How Do You Calculate Cable Tension in a Balancing System?

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To calculate cable tension in a balancing system, the mass of the cylinder is 10 kg, resulting in a gravitational force of 98.1 N. The system is in equilibrium, requiring the sum of the tension forces to equal this gravitational force. A free-body diagram is essential to set up two force equations for horizontal and vertical components. The equations involve using trigonometric functions to relate tensions in cables and the gravitational force. Proper algebraic manipulation of these equations will yield the correct tension values for cables CA and CB.
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Homework Statement



hello i have a problem with this question.

the cylinder shown in fig1 has a mass of 10 kg. if theta = 45 and the system is in eqlibrium determain the tension in cables ca, cb.

Homework Equations



all i want is the equations and a guide to the i.e what each component is m=mass or g=gravity. and it would be really helpful is some provided outher useful formula for similar questions. as i have 3 or 4 and one includes a spring but i don't want to be cheeky and ask all 3 questions so if i got a list of formula i think i could apply them ok providedi know each component.


The Attempt at a Solution



10x9.81 = 98.1 and because the system is in equlibrium the total tension must equal 98.1.

Tca= 30=w1=10x 9.81
98.1/sin30 =196.2N

Tcb = Fca cos 45 = 196.2 cos 45 = 138.59N
these two answers are far to big i think and i don't know of any way to find the tension
 

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markles said:

The Attempt at a Solution



10x9.81 = 98.1 and because the system is in equlibrium the total tension must equal 98.1.
Actually, no. That is only true for a mass hanging by a single, vertical cable.

Tca= 30=w1=10x 9.81
This makes no sense. In part, this equation says that 30=10x9.81?

98.1/sin30 =196.2N

Tcb = Fca cos 45 = 196.2 cos 45 = 138.59N
these two answers are far to big i think and i don't know of any way to find the tension

The approach, after drawing a free-body diagram, is to set up 2 separate force equations for the horizontal and vertical force components.
 
if i were to do

+> Fcb cos 45 = Fca cos 30

+^ Fca sin 30 = Fcb sin 45 = 98.1N

then

Fcb = Fca cos 30/ cos 45

then sub Fcb with:

Fa sin 30+ Fca cos(30) x sin 45 / cos 45

Fca [sin 30 + (cos 30 x sin 45/cos 45)]

Fca=98.1

sub in Fca into Fcb = 9.81 x cos 30 / cos 45


Would that be the right kind of theory? If not i am lost thanks for you help and any futher is greatly appriciated.

Thank You
 
markles said:
Fca [sin 30 + (cos 30 x sin 45/cos 45)]
Okay, that is the sum of the upward rope forces. If you set that whole expression equal to the downward 98.1 N force, you can then find Fca.

Fca=98.1
No.
 
How would i do that? Could you give me the equation please?

Thank you
 
Fca [sin 30 + (cos 30 x sin 45/cos 45)]

Fca=98.1

You have to modify this step.
Fca [sin 30 + (cos 30 x sin 45/cos 45)] = 98.1
Find Fca
 
i don't know how to do that could you show me please?
 
It's basic algebra. I would start by evaluating the expression,

(cos 30 x sin 45/cos 45)​
 
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