# Cable tension problem

1. Oct 16, 2009

### markles

1. The problem statement, all variables and given/known data

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.

2. Relevant 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 dont 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.

3. 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 dont know of any way to find the tension
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Oct 16, 2009

### Redbelly98

Staff Emeritus
Welcome to Physics Forums

Actually, no. That is only true for a mass hanging by a single, vertical cable.

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

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

3. Oct 16, 2009

### markles

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

4. Oct 16, 2009

### Redbelly98

Staff Emeritus
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.

No.

5. Oct 17, 2009

### markles

How would i do that? Could you give me the equation please?

Thank you

6. Oct 17, 2009

### rl.bhat

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

7. Oct 17, 2009

### markles

i dont know how to do that could you show me please?

8. Oct 17, 2009

### Redbelly98

Staff Emeritus
It's basic algebra. I would start by evaluating the expression,

(cos 30 x sin 45/cos 45)​