# Homework Help: Another Double Check ! FBD String Tension

1. Feb 3, 2013

### Inferior Mind

Calculate the tension in the cable connecting the two masses. Assume all surfaces are frictionless.

FBD

m1 = 5 kg
θ1 = 60°
m2 = 6 kg
θ2 = 70°

Equation 1 ~

FT - Fg = ma

FT - mgSinθ = ma

FT = 5a + 5(9.8)Sin60

FT = 5a + 42.44

Equation 2 ~

Fg - FT = ma

mgSinθ - FT = ma

6(9.8)Sin70 - 6a = FT

FT = 55.25 - 6a

~Set Equations Equal to Each Other ~

5a + 42.44 = 55.25 - 6a

11a = 12.81

a = 1.16 m/s2

~Sub into Eq to find Force Tension on da String Son !~

5(1.16) + 5(9.8)Sin60 = FT

FT = 48.2 N

2. Feb 3, 2013

### SammyS

Staff Emeritus
The results looks correct.

Comment: If I were grading this, I would be concerned that you are using the same variable names for mass 1 quantities and mass 2 quantities, particularly Fg and m .

3. Feb 3, 2013

### Inferior Mind

Thanks for the input, I will make amends in future questions.

4. Feb 3, 2013

### SammyS

Staff Emeritus
To be more specific: FT and a are the same in magnitude for both masses, so it's fine to use the same variable name for them.

Last edited: Feb 3, 2013