Solve for equilibrium ?

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3 Answers
Jun 24, 2016

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

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

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