Question

I need help with the following problem: Determine the tensions in wires cd, cb, and ba and the angle theta required for equilibrium of the 90lb cylinder E and the 30lb cylinder F

Answer 1

in the figure

Answer 2

I started by setting the sum of the forces in the x axis = 0 and then the sum of the y axis equal to 0 but end up with two unknowns and gets me stuck

Answer 3

Take the tensions to be T1, T2 and T3
\[T1\sin30=T2\sin \theta+90\]
\[T1\sin30=T2\cos \theta\]
\[T2\sin \theta+T3\sin45=30\]
\[T3\cos45=T2\cos \theta\]
Isnt that enough to solve for the unknowns?

Answer 4

why is T1sin30=T2cos\[\theta\]?

Answer 5

take( t =thetha,)which is given in Question,
apply equillibrium equations at point b first,which will give you
T(bc)cost=T(ba)cos(45)=T(ba)/sqrt2.....(1)
T(ba)sin45+T(bc)sint=30
T(ba)/sqrt2=30-T(bc)sint......(2)
from 1 and 2 you have
T(bc)cost=30-T(bc)sint
T(bc)=30/(sint+cost).............(A)
now take joint c
T(cd)cos30=T(bc)cost=T(cd)*sqrt(3)/2.....(3)
T(cd)sin30=T(bc)sint+90=T(cd)/2.......(4)
solving 3 and 4 gives
T(bc)(cost-sqrt(3)(sint)=90sqrt(3)
T(bc)=155.88/(cost-sqrt(3)Sint..........(B)
comparing A and B you will get...
155.88/(cost-sqrt(3)Sint=30/(sint+cost)
5.196cos(t)+5.196sin(t)=cost-sqrt(3)(sint)
tan(t)=-(4.196/6.92)
t=-31.23........................ans
substitute t in A..
which givesT(bc)=89.12N.......ans
sub. T(bc) in (1) gives T(ba)=107.7 N...............ans
sub.T(bc)in (3)T(cd)=87.99N.................ans

Answer 6

Thanks so much for all of your help. I understand is just hard to set up the problem to these values. Thanks again :)

Answer 7

What is the answer???

Answer 8

I got different values