3.4 Pully System - GAMSAT

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3.4 Pully System Post a reply 15 posts • Page 1 of 1

3.4 Pully System

by gem » Thu Jul 26, 2012 8:13 am

Hi, I do not understand how the solved for T to equal 2m1m2/m1+m2 multiplied by g? Please explain. gem Posts: 24Joined: Mon Mar 12, 2012 11:12 pm Top

by goldstanda3269 » Thu Jul 26, 2012 1:07 pm

You are definitely expected to do the math because the steps were described. Try the first step: 2 equations of motion where described in section 3.4 and then the first step is "Solve for a using the equation for motion" goldstanda3269 Posts: 1766Joined: Wed Aug 25, 2010 10:59 pm Top

Thanks

by gem » Thu Jul 26, 2012 3:12 pm

Thanks gem Posts: 24Joined: Mon Mar 12, 2012 11:12 pm Top

Re: 3.4 Pully System

by edwardmcd » Wed Feb 13, 2013 8:37 am

Hi,I can't see how you derive Newton's 2nd Law to get:F = T - m1g = m1aandF = T - m2g = -m2a.I am assuming there are some steps which have been left (which I can't fathom why).Now - Physics is not a strong point for me at all, but continuing with that question - I can't see how you solved for T either. It would be a really good idea to include this in the book itself - which is what most would expect.Sorry for the trouble. edwardmcd Posts: 1Joined: Mon Jan 14, 2013 10:33 pm Top

Re: 3.4 Pully System

by goldstanda3269 » Sun Feb 17, 2013 2:57 pm

First, there is no trouble at all!Second, ....Newton's Law: Sum of forces = mass x accelerationSo let's break it down:Sum of forces for m1 means: "what forces can be applied to m1?" = its weight downward (i.e. m1g) and the tension on the string upwards. Taking upwards as being positive, so T is positive, we get that the sum of forces about m1 is: T - m1gmass = m1acceleration = aso for Sum of forces = mass x accelerationWe getT - m1g = m1a goldstanda3269 Posts: 1766Joined: Wed Aug 25, 2010 10:59 pm Top

Re: 3.4 Pully System

by interwebme » Thu Jun 20, 2013 11:19 pm

I have spent two hours of my life looking at this and I can't understand how to solving the equations for motions for a then equating them leads to T = (2m1m2/m1+m2)gCan someone please spell this out for me. I can't seem to find anything on the net.Thank you so much interwebme Posts: 21Joined: Sun Apr 21, 2013 6:25 am Top

Re: 3.4 Pully System

by interwebme » Thu Jun 20, 2013 11:22 pm

The furthest I get is resolving the two equations for a which I believe to be T-m1g/m1=a and T-m2g/-m2=aWhere do I go from here? Or did I get this wrong?Cheers interwebme Posts: 21Joined: Sun Apr 21, 2013 6:25 am Top

Re: 3.4 Pully System

by goldstanda3269 » Sun Jun 23, 2013 1:54 pm

It's good for you to know that it is unlikely to get a question exactly like this on the real GAMSAT but it is very likely to get questions on the real test that require this level of equation manipulation once (i.e. you would never need to resolve for 'a' AND T in the same question).First, let's back up one step to this:T - m1g = m1aandT - m2g = -m2aIf we are looking for the acceleration then we can eliminate the tension T by subtracting the 2nd equation from the first:T - m1g = m1a-(T - m2g = -m2a)--------------------------0 - m1g + m2g = m1a + m2a (remember that a minus times a minus is positive)Now we resolve: g(-m1 + m2) = a (m1 + m2)So g (m2 - m1)/(m1 + m2) = aNow to resolve for tension T, you did the first step already which is to isolate the acceleration a:T-m1g/m1=aandT-m2g/-m2=aSince they are both equal to 'a' then we can equate the 2 equations:T-m1g/m1 = T-m2g/-m2 to get rid of the denominators, multiply through by (-m2m1)-m2(T-m1g) = m1(T-m2g)T(-m2) + m1m2g = T(m1) - m1m2g2m1m2g = T(m1) - T(-m2) = T(m1 + m2)2m1m2g / (m1 + m2) = T goldstanda3269 Posts: 1766Joined: Wed Aug 25, 2010 10:59 pm Top

Re: 3.4 Pully System

by nawaz » Mon Nov 03, 2014 10:39 pm

Now - Physics is not a strong point for me at all, but continuing with that question - I can't see how you solved for T either. It would be a really good idea to include this in the book itself - which is what most would expect nawaz Posts: 2Joined: Mon Nov 03, 2014 10:29 pm Top

Re: 3.4 Pully System

by goldstanda3269 » Wed Nov 05, 2014 8:31 pm

" I can't see how you solved for T either. " - I'm glad you are seeking clarification because these types of manipulations are regular GAMSAT fare (not many questions but certainly a few). However, from the steps laid out above, please copy/paste the point at which you would like more detail and I will do my best to clarify. " It would be a really good idea to include this in the book itself - which is what most would expect "- Honestly, we were surprised to get these questions because the solution for T and a has nothing to do with Physics, it's grade 9-10 math. So, I'm sure this won't impress you because of the timing, but we finally decided to add more than 300 extra pages to the new edition, about 1/2 of which is just math. goldstanda3269 Posts: 1766Joined: Wed Aug 25, 2010 10:59 pm Top

ENTIRE QUESTION REWIEW, 'REPOST'

by JordanS » Tue Mar 03, 2015 12:18 pm

I was directed here after posting for the same question elsewhere, and the reply may be of help to people so I am also posting it here. In Section 3.4, Pulley Systems on page PHY-23 of the GS textbook I have a question regarding the manipulation of the equations. For the first part, solving for a, that is a reasonably quick manipulation and I imagine would be possible within the time constraints of the exam. However, for the second part, solving for T, the textbook seems to insinuate that you find the manipulation and solution for T quite easily. Adding the two equations, one gets 2T=m1a - m2a + m1g + m2g. Taking a out as a factor of the first expression and then substituting the previously found value for a you get (m2-m1/m2+m1)g x (m1-m2) + m1g + m2g. Then taking g out as a factor you then proceed to expand the brackets to get g(m2m1+m1^2-m1^2+m1m2). Then taking a common denominator of m2+m1 you expand out the right side to equal the same expansion however with -m1^2 and - m2^2. Ultimately getting after couple more easy steps 2T=(g)x (2m1m2+2m1m2/m1+m2). Finally, solving for T and dividing by 2 you get the answer given of T=(g) x (2m1m2/m1+m2).So my question is, is this reasonable to expect as a question on the Gamsat? By cutting a couple of those steps out that aren't entirely necessary to write as I did above you can save yourself some time. But it seems to me that this manipulation is quite time consuming and would eat up a lot of your time in the exam merely to solve the equation for a value of T. Then you additionally have to substitute the values given (very simple, yes). So, is this too long-winded for the exam or are we expected to solve this <2 minsREPLY AS FOLLOWS from GS-ONLINE: 1) It is unlikely to get a question like that on the GAMSAT. Although it is possible to get a question that takes 2-3 minutes to resolve, usually it is followed be another 1-2 questions that are quickly resolved because of the work previously done. However, the reasoning that you applied is often fundamental to a handful of GAMSAT questions.2) Your work-through was done correctly. FYI, here is another discussion on the same topic: viewtopic.php?f=30&t=2028 JordanS Posts: 28Joined: Thu Aug 21, 2014 3:03 pm Top

Re: 3.4 Pully System

by ElleRT » Fri Jul 31, 2015 10:57 pm

I was just wondering for 3.4 pulley system. Could we use that equation a = T - m1g / m1 to find a? I suppose we would also need the second mass.... Thanks ElleRT Posts: 55Joined: Thu Jul 09, 2015 4:50 am Top

Re: 3.4 Pully System

by goldstanda3269 » Sat Aug 01, 2015 4:18 pm

"Could we use that equation a = T - m1g / m1 to find a? "Yes, definitely.Of course, it depends on the question; however, we try to show you how to arrive at equations like the one above while starting with something that is presumed knowledge for the GAMSAT (in this case, F = ma is presumed knowledge which should be known or memorized; the equation above is not something to memorize but something that you would be expected to derive). goldstanda3269 Posts: 1766Joined: Wed Aug 25, 2010 10:59 pm Top

Re: 3.4 Pully System

by brad.willo2670 » Thu Mar 21, 2019 5:36 pm

Hi there,Apologies if this has already been addressed. 3.4 Pulley Systems.Once a has been solved for and we now have values for M1, M2, g and a. Rather than isolating T with the formula: T = (2m1m2/m1+m2)g. Can we not just use T = m1a+m1g or T= m2g-m2a? Is there a particular rationale for using the combined formula?Thanks in advance! brad.willo2670 Posts: 38Joined: Mon Feb 18, 2019 1:26 pm Top

Re: 3.4 Pully System

by goldstanda3269 » Mon Apr 15, 2019 2:45 pm

[quote="brad.willo2670"]Hi there,Apologies if this has already been addressed. 3.4 Pulley Systems.Once a has been solved for and we now have values for M1, M2, g and a. Rather than isolating T with the formula: T = (2m1m2/m1+m2)g. Can we not just use T = m1a+m1g or T= m2g-m2a? Apologies for having missed your question!Let's try the former:T = m1a+m1g = (1)(5) + (1)(10) = 15 NThe latter:T= m2g-m2a = (3)(10) - (3)(5) = 30 - 15 = 15 NSo yes, both work seamlessly. If the acceleration 'a' is known, then your suggestion to calculate T is more efficient. Of course, ACER could ask for T in terms of the relevant variables instead of a numerical calculation so it's good to be flexible. And if 'a' is not known, then you could choose to first calculate T using the ungainly formula in PHY 3.4 and then plug that into one of the equations of motion you suggested above. goldstanda3269 Posts: 1766Joined: Wed Aug 25, 2010 10:59 pm Top Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending Post a reply 15 posts • Page 1 of 1

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