Casio

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.

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!

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.

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.

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!!

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?

While that maybe true in what you say, it however doesn't help with the orginal question asked!

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.

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!!

Why do scientist "think" they know everything?? Is that a personal opinion! I know scientists would truly admit they don't know everything. take our current world pandemic, the scientists have been battling away trying their best to find a vaccine for the world at large, now if they believed they knew everything then surely last December they would have just put the pan on and boiled up a vaccine and distributed it say January 2020 and we'd all be fine now.

I've no experience or training in this expertise area so please forgive if I don't seem to explain things quite right. Within the last few months I without warning got a sharp pain that just appeared in the right hand side of my chest and slowly seemed to moved down my body towards my stomach, but then disappeared for a short while before returning. The pain appears from the inside of my rib cage on the right hand side now just below my chest area to the side. After a nights sleep the pain is not present the following day until about lunch time or afternoon, and it seems depending how I move my body around will either cause the pain or stop the pain, hence standing a long period of time can bring the pain on, but if I sit down for a while the pain goes and then does not return straight away. If I press on my ribs there is no sensation or pain and the pain I do get has no effect on my breathing. Give the corona virus problems the world is facing my local doctors only do phone consultations and the hospitals are only seeing emergency type problems, which I'm not sure I qualify for that criteria? Any ideas? I have no known health issues other than this. Thanks

Infinity has an end? Infinity means many different things, depending on when it is used. The word is from a Latin word, which means "without end". Infinity goes on forever, so sometimes space, numbers, and other things are said to be 'infinite', because they never come to a stop. ... For example, adding 10 to a number repeatedly. Let me know when you reach the end...

I give up with this topic of God. My experiences show that when we get pandemics like the current one, nobody knows its existence initially, then later when medical science starts to recognise something not so normal, reports are made, feedback sent to the relevant authorities and then government put measures in place to stop the spread. Now worshipers go to church and sit side by side along with their colds and flu etc, spreading the diseases. Where did God write in the bible that CV-19 was life threatening and either God put control measures in place for his worshipers so they would not be affected, or write in the bible that social distancing must take place from 2019. All this bible prophecy appears well written but we must not forget that the bible is rewritten every fifty years. Today I listened to a video presentation from a religious group around the word using "zoom". The way the message was coming across to me seemed like these people in their religion was saying that they were aware of CV-19 before the scientists, the governments and the WHO, yet it was the governments we all know and can see that put social distancing in place and closed Gods places of worship. Last week I wrote an email to the religious group pointing out the facts about CV-19 and their churches and the way they go about conducting their religion. Normally when I write they have an answer for everything, but CV-19 has stopped then in their tracks, they have no answers for me this time. God either exists or does not exit. If he has a conflict with Satan then they need to both get together and sort themselves out, because blaming the human race for what the world is to me is not our fault. In a court case ALL facts must be presented not half a job. The so called Gods should reveal themselves to the human race and stop hiding.

Why would he need to? Look at it this way... When a woman is pregnant she is having a baby, God is not in there putting the baby together, but the power that he created to create the baby is fully at work doing its job. Now if you look at the earth and its functions, gravity holds all masses together on the planet. Another power he created to perform his function, but again he is not actually performing the work, his power is doing the work. I pretty sure that if God needed to block a signal in the mind of a human being to stop a process occurring, he is quite capable and does not require a persons permission to do so. He is the creator and we are the created. God created free will but that is not exactly what people think it is, which is another discussion...

Thank you for your help and advice. My sister opened her eyes today and acknowledged our presence.