Solve for equilibrium ?

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3 Answers
Jun 24, 2016

Answer:

#T_1~~3954.7kgf and T_2~~1797.6kgf#

Explanation:

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The situation given in the question has been shown in the figure.

  • #"weight of the shaft"=5.097Mgf=5097kgf#

  • # "O is the point of suspension, OP and OQ are chains of 4m "#

  • # "R is CG , OR is the vertical line along which total weight acts"#

  • PR =1.25m and RQ = 2.75m

  • #T_1 ="Tension along PQ" and T_2 =" Tension along OQ"#

  • In #Delta POQ, OP=OQ=QP=4m => Delta POQ " equilateral"#

  • So In #Delta POQ, " Each angle" = 60^@#

Let
#/_POR=x " then "/_QOR =60-x#

Now in #Delta OPR,(PR)/sinx=(OR)/sin60.......(1)#

And in #Delta ORQ,(QR)/sin(60-x)=(QR)/sin60....(2)#

Coparing (1) and (2) we get

#(PR)/sinx=(QR)/sin(60-x)#

#=>sin(60-x)/sinx=(QR)/(PR)=2.75/1.25=11/5 ......(3)#

Now considering the equilibrium of forces we can say that hrizontal components of #T_1 and T_2# are equal in magnitude.
So
#T_1sinx= T_2 sin(60-x)#

#=>T_1/T_2=sin(60-x)/sinx......(4)#

Comparing (3) and (4) we can write

#T_1/T_2=11/5=2.2=>T_1=2.2*T_2#

Now from equilibrium point of view the magnitude of Resultant of two tensions #T_1 and T_2# acting at angle #60^@# will be equal to
weight of the shaft i.e. #5097kgf#

So we can write

#T_1^2+T_2^2+2T_1*T_2cos60^@=5097^2#

Inserting #T_1=2.2T_2 and cos 60^@=1/2 # we get

#2.2^2T_2^2+T_2^2+2xx2.2*T_2^2*1/2=5097^2#

#=>2.2^2T_2^2+T_2^2+cancel2xx2.2*T_2^2*1/cancel2=5097^2#

#=>8.04T_2^2=5097^2#

#=>T_2=5097/sqrt8.04=1797.6kgf#
and
#T_1=2.2xxT_2=2.2xx1797.6=3954.7kgf#

Jun 24, 2016

Answer:

#abs (t_1)=3954.66# and #abs(t_3)=1797.57#

Explanation:

When in equilibrium, resultant weigth force passes across the shaft gravity center. The chain and the bar segment between anchored chains, form a equilateral triangle.

Let #p_1,p_2,p_3# be the triangle vertices, and #p_g# the point where the gravity center.

#p_1 = {0,0}#
#p_2 = {2, 2 sqrt[3]}#
#p_3 = {4,0}#
#p_g = {1.25,0}#

The weight passes along the line defined by #p_2, p_g# so if we have

#vec e_1 = (p_2-p_1)/norm(p_2-p_1)#
#vec e_3=(p_2-p_3)/norm(p_2-p_3)#
#vec f = (p_g-p_2)/norm(p_g-p_2)#

we can equate

#Mg vec f = t_1 vec e_1 + t_3 vec e_3#

and also

#Mg << vec f,vec e_1>> = t_1 + t_3 << vec e_3,vec e_1>>#
#Mg << vec f,vec e_2>> = t_1 << vec e_1,vec e_3>> + t_3#

Solving for #t_1, t_3# we obtain

#{t_1 = -(11 Mg)/sqrt[201], t_2 = -(5 Mg )/sqrt[201]}#

but #Mg = 5097# then #abs (t_1)=3954.66# and #abs(t_3)=1797.57#

Jun 24, 2016

Answer:

#T_1=3965.48#
#T_2=1802.49#

Explanation:

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#"All forces and their components"#

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#"torque according to the point A:"#

#mg*cos theta* 1.25=T_2.sin alpha*4#

#cos theta=0.98#

#T_2=(m*g*cos theta*1.25)/(4*sin alpha)=(5097*0.98*1.25)/(4*0.866)#

#T_2=(6243.825)/(3.464)#

#T_2=1802.49#

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#"torque according to the point B:"#

#T_1*sin alpha*4=mg*cos theta*2.75#

#T_1=(m*g* cos theta*2.75)/(4*sin alpha)#

#T_1=(5097*0.98*2.75)/(4*0.866)#

#T_1=(13736.415)/(3.464)#

#T_1=3965.48#