0
$\begingroup$

I would like to clarify if the equations that I got from this figure are correct.

enter image description here

$$-T_{CD}sin(30) + T_{DE}cos(0) = F_x = 0$$

$$T_{CD}cos(30) - \frac{W}{2} = F_y = 0$$

and

$$-T_{ED}cos(0) + T_{EG}sin(10) = F_x = 0$$

$$T_{EG}cos(10) - \frac{W}{2} = F_y = 0$$

$\endgroup$
0
$\begingroup$

if

  • the points D and E are pivot joints, or equivalently if all the connecting elements are wires/cables,
  • GE does not change in length

then you've written the right type of equations.

Some minor notes:

  • The equilibrium is $\sum F_x $ not $F_x$ (and $\sum F_y $) so for example:

$-T_{CD}sin(30) + T_{DE}cos(0) = F_x = 0$

is properly written:

$$-T_{CD}sin(30) + T_{DE}cos(0) = \mathbf{\sum F_x} = 0$$

  • Usually the symbol $T$ in this context is used to denote the tension on a wire in that case, regarding the tension of the cables in GE, I would have preferred to write:

$$-T_{ED}cos(0) + \mathbf{\color{red}2\cdot}T_{EG}sin(10) = \sum F_x = 0$$

$$\mathbf{\color{red}2\cdot}T_{EG}cos(10) - \frac{W}{2} = \sum F_y = 0$$

The reason for $\mathbf{\color{red}2}$ is that you have two cables between GE and therefore the tension on each cable will be $T_{EG}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.