I just have a question I'm having a bit of trouble with here, if anyone could help me that would be great! I'd really appreciate it!

A dog is pulling on two restraining chains exerting a maximum force of 500N. The chains, which are held by two steel stakes anchored in the ground, are 3m apart.
Determine the tension in the chains when the angles are 60° and 40° with the "vertical".

Also, is it possible for the dog to pull one of the stakes out if each is designed for a maximum direct force of 500N? Explain.

2. Originally Posted by Random-Hero-
I just have a question I'm having a bit of trouble with here, if anyone could help me that would be great! I'd really appreciate it!

A dog is pulling on two restraining chains exerting a maximum force of 500N. The chains, which are held by two steel stakes anchored in the ground, are 3m apart.
Determine the tension in the chains when the angles are 60° and 40° with the "vertical".

Also, is it possible for the dog to pull one of the stakes out if each is designed for a maximum direct force of 500N? Explain.
1. Draw a sketch (see attachment).

2. I assume that you mean with "vertical" the perpendicular direction to the base of 3 m length(?)

3. The two chains ( $c_1, c_2$) have to hold the dog which pulls with a force $|\overrightarrow{F_r}| = 500\ N$. The directions of the chains are determined by the given angles.

4. Calculate the components of $F_r$ using Sine rule:

$\dfrac{F_1}{500\ N}=\dfrac{\sin(40^\circ)}{\sin(80^\circ)} ~ \implies~ F_1\approx 326.35\ N$

Calculate $F_2$ similarly. I've got $F_2\approx 439.69\ N$

That means: The two stakes will hold the dog.

5. If the dog pulls directly in the direction of the base (see green arrow!) it will come free of the right stake, then it has to pull in the opposite direction to pull out the second stake.