3 blocks connected by line

Then we have to find the value of masses (m1, m2, m3) if system is on balance.In summary, the question asks for the limits of the masses of the blocks in a given configuration for the system to remain in balance. The values of friction and alpha are given, and the equation for determining the limits of the masses is mgsin(30°)+µmgcos(30°)+µmg=2mg. However, the question may be incomplete or missing some information.
  • #1
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Homework Statement


Last time I have had problem with one task:
http://img543.imageshack.us/i/cosc.jpg/
They questioned us to determine compartments in with that configuration will be on balance.
Informations:
u-Friction
alpha=30

Homework Equations



If that configuration must be on balance so acceleration must be 0. So we can compare forces working on that blocks.
-Qx1-T1-T2=Q3

The Attempt at a Solution



If we compare forces which working on that configuration we will go to moment that masses of that blocks will be reduced.
-mgsin(30)- umgcos(30)-umg=2mg
But this is not a solution of that task.
Sorry for my language. Thx for help:smile:
 
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  • #2
hi mojki1! :smile:

(have an alpha: α and a mu: µ and a degree: ° :wink:)

what exactly is the question?

are the blocks stationary, or are they moving with zero acceleration?

i'm guessing that you mean that the blocks are stationary, and that the question wants you to give the limits (not "compartments" … that doesn't mean anything in physics :wink:) (on µ) within which that is possible

if so, your answer should have "µ" on one side on its own (and perhaps should have a ≤ or ≥ rather than an =) :wink:
mojki1 said:
If that configuration must be on balance so acceleration must be 0. So we can compare forces working on that blocks.
-Qx1-T1-T2=Q3

If we compare forces which working on that configuration we will go to moment that masses of that blocks will be reduced.
-mgsin(30)- umgcos(30)-umg=2mg
But this is not a solution of that task.

your final equation looks correct to me (except that the minuses should be pluses, and you need to cancel the mgs and make it an equation for µ)

btw, "If we compare forces which working on that configuration we will go to moment that masses of that blocks will be reduced" …

i think you mean something like "If we equate the forces on that configuration (or system) at the instant that the blocks are about to move" :smile:
 
  • #3
You 're right tiny-tim. The blocks are stationary and question is about the limit of the masses of that blocks for which the system remains in balance. µ is given.
Final equation should looks:

mgsin(30°)+ µmgcos(30°)+µmg=2mg

But this isn't standard task that we have to designate limits for µ, we need designate limits for masses not for µ. µ is given. If i compare that equation, masses will be reduced, so i think that we have to look for this problem from another perspective
 
  • #4
hi mojki1! :smile:
mojki1 said:
mgsin(30°)+ µmgcos(30°)+µmg=2mg

But this isn't standard task that we have to designate limits for µ, we need designate limits for masses not for µ. µ is given. If i compare that equation, masses will be reduced, so i think that we have to look for this problem from another perspective

if µ is fixed (what is it, btw?), and if the ratio of the masses (m:m:2m) is also fixed, then the value of m makes no difference :confused:

there must be something missing from the question :redface:
 
  • #5
µ is constant for all blocks, only masses are your variables

EDIT:
data for this task were symbols
 
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