[img]http://uppix.net/c/a/4/82a907dfca0f0e40a4cd5fecedfaf.jpg[/img]
Okay for the first question, I say that the girl will have her swing break because the tensions of the rope are greater as compared to the tensions on the boy's swing. If I apply parallelograms to the girl's swing then that means the resultant of the two vectors of the string would be...greater?
And as for the second, it is impossible since the string will inevitably sag because the additional force will act upon the string?
So basically, trying to understand, the resultant has to be equal to the opposite weight of an object in order to have it in equilibrium. In order for that to occur, the two vectors are to be set at a wider angle (more tension) or a narrower angle (less tension)..?
And also, the length of a vector also displays the amount of magnitude it has right? So..longer the arrow, the more tension?
What?
The girl because she has more force pushing down on her at the bottom of the swing.
The second one is no because if you set up a system of vectors, the string is keeping the object up, and it has to counteract the weight somehow, even if the string were infinitely long, it would sag a little bit.
-snip-
Silly me.
[QUOTE=matark;23369046]The girl because she has more force pushing down on her at the bottom of the swing.
The second one is no because if you set up a system of vectors, the string is keeping the object up, and it has to counteract the weight somehow, even if the string were infinitely long, it would sag a little bit.[/QUOTE]
So that means you can't have an infinite tension???
Fuck, we are doing this in higher physics right now in school.
Scalars and Vectors and shit.
oh, the ropes are set differently
Well the answers are obviously 2.5.1 and 2.5.2
[QUOTE=CabooseRvB;23368955][B]So basically, trying to understand, the resultant has to be equal to the opposite weight of an object in order to have it in equilibrium. In order for that to occur, the two vectors are to be set at a wider angle (more tension) or a narrower angle (less tension)..?[/B][/QUOTE]
Ok, I haven't taken physics in awhile but I did take college level trig last semester and I can verify that the bolded statement is correct.
A vector carries two pieces of information, direction and magnitude. The "resultant" is a combination of two. If the vector of the force exerted by an object (W), is equal to the vector of whatever is holding it up at PI radians opposite W, then the two vectors will cancel each other out and the magnitude of the resultant will equal zero. Thus whatever object you have is in a state of equilibrium.
That's just what I remember, I'm sure I could dig up some of my notes if you really need help.
You taking summer school? I just got done taking Calculus I and now I'm taking CHEM 106 (Organic chemistry).
Oh, and I don't know how to explain it in terms of vectors. But it's obvious that the girls swing is more likely to break. (As the tension will be focused in the center of the ropes instead of the ends).
The boy. He looks fatter.
Woot sweet. I got it now.
The purple kid looks a bit fat, so I'm gonna put my money on him breaking the rope first.
But with today's safety standards, it's unlikely for them to break the swings at all.
Where the hell do you go to school that you don't have summer break?
Summer break is a waste of life.
I'm taking summer classes + working part time and I still have enough time to chill out, party, and enjoy the summer.
So some insight into the first problem.
There are three key forces involved in one of the rope. The tension force holding it to the frame, the normal force holding to the swing seat, and the force of gravity pulling it down. The tension force (Ft) is at an angle, so you can decompose it into its x and y components, Ftx and Fty. On the vertical cable, the force of gravity (Fg) is counteracted by Ft. On the angled cable, it's counteracted by Ft*sin(theta), the angle the cable makes with the frame/horizontal. You can visualize it as the shadow the cable makes on the y-axis. Thus, for an equally weighted person, the tension has to be greater in the angled cable to make its y-component force to be equal to the force of gravity.
Second problem: Draw a free body diagram. What forces have a vertical component, in other words, forces that counteract the force of gravity?
Man, I hated vectors when I was in Honors Physics.
The first is obviously the girl due to centrifugal force. [[i]Nevermind, misread the problem.[/i]]
The second, I don't know about all your tension talk but I see it in form of elasticity. In order to apply a reactive force to gravity, the string MUST bend to induce an elastic counterforce. No bending, no elastic force, no suspension.
In Linear Algebra you work with vectors, without any physics. Tough stuff.
[QUOTE=Emperorconor;23369112]Fuck, we are doing this in higher physics right now in school.
Scalars and Vectors and shit.[/QUOTE]
Really? I studied this shit in year 10.
The boy. He is moving at a greater velocity. The formula for Centripetal force is Force=Mv^2/r. The girl is moving at a relatively low velocity, while the boy is moving at the greatest velocity either of them will achieve in the swing.
Also, yeah the string can never be perfectly straight, but you got that right.
[QUOTE=redonkulous;23370440]The boy. He is moving at a greater velocity. The formula for Centripetal force is Force=Mv^2/r. The girl is moving at a relatively low velocity, while the boy is moving at the greatest velocity either of them will achieve in the swing.[/QUOTE]
Wrong, nothing is given in regards to their velocities. The question doesn't even state that they're moving.
This is true. I assumed they were both moving. The girl would have to be moving to be up there though.
[editline]11:13PM[/editline]
If the boy really is stationary then yeah the girl will snap the string since she is up in the air and shit, but it all depends on if they are moving or not. If they are moving and the boy is at the bottom of the arc then he is going to have more tension on the chain.
The boy, because his chains are taking f=mg while the girl's chains are only taking f=cos(a)*mg
The first question is bullshit because it doesn't tell you if they're moving. If they're not moving the whole demonstration picture seems useless because the physics of that girl sitting there at that angle without motion are nowhere near the same as her swinging.
[QUOTE=huntskikbut;23371211]The first question is bullshit because it doesn't tell you if they're moving. If they're not moving the whole demonstration picture seems useless because the physics of that girl sitting there at that angle without motion are nowhere near the same as her swinging.[/QUOTE]
You don't get it do you? We're talking about how we can describe the angles of the ropes as vectors and talk about the tension created on the ropes.
I.E. You're missing the point.
I'd say the girl. Because my inner voices said so. :tinfoil:
[QUOTE=redonkulous;23370580]This is true. I assumed they were both moving. The girl would have to be moving to be up there though.
[editline]11:13PM[/editline]
If the boy really is stationary then yeah the girl will snap the string since she is up in the air and shit, but it all depends on if they are moving or not. If they are moving and the boy is at the bottom of the arc then he is going to have more tension on the chain.[/QUOTE]
She isn't moving in the first question, her ropes are just placed further apart.
I can't stand how this stuff in school is taught so poorly and in a misleading fashion. If you work with this stuff on a daily basis, you understand it in a way that's so much more conceptual than procedural.
I gave that about 2 minutes before giving up. Good luck with that OP.
[QUOTE=Athena;23369890]So some insight into the first problem.
There are three key forces involved in one of the rope. The tension force holding it to the frame, the normal force holding to the swing seat, and the force of gravity pulling it down. The tension force (Ft) is at an angle, so you can decompose it into its x and y components, Ftx and Fty. On the vertical cable, the force of gravity (Fg) is counteracted by Ft. On the angled cable, it's counteracted by Ft*sin(theta), the angle the cable makes with the frame/horizontal. You can visualize it as the shadow the cable makes on the y-axis. Thus, for an equally weighted person, the tension has to be greater in the angled cable to make its y-component force to be equal to the force of gravity.
Second problem: Draw a free body diagram. What forces have a vertical component, in other words, forces that counteract the force of gravity?[/QUOTE]
I believe this is correct.
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