Gareth56 Posted April 21, 2010 Share Posted April 21, 2010 (mg - ma)/g = 1 - a/g Reasoning = (g - a)/g (because mg - ma = g - a) = g/g - a/g = 1 - a/g Is that correct? Link to comment Share on other sites More sharing options...
timo Posted April 21, 2010 Share Posted April 21, 2010 (mg - ma)/g = (g - a)/g That's wrong. Test it for m=5, g=1 and a=0, for example. Link to comment Share on other sites More sharing options...
Gareth56 Posted April 21, 2010 Author Share Posted April 21, 2010 Should it be (mg - ma)/mg = 1 - a/g ? If so then there's a typo in the physics book I'm reading. Link to comment Share on other sites More sharing options...
D H Posted April 21, 2010 Share Posted April 21, 2010 The equations in the original post are obviously incorrect from a dimensional analysis perspective. The left-hand side of (mg-ma)/g=1-a/g has units of mass while the right hand side is unitless. Are you sure the text doesn't have (mg-ma)/g = m(1-a/g), which is trivially correct? Link to comment Share on other sites More sharing options...
Gareth56 Posted April 21, 2010 Author Share Posted April 21, 2010 The question is asking for a ratio of force to a persons weight. The solution in the book states the following:- F(rope) = mg - ma and the weight of person = mg therefore the ratio of F(rope) to their weight = (mg - ma)/g = 1.0 - a/g So my question was how do you get from (mg - ma)/g to 1.0 - a/g Link to comment Share on other sites More sharing options...
triclino Posted April 21, 2010 Share Posted April 21, 2010 The question is asking for a ratio of force to a persons weight. The solution in the book states the following:- F(rope) = mg - ma and the weight of person = mg therefore the ratio of F(rope) to their weight = (mg - ma)/g = 1.0 - a/g So my question was how do you get from (mg - ma)/g to 1.0 - a/g The ratio of F(force) to their weight is : (mg-ma)/mg = 1-a/g Link to comment Share on other sites More sharing options...
Gareth56 Posted April 21, 2010 Author Share Posted April 21, 2010 Yes it is but in the book the 'm' is missing from the denominator so I am assuming that there is a typo in the book as without the 'm' being in the denominator the answer cannot be 1 - a/g. If there was a mechanism to display an image of the page from the book then I think it would be easier to show what the problem is. Link to comment Share on other sites More sharing options...
triclino Posted April 21, 2010 Share Posted April 21, 2010 Yes it is but in the book the 'm' is missing from the denominator so I am assuming that there is a typo in the book as without the 'm' being in the denominator the answer cannot be 1 - a/g. If there was a mechanism to display an image of the page from the book then I think it would be easier to show what the problem is. Yes there is a typo in the book. No doubt about it Link to comment Share on other sites More sharing options...
remodiyaz Posted July 28, 2010 Share Posted July 28, 2010 it should be (mg - ma)/mg then = (1 - a)/g it's wright anwer Braun 790 Shaver Link to comment Share on other sites More sharing options...
alpha2cen Posted December 4, 2010 Share Posted December 4, 2010 Its conditionally correct. if mg-ma = g-a , m should be 1, 0 if m=1 (mg-ma)/g =(g-a)/g =1 -a/g if m=0 g-a =0 (mg-ma)g= 0/g = g-a/g = 1- a/g a = g, not 0 Link to comment Share on other sites More sharing options...
Bignose Posted December 4, 2010 Share Posted December 4, 2010 but, considering that it is a physics equation, the terms have units. 0 kg does not equal 0 with no units. Its conditionally correct. if mg-ma = g-a , m should be 1, 0 if m=1 (mg-ma)/g =(g-a)/g =1 -a/g if m=0 g-a =0 (mg-ma)g= 0/g = g-a/g = 1- a/g a = g, not 0 Link to comment Share on other sites More sharing options...
khaled Posted January 19, 2011 Share Posted January 19, 2011 (edited) I think (based on your initial post), equation: (mg - ma)/g = m(g/g) - m(a/g) = m - m(a/g) given: mg-ma = g-a, m = 1 OR 0 for m=0: m - m(a/g) = 0 for m=1: m - m(a/g) = 1 - (a/g) Edited January 19, 2011 by khaled Link to comment Share on other sites More sharing options...
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