Calculating the Fourth Mass: A Rope and Four Masses

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The discussion centers on calculating the fourth mass in a system of four hanging masses connected by a rope. The relationship between the tensions and the known masses leads to the equation m4 = [m1T2/(T1 − T2)] − m2 − m3. Participants express difficulty in solving the problem without a visual reference of the setup. One user claims to have solved the problem independently despite initial challenges. The conversation highlights the importance of visual aids in understanding physics problems involving multiple forces and tensions.
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Homework Statement


2. There are four masses hanging by a rope from the
ceiling, as shown in the figure. Two of the tensions
and three of the masses have been measured.
Show that the fourth mass can be expressed as
m4 =[m1T2/(T1 − T2)]− m2 − m3.



Homework Equations


F=ma, F1=-F2, T=ma


The Attempt at a Solution


I think that (m1+m2+m3+m4)g=T1 and (m2+m3+m4)g=T2
I tried solving for g and substituting and try and solve for m4 but I just find it impssible to do.
 
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We can't help you unless we see the figure that goes with this question.
 
I tried copying and pasting from a pdf file but it didn't work but it's really easy actually I already figured it out, :). Thanks anyways.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
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