tag:blogger.com,1999:blog-5088860151651047897.post8158339822762648667..comments2017-06-23T05:35:34.263-04:00Comments on Jacobs Physics: Three masses connected over a pulleyGreg Jacobshttp://www.blogger.com/profile/03854009948036330746noreply@blogger.comBlogger13125tag:blogger.com,1999:blog-5088860151651047897.post-90224005468032518252017-05-18T15:38:11.640-04:002017-05-18T15:38:11.640-04:00the sum is a little tricky and I had forgotten eve...the sum is a little tricky and I had forgotten everything so it was a mind boostAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-46632611426630274282017-02-01T08:59:18.521-05:002017-02-01T08:59:18.521-05:00Thanks...I spent hours for thisThanks...I spent hours for thisAlperen Bulgurhttp://www.blogger.com/profile/07579442787256942270noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-73951020800707602162016-09-28T12:17:33.027-04:002016-09-28T12:17:33.027-04:00The tensions are different because the top rope ha...The tensions are different because the top rope has to hold back both of the masses or a total of 6m, the bottom rope is only holding back 4m. This is consistent with the answer as the top rope holds 1.5 the mass of the bottom and so has 1.5(with rounding) the tension of the bottom rope Matthew Wagonerhttp://www.blogger.com/profile/04337364726166416467noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-49639125022165792252016-05-09T23:46:59.467-04:002016-05-09T23:46:59.467-04:00why are the tensions different?
why are the tensions different?<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-64450464627009717142013-06-11T10:00:54.925-04:002013-06-11T10:00:54.925-04:00Derive an expression for tension & acc.Derive an expression for tension & acc.malik Baberhttp://www.blogger.com/profile/00760900958294515241noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-34194640651883044292012-05-13T18:05:26.805-04:002012-05-13T18:05:26.805-04:00Well that's three hours of review time i'l...Well that's three hours of review time i'll never get back..<br />Thanks for the heads up!Angelohttp://www.blogger.com/profile/13477097052010322960noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-16929910845915756652011-08-24T12:57:52.522-04:002011-08-24T12:57:52.522-04:00Thank you Master Jacobs, I have spent hours workin...Thank you Master Jacobs, I have spent hours working this problem every which way and I kept getting your answer; not the books. Thanks for validating my work.Dananoreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-88666762252774526702011-05-09T00:00:11.918-04:002011-05-09T00:00:11.918-04:00oh i just read that comment above - the answer for...oh i just read that comment above - the answer for #6 in the 2010-2011 edition, the answer is a = 8.6 m/s^2, T1=19 N, and T2=9 NBeahttp://www.blogger.com/profile/09281061236059408771noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-37313833750990432862011-05-08T23:57:14.391-04:002011-05-08T23:57:14.391-04:00whoa crazy, i looked this up on google because i w...whoa crazy, i looked this up on google because i was having trouble with this exact problem!! thank you so much :)Beahttp://www.blogger.com/profile/09281061236059408771noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-27924671972621654062011-02-20T16:11:32.913-05:002011-02-20T16:11:32.913-05:00I had forgotten how to do this kind of problem, so...I had forgotten how to do this kind of problem, so this was so incredibly helpful!!!! Thanks!!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-1827244723267715412010-05-09T19:52:21.972-04:002010-05-09T19:52:21.972-04:00Rachel, that error has been corrected in the 2010-...Rachel, that error has been corrected in the 2010-11 edition. For that 2010-11 edition, we had someone go through every question and solution thoroughly. You're right that there were (and are) errors, but many were rooted out unmercifully.<br /><br />GCJGreg Jacobshttp://www.blogger.com/profile/03854009948036330746noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-40770553117516961452010-05-09T17:49:10.833-04:002010-05-09T17:49:10.833-04:00I'm not a physics expert, but I think there ar...I'm not a physics expert, but I think there are more errors than you realize...For example, on page 127 in the 08-09 book, the acceleration of a block on an inclined plane is written as = mu*g*cos(theta).<br /><br />Where did the "mgsin(theta)" go (from the original horizontal force summation? I keep getting that a = g(sin(theta)-mu*cos(theta))<br /><br />If the answer in the book is right, can you let me know what I'm doing wrong? Thanks.Rachelhttp://www.blogger.com/profile/04311287939533522470noreply@blogger.comtag:blogger.com,1999:blog-5088860151651047897.post-49860835696896806762010-05-08T16:11:54.399-04:002010-05-08T16:11:54.399-04:00I'm so glad I found this...I've been worki...I'm so glad I found this...I've been working on this problem for at least an hour and I wasn't getting the answer in the book! Thanks!!Rachelhttp://www.blogger.com/profile/04311287939533522470noreply@blogger.com