• Can facepunch help me with a Physics problem?
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[IMG]http://i.imgur.com/ROJG1.png[/IMG] So far I was able to figure out that the tension on T3 is 30N of force. I'm having trouble figuring out the rest.. If anyone is unfamiliar with the problem it's determining magnitude of force, if anyone could help me i'd be very appreciative. Oh god forgot to add it, there is a 60 degree angle next to T1 On the left Forgot the question too, sorry guys : Draw a quantitive force vector map for the junction of the cords in the diagram Find the values of T1 T2 and T3
Assuming it's in equilibrium, T2 is the negative x and T3 is the negative y direction, blah, blah T3 = T1y T2 = T1x By "there is a 60 degree angle next to T1 On the left" I'm assuming you mean from the y-axis? Adjust the trig if that is incorrect. So T2 = tan60(T3), then use Pythagorean theorem to find T1. [editline]18th October 2011[/editline] Sorry for any errors, but I am on the fly.
break it apart into vectors
Our teacher doesn't want to teach us the way to solve it with Trig. [editline]19th October 2011[/editline] Also I need a force vector map, If I know what it looks like I can solve the problem on my own :/
Why would you not solve it with trig, that's such a good way to.
I feel like answering this, so I will. If I make ANY mistake, [B][U]PLEASE[/U][/B], tell me. Nothing worse than being mistaken. SO... A good way to start is having a good schema of your problem. I made one (I noted the weight of the block in [B]g[/B], not in [B]kg[/B], my bad. Always work in [B]kg[/B]) : [IMG]http://s3.noelshack.com/uploads/images/9848227108797_shema1.png[/IMG] First thing to do is to spot every "object". I usualy circle them. It's easier to spot and it helps. [IMG]http://s3.noelshack.com/uploads/images/11910664234458_shema2.png[/IMG] Now, as people told you, effin' draw your vectors. ALWAYS draw them. It may sound stupid, but it's the best way to go. Also, don't forget your axis. I'm sure your teacher will mark this as a mistake if you forget it. [IMG]http://s3.noelshack.com/uploads/images/11603728091142_shema3.png[/IMG] Zing. Now that we have a good plan, we're ready to go. We’re going to start out with the 3kg block since it’s the easiest. Now it’s time to apply Newton’s second law. Remember: [B]∑F = ma[/B] Which can be separated like this : [B]∑Fy = may[/B] and [B]∑Fx = max[/B] Fortunately, as shown in our schema, there are NO forces what so ever in the ‘x’ axis. That saves us some trouble. Now, we just gotta plug in our values. We’re also lucky, there are no acceleration, because the block is hanging. Warning : Everything inverted to the ‘y’ axis becomes negative. [B]∑Fy = may T – Fg = 3(0) T – Fg = 0[/B] We know that Fg = m(9,81). [B]T – m(9,81) = 0 T – 3(9,81) = 0 T – 29,43 = 0 T = 29,43N[/B] Allright, looks like we found our first tension. Now, let’s go onto the small round thing. Here it’s a little harder, but we will do the same thing as previous. Let's start with the 'y' axis. [B]∑Fy = may T3sin30° + T’ = may[/B] We know for a fact that T’ is the exact opposite of T. That’ll help. [B]T3sin60° + -29,43 = 3(0) T3(0,866) - 29,43 = 0 T3 = 29,43 / 0,866 T3 = 33,98N[/B] Neat, only one more tension to find. Let’s move on onto the ‘x’ axis. [B]∑Fx = max T3cos60° - T2 = 3000(0) T3(0,5) – T2 = 0 T3(0,5) = T2[/B] Easy as pie, we just gotta replace T3 and voilà ! [B]33,98(0,5) = T2 16,99N = T2[/B] [B][U]RESULTS :[/U][/B] [I]T = 29,43N ≈ 30N T2 = 16,99N ≈ 17N T3 = 33,98N ≈ 34N [/I] Done. I really enjoyed making this, it helped me too to be honest. :v:
[QUOTE=X-tra;32853346]I feel like answering this, so I will. If I make ANY mistake, [B][U]PLEASE[/U][/B], tell me. Nothing worse than being mistaken. SO... A good way to start is having a good schema of your problem. I made one (I noted the weight of the block in [B]g[/B], not in [B]kg[/B], my bad. Always work in [B]kg[/B]) : [IMG]http://s3.noelshack.com/uploads/images/9848227108797_shema1.png[/IMG] First thing to do is to spot every "object". I usualy circle them. It's easier to spot and it helps. [IMG]http://s3.noelshack.com/uploads/images/11910664234458_shema2.png[/IMG] Now, as people told you, effin' draw your vectors. ALWAYS draw them. It may sound stupid, but it's the best way to go. Also, don't forget your axis. I'm sure your teacher will mark this as a mistake if you forget it. [IMG]http://s3.noelshack.com/uploads/images/11603728091142_shema3.png[/IMG] Zing. Now that we have a good plan, we're ready to go. We’re going to start out with the 3kg block since it’s the easiest. Now it’s time to apply Newton’s second law. Remember: [B]∑F = ma[/B] Which can be separated like this : [B]∑Fy = may[/B] and [B]∑Fx = max[/B] Fortunately, as shown in our schema, there are NO forces what so ever in the ‘x’ axis. That saves us some trouble. Now, we just gotta plug in our values. We’re also lucky, there are no acceleration, because the block is hanging. Warning : Everything inverted to the ‘y’ axis becomes negative. [B]∑Fy = may T – Fg = 3(0) T – Fg = 0[/B] We know that Fg = m(9,81). [B]T – m(9,81) = 0 T – 3(9,81) = 0 T – 29,43 = 0 T = 29,43N[/B] Allright, looks like we found our first tension. Now, let’s go onto the small round thing. Here it’s a little harder, but we will do the same thing as previous. Let's start with the 'y' axis. [B]∑Fy = may T3sin30° + T’ = may[/B] We know for a fact that T’ is the exact opposite of T. That’ll help. [B]T3sin60° + -29,43 = 3(0) T3(0,866) - 29,43 = 0 T3 = 29,43 / 0,866 T3 = 33,98N[/B] Neat, only one more tension to find. Let’s move on onto the ‘x’ axis. [B]∑Fx = max T3cos60° - T2 = 3000(0) T3(0,5) – T2 = 0 T3(0,5) = T2[/B] Easy as pie, we just gotta replace T3 and voilà ! [B]33,98(0,5) = T2 16,99N = T2[/B] [B][U]RESULTS :[/U][/B] [I]T = 29,43N ≈ 30N T2 = 16,99N ≈ 17N T3 = 33,98N ≈ 34N [/I] Done. I really enjoyed making this, it helped me too to be honest. :v:[/QUOTE] I must say that was quite helpful, and I really appreciate the time you set aside to help me! Unfortunatly we aren't using trig to solve it, rather than measuring the angles and drawing lines until they intersect and then measuring the force in form of centimeters, for example 1 CM = 10N. My teacher set aside a margin of "Human error" of 5 above and below. I'm going to apply your reasoning so it will help me draw my angles. Thank you very much, really appreciate this! If I were going to draw a force vector map what would it look like? Would I still apply the force of gravity if the object is suspended? And since T3 was practically given would I still include this? [editline]19th October 2011[/editline] Also sorry for mentioning this, though the various T's need to be in place so I get the measurements of tension on the cords correct, as ive ruled so far T3 is equal to 30N, since it's attached to the 3KG weight. I apologize for my rather crude drawing.
[QUOTE=DerpHurr;32854002]I must say that was quite helpful, and I really appreciate the time you set aside to help me! Unfortunatly we aren't using trig to solve it, rather than measuring the angles and drawing lines until they intersect and then measuring the force in form of centimeters, for example 1 CM = 10N. My teacher set aside a margin of "Human error" of 5 above and below. I'm going to apply your reasoning so it will help me draw my angles. Thank you very much, really appreciate this! If I were going to draw a force vector map what would it look like? Would I still apply the force of gravity if the object is suspended? And since T3 was practically given would I still include this? [editline]19th October 2011[/editline] Also sorry for mentioning this, though the various T's need to be in place so I get the measurements of tension on the cords correct, as ive ruled so far T3 is equal to 30N, since it's attached to the 3KG weight. I apologize for my rather crude drawing.[/QUOTE] What a shit way to do it. Trig is the way to go. You would apply gravity if it is suspended, yes. Gravity always applies. However, Newton's laws dictate that for every action there is an equal and opposite reaction. If it'll make more sense: consider a diagram for a brick sitting on a table, there would be gravity as a force going down, and an [I]equal[/I] force going up (the table pushing on the brick, resisting motion). In that case (but not this case) the force is called the Normal Force. "Normal" doesn't mean "not odd", it's just a name. [img]http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Dynamics/gifs/Forces10.gif[/img] Anyway, as far as how you'd do what your teacher is instructing you to do: The object is in equilibrium so the sum of the forces is equal to zero. So if we draw a force vector map, it will end up at the origin 0,0. [img]http://i.imgur.com/WX20y.png[/img] [B]Don't use that cause it's shit done in MSPaint.[/B] Create your own. Use a protractor to create proper angles, off of the original 30cm, 30N line we know (the vertical one on the left). The lines will intersect, forming a triangle. Then measure the sides. [editline]18th October 2011[/editline] Also what class are you in? Honors/General Physics? Physical Science?
[QUOTE=DerpHurr;32854002]I must say that was quite helpful, and I really appreciate the time you set aside to help me! Unfortunatly we aren't using trig to solve it, rather than measuring the angles and drawing lines until they intersect and then measuring the force in form of centimeters, for example 1 CM = 10N. My teacher set aside a margin of "Human error" of 5 above and below. I'm going to apply your reasoning so it will help me draw my angles. Thank you very much, really appreciate this! If I were going to draw a force vector map what would it look like? Would I still apply the force of gravity if the object is suspended? And since T3 was practically given would I still include this? [editline]19th October 2011[/editline] Also sorry for mentioning this, though the various T's need to be in place so I get the measurements of tension on the cords correct, as ive ruled so far T3 is equal to 30N, since it's attached to the 3KG weight. I apologize for my rather crude drawing.[/QUOTE] Not a problem dude. As I said, it helped me too (as I'm currently studying the same thing as you), so it's win-win. But like Elecbullet said, I don't know why your teacher want you to use another method, the one I showed you proved to be the more efficient. But oh well, if that's the case, I can't really provide any help because I've never done it that way.
[QUOTE=Elecbullet;32854383]What a shit way to do it. Trig is the way to go. You would apply gravity if it is suspended, yes. Gravity always applies. However, Newton's laws dictate that for every action there is an equal and opposite reaction. If it'll make more sense: consider a diagram for a brick sitting on a table, there would be gravity as a force going down, and an [I]equal[/I] force going up (the table pushing on the brick, resisting motion). In that case (but not this case) the force is called the Normal Force. "Normal" doesn't mean "not odd", it's just a name. [img]http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Dynamics/gifs/Forces10.gif[/img] Anyway, as far as how you'd do what your teacher is instructing you to do: The object is in equilibrium so the sum of the forces is equal to zero. So if we draw a force vector map, it will end up at the origin 0,0. [img]http://i.imgur.com/WX20y.png[/img] [B]Don't use that cause it's shit done in MSPaint.[/B] Create your own. Use a protractor to create proper angles, off of the original 30cm, 30N line we know (the vertical one on the left). The lines will intersect, forming a triangle. Then measure the sides. [editline]18th October 2011[/editline] Also what class are you in? Honors/General Physics? Physical Science?[/QUOTE] I'm just in regular Physics. It gets difficult at times I will admit, but i'm not that bad at it. Also, thank you very much for the force vector map, that's the thing I mainly have trouble with, for some reason it's easy for me to identify the forces at hand but when I put them on a vector map it gets hard. By the way everyone, I really appreciate all your help. Seriously, it's nice to have a place to go to when you need help with things and you guys provided it.
In the future OP, we have a thread for this kind of stuff [url]http://www.facepunch.com/threads/1091604[/url] People are almost always willing to help out especially if you're polite
[QUOTE=Turnips5;32857568]In the future OP, we have a thread for this kind of stuff [url]http://www.facepunch.com/threads/1091604[/url] People are almost always willing to help out especially if you're polite[/QUOTE] Ah cool, thank you very much - I will perhaps use this in the future
Draw a scale model, representing the forces in the digram of a triangle. Then take your measurements. Use trigo if possible/necessary [editline]19th October 2011[/editline] Never mind, I am late
This thread makes me sad that I will never be this smart
[QUOTE=Don Knotts;32862274]This thread makes me sad that I will never be this smart[/QUOTE] Don't say that. You just have to learn with a teacher in order to understand proprely how this works. This problem wasn't even that hard, trust me. However, sometimes in physics, shit can get waaaay out of hand and become over-complicated easily.
is it 4? YES, yes, the answer must be 4.
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