direction along BC, which is the hypotenuse, which then changes direction again by 37.59 degrees along CA to the 90 degree start point This is just incorrectly imagined nonsense, as written above. It certainly has little or nothing to do with the 'triangle of forces' Link to comment Share on other sites More sharing options...

Casio Posted June 26 Author Share Posted June 26 4 hours ago, studiot said: Forgive me for knowing what i am talking about. You posted Applied Mathematics in the Analysis and Calculus subforum where it is certainly misplaced, but i thought it a small point, easily rectifyable. Perhaps @swansont would be kind enough to move it to Applied Maths where it truly belongs. So what is the triangle if it is not part of the physical world? This is just incorrectly imagined nonsense, as written above. It certainly has little or nothing to do with the 'triangle of forces' Ok thanks for your input anyway. I'll need to find an engineering forum that has experience in these areas of expertise. Ok I've managed to work out how to do it. I've got a force of 102.31 N acting along AB. I wanted to work out the force acting along BC and CA. I knew that some losses must occur due to the angles the forces had to pass round. This is how it is done... F Cos theta = 102.31 N x cos 37.59 = 81.1 N, and F cos theta = 102.31 N x cos 52.41 = 62.4 N Now using Pythagorus we have... c = square root a^2 + b^2 c = square root 81.1^2 + 62.4^2 c = 102.32 N In conclusion... Force AB applied was 102.31 N, the two remaining forces acting after the acute angles are; 81.1 N and 62.4N That's good enough for me to give me some guidance, and I'm seriously NO expert in this. Link to comment Share on other sites More sharing options...

studiot Posted June 28 Share Posted June 28 On 6/26/2024 at 1:09 AM, Casio said: Ok I've managed to work out how to do it. Oh! On 6/26/2024 at 1:09 AM, Casio said: I've got a force of 102.31 N acting along AB. I wanted to work out the force acting along BC and CA. I knew that some losses must occur due to the angles the forces had to pass round. This is how it is done... F Cos theta = 102.31 N x cos 37.59 = 81.1 N, and F cos theta = 102.31 N x cos 52.41 = 62.4 N Now using Pythagorus we have... c = square root a^2 + b^2 c = square root 81.1^2 + 62.4^2 c = 102.32 N In conclusion... Force AB applied was 102.31 N, the two remaining forces acting after the acute angles are; 81.1 N and 62.4N That's good enough for me to give me some guidance, and I'm seriously NO expert in this. You may be about half right. Unfortunately this includes some serious misconceptions which could lead you far astray if this not not a one off. What a pity you didn't want to discuss the problem from start to finish. A good place to start would be to answer the one simple question I asked, or ask for more information about it. Link to comment Share on other sites More sharing options...

Casio Posted June 28 Author Share Posted June 28 I didn't think the conversation being carried on was necessary. You pointed out that what I said was nonesense. I thought that if you'd ofunderstood what I was trying my best to explain, then you might of returned with some constructive suggestions and or applied math to show how its done, but you didn't so I took it you didn't understand what I was trying to do. I appreciate it difficult to interpret people and ideas on such forums as these, but at least I tried. If it loads I've now included a diagram. Link to comment Share on other sites More sharing options...

swansont Posted June 28 Share Posted June 28 1 hour ago, Casio said: I didn't think the conversation being carried on was necessary. You pointed out that what I said was nonesense. I thought that if you'd ofunderstood what I was trying my best to explain, then you might of returned with some constructive suggestions and or applied math to show how its done, but you didn't so I took it you didn't understand what I was trying to do. I appreciate it difficult to interpret people and ideas on such forums as these, but at least I tried. If it loads I've now included a diagram. The diagram doesn’t agree with the Pythagorean theorem. The distances do, but the forces do not; if there is no acceleration, the forces have to add to zero. If the force from BC is 102.3 N, and AB is 81.3 N, it works. Link to comment Share on other sites More sharing options...

studiot Posted June 29 Share Posted June 29 In Engineering we deal with real world objects called Bodies. Bodies can be affected by agents we call Forces according to specific rules. Individual forces cannot combine directly, but many forces can act on a single body - with the overall effect being a specific combination of the effect of each force acting individually. The same effect as all these combined forces can also be caused by a single suitably applied force called the resultant. Because both bodies and forces exist in the same geometrical universe or framework, there exists a correspondence between the geometry of the lengths and positions of the bodies and the geometry of the diagrams describing the forces. In fact one is a scale diagram of the other. For our present purposes forces acting on a by may be considered as a) Externally Imposed - These are called Loads b) Constraints on the Body by other bodies or forces - These are called Reactions c) Forces generated internally witin the body by the actions of (a) and (b) We will only need to examine (a) and (b). Rules for the actions of Forces on Bodies A force is a push or a pull All forces act only in straight lines, called their line of action. Forces cannot "turn corners" or "change direction" Interaction with a body may produce a new force in a different direction. Individual forces generally act on bodies at a single point called the point of application. A body for which the resultant of all acting forces is zero is said to be in equilibrium. A consequence of (5) is that a body with only a single non zero force acting on it cannot be in equilibrium. If a body is under the action of two forces it can only be in equilibrium if the two forces are acting along the same line. A body under the action of two or more (non zero) forces may always be brought into equilibrium by the application of an extra single force whcih is equal in magnitude but opposite in direction to the resultant of the original forces. This is callant the equilibrant. (This forms the force basis of the triangle of forces). The nature of the geometrical link between the configuration of the points of application and the directions and magnitudes of the applied forces enables a diagram called "The Polygon of Forces" to be either drawn or calculated. Scale drawing alone was once a popular method of obtainingt the Resultant/Equilibriant without calculation. The Triangle of Forces is simplest such polygon and uses three forces, two applied plus the resultant/equilibrant In the next post I will show how this is done using some simple example diagrams. I will also comment on where and why your information is correct or incorrect. I also think you are perhaps confusing the obtaining of a resultant by the triangle of forces with resolving a single force into components in specific directions, which the triangle can do for you because of the aforementioned geometric relationship. But this is a different calculation altogether. 1 Link to comment Share on other sites More sharing options...

Casio Posted July 2 Author Share Posted July 2 On 6/29/2024 at 11:43 PM, studiot said: In Engineering we deal with real world objects called Bodies. Bodies can be affected by agents we call Forces according to specific rules. Individual forces cannot combine directly, but many forces can act on a single body - with the overall effect being a specific combination of the effect of each force acting individually. The same effect as all these combined forces can also be caused by a single suitably applied force called the resultant. Because both bodies and forces exist in the same geometrical universe or framework, there exists a correspondence between the geometry of the lengths and positions of the bodies and the geometry of the diagrams describing the forces. In fact one is a scale diagram of the other. For our present purposes forces acting on a by may be considered as a) Externally Imposed - These are called Loads b) Constraints on the Body by other bodies or forces - These are called Reactions c) Forces generated internally witin the body by the actions of (a) and (b) We will only need to examine (a) and (b). Rules for the actions of Forces on Bodies A force is a push or a pull All forces act only in straight lines, called their line of action. Forces cannot "turn corners" or "change direction" Interaction with a body may produce a new force in a different direction. Individual forces generally act on bodies at a single point called the point of application. A body for which the resultant of all acting forces is zero is said to be in equilibrium. A consequence of (5) is that a body with only a single non zero force acting on it cannot be in equilibrium. If a body is under the action of two forces it can only be in equilibrium if the two forces are acting along the same line. A body under the action of two or more (non zero) forces may always be brought into equilibrium by the application of an extra single force whcih is equal in magnitude but opposite in direction to the resultant of the original forces. This is callant the equilibrant. (This forms the force basis of the triangle of forces). The nature of the geometrical link between the configuration of the points of application and the directions and magnitudes of the applied forces enables a diagram called "The Polygon of Forces" to be either drawn or calculated. Scale drawing alone was once a popular method of obtainingt the Resultant/Equilibriant without calculation. The Triangle of Forces is simplest such polygon and uses three forces, two applied plus the resultant/equilibrant In the next post I will show how this is done using some simple example diagrams. I will also comment on where and why your information is correct or incorrect. I also think you are perhaps confusing the obtaining of a resultant by the triangle of forces with resolving a single force into components in specific directions, which the triangle can do for you because of the aforementioned geometric relationship. But this is a different calculation altogether. In the next post I will show how this is done using some simple example diagrams. I will also comment on where and why your information is correct or incorrect. Also, just in case some information NOT INCLUDED by me was overlooked, the Right Triangle is ONLY showing one force and half the calculations. In Vectors there must be two forces to find a resultant etc. I only included one. In a brake drum assembly, there are two shoes, one leading and one trailing. I only ever talked about forces acting on the trailing shoe using a Right Triangle. I also agree that force cannot turn corners, but maybe the confusion was around two areas of my concern that may have been misinterpreted? 1/ The applied force of 102.3 N travelled along AB ONLY. There then is a frictional force created (uP) This frictional force amounts to 32 N. I didn't know whether in my calculations to include it or dismiss it? 2/ The force (I said) along BC is NOT redirected from AB. 3/ The force I calculated along BC being 81.1 N I am referring that force to or calling that force the "Component" force. What do I mean by component force! The brake shoe is metal, the lining material is a mixture of compounds and resin. The frictional force acting is 32 N. This force is the force on the shoe material and the rubbing surface of the brake drum. This frictional force 32 N being pushed on the brake drum is a calculated force from the 102.3 N appllied force, but I think is a seperate force. The force acting along BC I refer to as the component force. This is what I am trying to establish as the force building up inside the brake shoe itself. So Newton said, to every action there is an equal and opposite reaction, hence, the force applied at B towards C is the component force and I calculated that as 81.1 N. Now to achieve that force, the pivot at the heel of the shoe must maintain the shoe in that fixed position, therefore a reverse reaction is created equal and opposite the applied force. The shoe has now got an internal force present of 81.1 N. So, how dense does the material a shoe is made from have to be? If the material is too thin the shoe will distort and break. The component force calculated is to help interpret what force what thickness material can cope with without distortion and damage. Hence why I was thinking about vectors and forces to calculate the component force! Link to comment Share on other sites More sharing options...

Joatmon Posted July 2 Share Posted July 2 I'm a bit puzzled scanning through this thread but there are a few points I'd like to add. The first is obvious - Right angled triangles have sides that comply with Pythagoras theorem. The second, once you have your triangle you cannot multiply different sides by different amounts - in effect you would be altering the shape of the triangle. Each side must be multiplied by the same amount. Lastly such a triangle represents either constant forces or if forces are changing it can only represent a particular instant in time. I'm retired now but my speciality was electronics and our main complication concerned sinusoidal forces (voltages) where appreciating the difference between instantaneous value and a steady state equivalent was necessary. I don't know if this helps, but it's my two pennyworth 1 Link to comment Share on other sites More sharing options...

Casio Posted July 5 Author Share Posted July 5 On 6/29/2024 at 11:43 PM, studiot said: In Engineering we deal with real world objects called Bodies. Bodies can be affected by agents we call Forces according to specific rules. Individual forces cannot combine directly, but many forces can act on a single body - with the overall effect being a specific combination of the effect of each force acting individually. The same effect as all these combined forces can also be caused by a single suitably applied force called the resultant. Because both bodies and forces exist in the same geometrical universe or framework, there exists a correspondence between the geometry of the lengths and positions of the bodies and the geometry of the diagrams describing the forces. In fact one is a scale diagram of the other. For our present purposes forces acting on a by may be considered as a) Externally Imposed - These are called Loads b) Constraints on the Body by other bodies or forces - These are called Reactions c) Forces generated internally witin the body by the actions of (a) and (b) We will only need to examine (a) and (b). Rules for the actions of Forces on Bodies A force is a push or a pull All forces act only in straight lines, called their line of action. Forces cannot "turn corners" or "change direction" Interaction with a body may produce a new force in a different direction. Individual forces generally act on bodies at a single point called the point of application. A body for which the resultant of all acting forces is zero is said to be in equilibrium. A consequence of (5) is that a body with only a single non zero force acting on it cannot be in equilibrium. If a body is under the action of two forces it can only be in equilibrium if the two forces are acting along the same line. A body under the action of two or more (non zero) forces may always be brought into equilibrium by the application of an extra single force whcih is equal in magnitude but opposite in direction to the resultant of the original forces. This is callant the equilibrant. (This forms the force basis of the triangle of forces). The nature of the geometrical link between the configuration of the points of application and the directions and magnitudes of the applied forces enables a diagram called "The Polygon of Forces" to be either drawn or calculated. Scale drawing alone was once a popular method of obtainingt the Resultant/Equilibriant without calculation. The Triangle of Forces is simplest such polygon and uses three forces, two applied plus the resultant/equilibrant In the next post I will show how this is done using some simple example diagrams. I will also comment on where and why your information is correct or incorrect. I also think you are perhaps confusing the obtaining of a resultant by the triangle of forces with resolving a single force into components in specific directions, which the triangle can do for you because of the aforementioned geometric relationship. But this is a different calculation altogether. Why only AB? The two problems I forsaw were... 1/ I only have one force N acting, 2/ The whole point of the excercise was to establish the component force, which you've pointed out as the internal force. That is what I was always trying to establish. The force N acting along AB was always known, the problem was understanding how to work out the force internally (or along) BC to establish the internal (component) force. Hope I've made that clearer, sorry if I've confused everyone. Link to comment Share on other sites More sharing options...

exchemist Posted July 5 Share Posted July 5 46 minutes ago, Casio said: Why only AB? The two problems I forsaw were... 1/ I only have one force N acting, 2/ The whole point of the excercise was to establish the component force, which you've pointed out as the internal force. That is what I was always trying to establish. The force N acting along AB was always known, the problem was understanding how to work out the force internally (or along) BC to establish the internal (component) force. Hope I've made that clearer, sorry if I've confused everyone. The mistake you made at the beginning was not to realise that the length of a force vector is proportional to the magnitude of the force. That is how a force vector is defined in the first place. So multiplying the length of sides of the triangle of force vectors by the size of the force makes no sense. The triangle already is the forces. Any vector can be expressed as 2 components at right angles, an "x" and a "y" component if you like, which you add by means of constructing a right angled triangle out of them. Which you can draw as (You can ignore the "sin" and "cos" bits of trigonometry if you like: they were just there on the drawings I found on the web.) Link to comment Share on other sites More sharing options...

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