Question

Solve for equilibrium ?

Answer 1

`T_1~~3954.7kgf and T_2~~1797.6kgf`

Explanation:

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`

Answer 2

`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`

Answer 3

`T_1=3965.48`
`T_2=1802.49`

Explanation:

`"All forces and their components"`

`"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`

`"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`