## anonymous 5 years ago physics question, suppose two people pull on rope. person A is 50 kg and pulls towards the right (+) with 30 N force. person b who is 40 kg pulls towards the left with a 20 N force. assume the rope is massless. I dont get why the tension in the rope is the same magnitude. For the rope the forces on it are F1 - F2 = 0*a. but F1 and F2 should be the 30N and 20N , so it should be not zero but 10 N to the right for tension. something doesnt make sense here

1. anonymous

You are suggesting that at one part of the rope, there is 10 newtons of tension, and at the other end there is 0 tension (i.e. slack)?

2. anonymous

I'm not sure why you're looking at the forces on the rope. You're assuming it's massless, so you can't really talk about the rope's acceleration.

3. anonymous

hi

4. anonymous

well i was using my book. it says the forces on the rope are F1 - F2 = 0. isnt tension the force on the rope?

5. anonymous

What are you being asked to find?

6. anonymous

A rope is one element. It can't have variable force along it. So whatever the tension is in the rope, it has to be the same all along the rope.

7. anonymous

Here is the full paragraph. "Suppose two people pull the ends of a rope with oppositely directed forces F1 and F2 (in bold). The rope also exerts forces -F1 and -F2 on the people (in bold)From the second law, we know F1 - F2 = ma , where m is the mass of the rope and a is its acceleration. If that acceleration is zero , or if the mass of the rope is so small that we can idealize the rope as having zero mass, then F1 - F2 = 0. In this special situation the forces exerted by the rope on the two people are equal in magntitude

8. anonymous

I agree with that.

9. anonymous

well im wondering, what is the force on the rope , if it says F1 - F2 = 0, then F1 = F2, but 30 N does not equal 20 N

10. anonymous

What it's actually saying is that it's impossible for person A to be pulling with 30N of force and person B to be pulling with 20N of force and for there to be zero acceleration. But you know that already since: $$F=ma$$ If F were not 0, you'd have an acceleration.

11. anonymous

in my example, one guy pulls 30 N right and the other guy pulls 20 N left

12. anonymous

I don't see where you got 30N and 20N

13. anonymous

14. anonymous

In your example, there would be acceleration towards the guy pulling with 30N.

15. anonymous

you mean 10 N

16. anonymous

There would be a net force of 10N which would result in an acceleration in the direction of the guy pulling with 30N force.

17. anonymous

Yes, but you are contradicting the hypothisis

18. anonymous

my question is, what is the force on the rope? and why does it say the force exerted by the rope on the two people is equal in magnitude. and which force is the tension. so there are 3 questions

19. anonymous

They are saying that If the rope is not accelerating, we know the forces on both ends are equal

20. anonymous

how do you get the forces at both ends are equal, and what is this force? the tension?

21. anonymous

and which force is it, what is exerting on what

22. anonymous

you said the forces at both ends are equal. what do you mean by that. the force that the rope exerts on the two people are equal?

23. anonymous

No, the force on the rope.

24. anonymous

but there are different forces, one is 30 N and one is 20 N

25. anonymous

which we stipulated

26. anonymous

Yes, you made up that example, where the conclusion is false, and trying to make a claim about the implication.

27. anonymous

the conclusion is false?

28. anonymous

you said the force on the rope at both ends is equal. clearly this is false, it is 20 N and 30 N , they are not equal

29. anonymous

So the acceleration is not 0

30. anonymous

wait, youre saying the net force on both ends is equal in magnitude?

31. anonymous

but my rope is massless

32. anonymous

I'm saying IF the acceleration is 0, then the NET force on the rope is 0.

33. anonymous

why does the book say F1-F2 = 0, so F1 = F2,

34. anonymous

You can't talk about force acting on massless things.

35. anonymous

what is F1 and F2 ?

36. anonymous

It says F1-F2 is 0 BECAUSE it claimed already that the acceleration is 0

37. anonymous

so then F1 and F2 are zero?

38. anonymous

No.

39. anonymous

hmmm

40. anonymous

is tension the F1 and F2

41. anonymous

$\sum F = ma$

42. anonymous

If a =0 then $\sum F = F1-F2 = 0 \implies F1 = F2$

43. anonymous

But if the forces do not sum to 0 than the acceleration of the rope will not be 0.

44. anonymous

yes but in my example, what is F1 and F2

45. anonymous

is the acceleration equal to 0?

46. anonymous

nomo, if you cant type , you should restart.

47. anonymous

www.openstudy.com start over

48. anonymous

A review of massless rope mechanics: A rope carries a tension. This tension is the force felt at the endpoints of the rope by whatever the rope is attached to. So, if you pull against a rope caught tied to a wall: |------ -> 30N the tension in the rope is 30N. But, by action/reaction, the wall is also pulling on the rope. And since there is no acceleration, it is pulling with an equal and opposite force of -30N. Now let's look at your example. You now have two people, pulling on the rope. One is pulling harder than the other. Without doing the math, it's obvious that whoever is pulling harder is going to move the other person. That's exactly what happens: 20N <--------> 30N --> everything moves this way So your next question is, what's the tension in the rope? Well, the tension in the rope has to be 20N, because it's the force the weaker puller is pulling with. The rest of the force of the stronger puller, 10N, goes into acceleration. So: 1. The tension in the rope is 20N. 2. The net external force on the rope is 10N. Hope this helps.

49. anonymous

lets see

50. anonymous

51. anonymous

30 N - T = 50kg * a -20 N + T = 40 kg * a

52. anonymous

Heh. Cantorset, you are in fact correct.

53. anonymous

Those are the equations of motion, and I should have written them out before posting.

54. anonymous

so 10 N = 90 kg *a , so a = 1/9 m*s^-2

55. anonymous

Yes, I agree.

56. anonymous

So what is T?

57. anonymous

one sec, my browser is crashing

58. anonymous

weird, one sec

59. anonymous

plug back in

60. anonymous

so 30 - 50/9 N = T

61. anonymous

24.444 N is the tension

62. anonymous

so natho was wrong

63. anonymous

The problem with this exercise is that in the absence of any frictional forces, you 're going to have that both people accelerate toward each other which will remove the tension.

64. anonymous

so what is F1 and F2, i still dont see

65. anonymous

So your acceleration is not constant.

66. anonymous

F1 and F2 are the tension forces?

67. anonymous

F1 is the force of one person pulling, F2 is the force of the other.

68. anonymous

but thats false, because 30 - 20 is not 0

69. anonymous

Who said it was = 0?

70. anonymous

is the a = 0?

71. anonymous

The equations of motion that cantorset wrote are valid at the instant when both people start pushing, as are the results. However, polpak is correctly pointing out that this is a dynamic problem, not a static problem. Odds are you'll never actually find a problem like this in a problem set, because there is too much information missing to be able to give a complete answer.

72. anonymous

nomo, so do you agree that 24.444 is the tension?

73. anonymous

74. anonymous

In the instant when they both pull, (and in the absence of any friction) The tension is 24.444

75. anonymous

Is the acceleration given to be 0?

76. anonymous

"Suppose two people pull the ends of a rope with oppositely directed forces F1 and F2 (in bold). The rope also exerts forces -F1 and -F2 on the people (in bold)From the second law, we know F1 - F2 = ma , where m is the mass of the rope and a is its acceleration. If that acceleration is zero , or if the mass of the rope is so small that we can idealize the rope as having zero mass, then F1 - F2 = 0. In this special situation the forces exerted by the rope on the two people are equal in magnitude, and the rope can be thought of as simply transmitting a force from one person to the otherl the force at any point in the rope is referred to as tension. It is the same everywhere in the rope only if the rope is unaccelerated or if is idealized as massless

77. anonymous

No, but the rope is given to be massless

78. anonymous

So the acceleration is 0.

79. anonymous

But that's not a problem, it's a lesson.

80. anonymous

You CANNOT have forces acting on massless things.

81. anonymous

Your example is a problem, and it's a problem not covered by this lesson, since it clearly stipulates that the acceleration is 0; in your case, it isn't.

82. anonymous

hmmm, so the book is contradicting itself?

83. anonymous

No.

84. anonymous

You are trying to contradict the book

85. anonymous

But not making a reasoned argument.

86. anonymous

No, you're contradicting the book. You are constructing a situation where the acceleration cannot be zero, and trying to analyze the problem as if it could.

87. anonymous

as if it were 0.

88. anonymous

ok , i know the acceleration is not zero

89. anonymous

So then the book doesn't apply anymore.

90. anonymous

Then the forces cannot be the same.

91. anonymous

ok lets look at the problem again. the forces on the rope is F1 - F2 = ma , the mass of the rope is neglible. so we have F1 - F2 = 0

92. anonymous

Therefore the F1 = F2

93. anonymous

If you contradict by saying $F1 \ne F2$ then $ma \ne 0.$

94. anonymous

the end.

95. anonymous

but didnt we stipulate earlier that F1 = 30 N and F2 = 20 N

96. anonymous

You cannot say that ma = 0 if f1 not equal f2

97. anonymous

The sum of the forces IS ma.

98. anonymous

ok, hmmmm

99. anonymous

If the sum is non-zero ma is non-zero. If ma is 0 the sum is 0. If the sum is 0 ma is 0. If ma is non-zero the sum is non-zero.

100. anonymous

but in my problem, i started with 30 N and 20 N , and they applied the forces at either end, correct?

101. anonymous

You can *only* assume the rope is massless *IF* F1=F2. Otherwise, the discussion in the book doesn't apply.

102. anonymous

So while there are real-world situations where F1 is not equal to F2, you can't analyze them using what the book says. It *only* applies for the special case where F1=F2.

103. anonymous

but didnt we assume that the two guys are applying two different forces

104. anonymous

We did. But then you tried to use what the book told you. You can't do that.

105. anonymous

If forces are non-zero then ma is non-zero.

106. anonymous

Pick one or the other, but you can't just pick both to be different.

107. anonymous

ok let me try again, one sec

108. anonymous

person A is 50 kg and pulls towards the right (+) with 30 N force. person b who is 40 kg pulls towards the left with a 20 N force. assume the rope is massless.

109. anonymous

You can't do that.

110. anonymous

You cannot say the sum of the forces is not zero and ma is 0. You cannot say that ma is not zero but the sum of the forces is 0. Because the sum of the forces IS ma. If you pick one you are also choosing the other.

111. anonymous

You can't assume the rope is massless.

112. anonymous

they do that in problems though, i thought

113. anonymous

No.

114. anonymous

They do that when the forces are equal.

115. anonymous

You are trying to do something like: Let $$1 = 0 \implies 2+1 = 2$$ And then complaining because you know 2+1 should be 3.

116. anonymous

i dont understand , what is wrong with my problem

117. anonymous

You have ma = 0, but the sum of the forces is not 0.

118. anonymous

You cannot do that.

119. anonymous

If it is not accelerating then there is no net force on the object.

120. anonymous

Things do not accelerate in the absence of a net force.

121. anonymous

im confused

122. anonymous

Things with net forces acting on them MUST accelerate.

123. anonymous

so what exactly is the contradiction in the question

124. anonymous

nothing wrong with assuming the rope is massless person A is 50 kg and pulls towards the right (+) with 30 N force. person b who is 40 kg pulls towards the left with a 20 N force. assume the rope is massless.

125. anonymous

You try to claim $$m = 0 \implies ma=0$$ But $\sum F = ma$ And you pick two forces also that don't sum to 0.

126. anonymous

We are telling you THERE IS SOMETHING WRONG with that assumption.

127. anonymous

ok

128. anonymous

so what is the force on the rope

129. anonymous

Because what happens in reality is that ma is not 0, You will get a tension, the tension will pull the two people together

130. anonymous

oh you said there is no force on the rope, since its massless

131. anonymous

I'm not saying anything. You are choosing two sides of the equation that are non-equal

132. anonymous

wait, you cant have tension with a massless rope?

133. anonymous

Which do you want, a rope that doesn't accelerate? or a net force?

134. anonymous

Ive seen many tension problems where we assume the mass of the rope is negligible

135. anonymous

Was the rope accelerating?

136. anonymous

i dont know

137. anonymous

I do.

138. anonymous

one sec, let me get a problem

139. anonymous

Good.

140. anonymous

An important property of massless ropes is that the total force on the rope must be zero at all times. To prove this, we go back to Newton's Second Law. If a net force acts upon a massless rope, it would cause infinite acceleration, as a = F/m , and the mass of a massless rope is 0.

141. anonymous
142. anonymous

Right. So if you assume the rope is massless you have to have it not accelerating. If it is not accelerating the net force must be 0. If the net force is zero, each force acting on the rope must be cancelled by another foce.

143. anonymous

so youre saying, if you have a 40 kg guy and a 50 kg guy and they are pulling on a rope in opposite directions, and the rope is massless, then

144. anonymous

then the force exerted by the rope on the two people must be equal

145. anonymous

yes

146. anonymous

but then the two people, what about the force of the two people exerting on the rope , or the forces on the rope

147. anonymous

The net force they exert on the rope must be 0.

148. anonymous

but they can have different forces as well

149. anonymous

Yes

150. anonymous

but then by newtons third law, then that would force them to have the same force

151. anonymous

the reaction force

152. anonymous

Wait, no

153. anonymous

They cannot be exerting different forces.

154. anonymous

If they were then the sum would not be 0.

155. anonymous

ok lets say we have a pulley and two masses

156. anonymous

Ok.

157. anonymous

one mass is 50 kg, the other mass is 30 kg

158. anonymous

m1 = 50 kg, m2 = 30 kg

159. anonymous

they go up and down, like vertically

160. anonymous

yes

161. anonymous

simple pulley. the force on m1 are 9.8 *50 - T , where + is down direction

162. anonymous

Certainly.

163. anonymous

the force on m2 is T - 30 *9.8

164. anonymous

but the problem says the mass of the cable and the pulley is neglible

165. anonymous

that means don't try to use it for finding the mass

166. anonymous

, wait, you said then the acceleration is zero

167. anonymous

but the pulley is accelerating, i mean the cable is accelerating

168. anonymous

You just keep talking in circles. What exactly is the question?

169. anonymous

the force on the cable should be zero

170. anonymous

since it is massless, but there is force on the cable

171. anonymous

It's not massless.

172. anonymous

thats what the problem states

173. anonymous

It says ignore the mass, not that it's massless.

174. anonymous

In all these cases you are going to assume that the mass of the rope is ignorable, and that the tension in the rope is uniform.

175. anonymous

i think it says the mass is zero

176. anonymous

It's telling you to ignore it because otherwise it becomes impossible to solve with your current tools. IT CANNOT BE 0 everything (even light) has mass. Forces cannot act on objects with no mass.

177. anonymous

it says let it be zero

178. anonymous

I'm not going to continue this further. If you want to argue further you can contact the authors, or your professor.

179. anonymous

If the rope has mass m, then Newton’s second law applied to the rope gives T - T'= ma. If the mass m is taken to be zero, however, as in the upcoming examples, then T = T'.

180. anonymous

What is the acceleration of a massless object with a force of 1 newton acting on it?

181. anonymous

here i will show you a link , one sec http://www.wadsworthmedia.com/marketing/sample_chapters/0495106194_ch04.pdf page 70

182. anonymous

I don't disagree with what they state there.

183. anonymous

All they are saying with the 'massless rope' thing is that the tension in both ends is the same. It's a trick used to simplify your problems

184. anonymous

in reality that cannot happen.

185. anonymous

But if you violate the precepts of the 'trick' then you cannot use the rules they infer from it.

186. anonymous

but if the tension is the same, then it wont move

187. anonymous

I'm not talking about the external force on the rope.

188. anonymous

the tension is the internal force.

189. anonymous

The point is that it is not slack, it does not break, and it is not elastic.

190. anonymous

You can have it move, but you cannot try to find it's acceleration in the absence of other objects because it has no mass.

191. anonymous

It's a tool, it is not real.

192. anonymous

It is a way to magically give direction to forces acting on objects.

193. anonymous

here is a diagram of a pulley http://www.sparknotes.com/physics/dynamics/newtonapplications/section3.rhtml

194. anonymous

it says specifically the pulley and cable are massless

195. anonymous

Do you have a question, or are you just going to keep repeating the same thing and pasting examples from the book.

196. anonymous

I have read the book. I don't need to see it again.

197. anonymous

ok if the pullies and cable are massless then

198. anonymous

how can it have acceleration

199. anonymous

I'm done. Until you actually read what I have written I have no reason to continue to write.

200. anonymous

so your saying the cable experiences no acceleration, ok i agree

201. anonymous

I'm not saying that.

202. anonymous

well if it is massless then it experiences no acceleration. the cable i mean

203. anonymous

I'm saying massless cables do not goddamn exist. If you want to talk about them then you have to obey very specific rules.

204. anonymous

right, its idealized

205. anonymous

well actually kevlar ropes exist which can carry so much weight, that in comparison to the weight it is massless

206. anonymous

They are a figment of your imagination. A tool to make it easy for you to solve these problems without a hell of a lot of integrals.

207. anonymous

ok , so then my original problem was impossible, having a massless rope and 20 N left and 30 N right, force

208. anonymous

On a rope by itself, yes.

209. anonymous

Because F = ma therefore F/m = a and m = 0 so F/m is not defined.

210. anonymous

right i agree

211. anonymous

the accelleration is not 0, it's exploding.

212. anonymous

The only time it is ok to talk about external forces on massless objects is when those forces cancel out (a = 0)

213. anonymous

ok then can you think of an example where we want a massless rope?

214. anonymous

The pulley example is fine.

215. anonymous

As long as you talk about the a of the masses, not the a of the rope.

216. anonymous

ok

217. anonymous

All you need to consider about the rope is that the tension is uniform.

218. anonymous

so the point of having a massless rope is to make or force the tension to be the same throughout , it simplifies the problem

219. anonymous

yes

220. anonymous

even though intuitively, there are different forces on the ends of the rope

221. anonymous

so you might think there are two different tensions on the ends of the rope

222. anonymous

like in my example, there would be a tension of 20 N opposite to the guy pulling left 20N , and a tension of 30 opposite to the guy pulling to the right 30 N ... well i guess in this case we need to know the mass of the rope to determine the tension err, i mean .... the tension does not hold because it is accelerating , so there isnt tension , in the case where there is mass in the rope

223. anonymous

look at the pully example. You have that 50*9.8 - T = (50+30)a 30*9.8 + T = (50+30)a So we can find T, and we can find a because we assume that T and a are the same (one block accelerates up while the other accelerates down. We know the rope is accelerating at the same rate as the blocks. So there is a net force on the rope but the tension is uniform. Also the rope is not massless, we just ignore it.

224. anonymous

supposing that in my example the mass of the rope is 1 kg. so a guy is pulling 20 N to the left and 30 N to the right someone else pulls.

225. anonymous

what is the tension in the rope

226. anonymous

you're making up impossible scenarios again.

227. anonymous

just to make it simple , 1 kg rope

228. anonymous

Are you assuming the rope stays taut?

229. anonymous

yes

230. anonymous

oh i see what you mean

231. anonymous

Are you assuming that the guys are standing on the ground and don't fall over?

232. anonymous

yes

233. anonymous

Does the ground have friction?

234. anonymous

or with a pulley example , even

235. anonymous

yes

236. anonymous

so its much more complicated

237. anonymous

ok pulley, 50 kg moving down, 40 kg moving up. the cable is 1 kg

238. anonymous

what is the tension in the cable,

239. anonymous

this will be my last example, sorry for keeping ya

240. anonymous

i was wondering if that made sense

241. anonymous

Yes, that's fine.

242. anonymous

also ignore gravity for the cable,

243. anonymous

yes

244. anonymous

how do you calculate the tension of the rope , if it has mass 1 kg

245. anonymous

The same was you would if it had no mass.

246. anonymous

ok so the forces are 50* 9.8 - 40 * 9.8 , the forces on the cable

247. anonymous

No.

248. anonymous

50* 9.8 - 40 * 9.8 = 1kg * a

249. anonymous

Stop that

250. anonymous

the net force on the cable

251. anonymous

No.

252. anonymous

If you do that the rope will fly away.

253. anonymous

Stop obsessing about the stupid rope. It doesn't work like that in reality and you don't have the tools to look at it that way. Look at the masses.

254. anonymous

We MUST assume that they all accelerate at the same rate. Otherwise we have a dynamics system and you don't have the PDE's to solve that.

255. anonymous

The blocks, and the ropes all move with the same a.

256. anonymous

ok

257. anonymous

ok so dont treat the rope independently of the masses

258. anonymous

Yes, because if you have a 40N force acting on a 1kg rope it will go shooting off into space.

259. anonymous

And if the mass of the rope is smaller, it will fly away faster.

260. anonymous

They just want you to ignore the rope, not make any calculations based on its mass or acceleration.

261. anonymous

oh actually its (50-40)* 9.8=98 Newtons force on the cable, ok . bad way of looking at it, also for my other example, two guys 30 N and 20 N right and left respectively, rope is 1 kg. the tension is still 24.4444, doing it the right way. dont treat the cable seperately

262. anonymous

ok got it. ignore the rope

263. anonymous

Yes, exactly.

264. anonymous

how come its not obvious the tension is 24.444

265. anonymous

the other guy assumed the tension was 20N.

266. anonymous

so with my other rope example, it would be wrong to do (30 - 20) = 1kg * a , 10N/1kg = a , or 10 m/s^2. it will fly off , ok i see

267. anonymous

it would shoot off in the direction of the stronger person

268. anonymous

right.

269. anonymous

interesting

270. anonymous

so this is why we dont deal with the rope directly

271. anonymous

If you say something like Two men are having a tug of war. One man (weighs 50kg) is pulling with a force of 20N, the other weighing 40kg is pulling with a force of 10N. What is the acceleration? and the tension in the rope.

272. anonymous

ok good question,

273. anonymous

Again you must assume the acceleration is a constant for the whole 2 men system

274. anonymous

If it werent then the tension in the rope would change

275. anonymous

so 10 N in the right direction?

276. anonymous

you can do 10 N = ( 40 + 50) * 1kg ?

277. anonymous

errr, 10 N = (40 + 50) kg * a

278. anonymous

Actually that's still a bad example because we have to assume some friction under their feet or they would just go toward each other. (and lose tension in the rope)

279. myininaya

ok i don't know anything about physics but it sounds like cantorset is trying to use a theorem. And it sounds like you guys are saying just because the conclusion is true doesn't mean the is. p->q is not equivalent to q->p or are you guys talking about somtehing different. You might have already figure this out so sorry if you have. if you don't have any clue what i'm talking about ignore this post

280. anonymous

so a = 1/9 or .1111 m/s^2

281. anonymous

yeah, that's what I was complaining about earlier myininaya. But I think we worked through it.

282. anonymous

but i ignored the tension

283. anonymous

in fact i ignored the rope altogether

284. anonymous

I like the pully problem better ;)

285. anonymous

This one still has too many unknowns.

286. anonymous

you can do 20 N - T = 50 * a T - 10 N = 40 * a

287. anonymous

oh right, we solved for a , so 20 N - T = 50 * 1/9

288. anonymous

T = 14.4444

289. anonymous

Sure, that will tell you how fast they fly together.

290. anonymous

hmmm

291. anonymous

And how tight the rope was at the instant they both pulled (before they start moving)

292. anonymous

oh right, so we would have friction , and...

293. anonymous

Though it's not quite right because you did the forces wrong

294. anonymous

yes this only works for an instant at the beginning before they actually start moving

295. anonymous

i did ?

296. anonymous

Two men are having a tug of war. One man (weighs 50kg) is pulling with a force of 20N, the other weighing 40kg is pulling with a force of 10N. 20 N - T = 50 * a (force on bigger guy) T - 10 N = 40 * a (force on smaller guy )

297. anonymous

The net acceleration of the system will be 0, and you'll do stuff like this when you get to center of mass, but the system you have there doesn't correctly model what happens.

298. anonymous

you dont agree the tension is 14.444 N , in the idealized case

299. anonymous

what do you mean net acceleration is zero. oh you mean that the people are staying in their places and not moving (so the string is taut)

300. anonymous

Imagine for a moment that it's an astronaut(75kg) in space, pulling (20N) on a cord attached to a free floating 50 kg backpack. The astronaut pulls the rope. The rope pulls the astronaut.

301. anonymous

This will work.

302. anonymous

why not 2 different astronauts in space

303. anonymous

304. anonymous

ok so whats the question?

305. anonymous

what is tension in rope

306. anonymous

What is the acceleration of the backpack?

307. anonymous

isnt it just 20 N = 50kg * a

308. anonymous

20 /50 m s^-2

309. anonymous

yep.

310. anonymous

ok good

311. anonymous

And what is the acceleration of the astronaut?

312. anonymous

we never said the astronaut is accelerating

313. anonymous

hes just pulling, so maybe zero

314. anonymous

he's floating in space, pulling on a cord attached to a 50kg backpack.

315. anonymous

he's definitely gonna accelerate.

316. anonymous

because the bookbag pulls back on him ?

317. anonymous

is he exerting a force?

318. anonymous

he is pulling on a rope

319. anonymous

then the rope must also be pulling on him

320. anonymous

right with equal and opposite force

321. anonymous

so 20/75 m*s^-2

322. anonymous

in the opposite direction to bookbag

323. anonymous

20 N = 75 kg * a

324. anonymous

Right, so they start flying toward eachother.

325. anonymous

right

326. anonymous

And the tension in the rope?

327. anonymous

hmmm

328. anonymous

(for one instant anyway)

329. anonymous

oh

330. anonymous

one sec

331. anonymous

20 N - T =

332. anonymous

20 N - T = 75 * a , and a = 20 /75

333. anonymous

so T = 40 N ?

334. anonymous

Hold up.

335. anonymous

You are listing the forces on the astronaut?

336. anonymous

yes, but there is tension in that force, so im solving for the tension

337. anonymous

What forces are on the astronaut?

338. anonymous

oh ,

339. anonymous

lets see

340. anonymous

we know astronaut is accelerating 20/75 m * s^-2

341. anonymous

20 N

342. anonymous

duh, equal and opposite force

343. anonymous

right.

344. anonymous

Now lets do the same thing, but with 2 astronauts.

345. anonymous

so 20 N - T = 75* a

346. anonymous

wait, whats the tension?

347. anonymous

No.

348. anonymous

Name the forces on the astronaut. (don't bother with the number)

349. anonymous

the forces on the astronaut are 20 N ....

350. anonymous

yes. And what is exerting that force?

351. anonymous

ummm, just the tension ?

352. anonymous

yes

353. anonymous

and the tension comes from the equal and opposite force

354. anonymous

yes

355. anonymous

ok so the tension is , instantaneously 20 N

356. anonymous

yes

357. anonymous

we dont even have to solve for that

358. anonymous

right.

359. anonymous

So now lets analyze two astronauts. If one is just floating and the other pulls, we have the same situation as we do in the backpack case right?

360. anonymous

ok 2 astronauts

361. anonymous

yes

362. anonymous

so lets do both pulls >

363. anonymous

Will the tension in the rope be greater, the same, or less if both astronauts pull?

364. anonymous

i dont know

365. anonymous

hmmm, less ?

366. anonymous

well if they both pull 20 N say

367. anonymous

Hold up.

368. anonymous

Imagine only one pulls.

369. anonymous

ok

370. anonymous

They both fly together right?

371. anonymous

yes

372. anonymous

like our bookbag case

373. anonymous

Will they fly together just as fast if they both pull, or will it be slower, or faster?

374. anonymous

dont know

375. anonymous

without thinking about it, faster >?

376. anonymous

it depends on the pulls, are they equal

377. anonymous

even if they are equal.

378. anonymous

oh much faster

379. anonymous

indeed

380. anonymous

If they accelerate faster than they would if just one pulled, what can we say about the tension?

381. anonymous

then the tension must be greater

382. anonymous

Also correct.

383. anonymous

so lets say that they both pull 20 N force

384. anonymous

ok.

385. anonymous

20 N - T = 75 * a . and

386. anonymous

hold on

387. anonymous

What is that 20N?

388. anonymous

they are both pulling on each other 20 N

389. anonymous

No, they are both pulling on the rope with 20N

390. anonymous

so the m1 = 75 kg, and m2 = 50 kg , the astronauts

391. anonymous

yes but i will examine the forces on the astronauts, and glean out the tension

392. anonymous

indirectly

393. anonymous

True, but the force on the astronaut from the rope is not 20N.

394. anonymous

If it were, they would accelerate just as if only one was pulling.

395. anonymous

the 20 N is from the opposite reaction ,

396. anonymous

ohhh

397. anonymous

theres only tension then

398. anonymous

i keep thinking of the pulley example

399. anonymous

yes For astronaut 1, the only force on them is the tension.

400. anonymous

right theres no gravity

401. anonymous

so if we assume the tension is the same throughout, dunno

402. anonymous

the rope doesn't fly away.

403. anonymous

how to calculate tension

404. anonymous

well we add the forces and the reaction forces

405. anonymous

yes.

406. anonymous

so 2 * 20 + 20/75 + 50/75 ?

407. anonymous

no it's easier than that.

408. anonymous

Are you familiar with hanging scales?

409. anonymous

so i did it wrong >

410. anonymous

yeah

411. anonymous

you know like hanging scales in a grocery store?

412. anonymous

yes

413. anonymous

what was wrong with my analysis

414. anonymous

Ok well it tells how much the thing in the basket weighs by the tension pulling down.

415. anonymous

right

416. anonymous

If I put a 1kg mellon in the basket the scale will read 1kg. If I push down on the scale with a 9.8N of force what will the scale read?

417. anonymous

9.8 kg ?

418. anonymous

2kg actually. but that's ok. Point is my force is added to the tension from the mellon

419. anonymous

you mean if you push down with the melon in the scale, oh

420. anonymous

right, 1 kg is getting pushed down with 9.8 N of force

421. anonymous

and if you push down 9.8 N and keep the force there, its like adding another kg

422. anonymous

right.

423. anonymous

so how does that help us with the tension when two astronauts are pulling on each other 20 N

424. anonymous

If I accelerate the scale upwards at a rate of 9.8m/s^2 the scale will also read 2kg

425. anonymous

how can you accelerate a scale upwards?

426. anonymous

a rocket? my arms?

427. anonymous

a pully?

428. anonymous

we're talking about a scale in a store. it doesnt register upward forces

429. anonymous

oh you mean downwards, same thing, ok

430. anonymous

upwards meaning pushing towards the ground, ok

431. anonymous

No I mean upwards because accelerating the scale up will change the tension in the chain

432. anonymous

no I mean blasting the whole apparatus into the sky.

433. anonymous

and the melon is inside the scale ?

434. anonymous

I think I'm just confusing you further here. The point with the astronauts is this. If they both pull, the tension in the rope will be the sum of their individual tensions.

435. anonymous

and its not 20 + 20 + 20/75 + 20 / 50 ?

436. anonymous

No, it's 20 + 20 = 40

437. anonymous

438. anonymous

the newtons third law pairs

439. anonymous

oh

440. anonymous

that effects the astronauts

441. anonymous
442. anonymous

It's no good without pictures

443. anonymous

so the astronauts are accelerating, one is accelerating 20/75 and other is accelerating 20/50 m/s^2

444. anonymous

40

445. anonymous

40 , yes

446. anonymous

40/75 and 40/50, because the reaction to tension

447. anonymous

yes.

448. anonymous

hmm, and why do you add the tensions to get 40

449. anonymous

Hrm?