# Calculating restorative force with Hooke's Law [duplicate]

Possible Duplicate:
Mass points of a Mass-spring model

Say I have a spring like the one in the picture below:

The point at the top is fixed to a ceiling.

The red coloured arrow is the direction in which I pull the spring to, and its vector is $(3, 6)$. Let's say the triangluar spring is equilateral and gravitational force is considered to be as $mg$.

Now, I want to find its restorative force in the y-axis direction. Based on Hooke's Law,

where $k$ is stiffness.

But I am not sure if I did this correctly.

Do I have to include the force of $-6k$, which is the change in length from the pulled direction vector in the y-axis?

• Related question by OP: physics.stackexchange.com/q/44967/2451 Nov 24, 2012 at 6:04
• What is preventing the three springs from rotating about the top point such that equilibrium is reached? Nov 24, 2012 at 7:06
• I wish you had labeled the nodes A, B and C so we can be clear on our answers. Nov 24, 2012 at 18:17
• I do not see how this differs from the earlier question and will likely merge them. On the whole reposting questions because you didn't get the answer you wanted is discouraged as is minutely detailed iterative problem solving by StackExcahnge. Edit you original question for corrections and clarification, take it to chat, or work on it your self for a while. Nov 25, 2012 at 16:45

First thing--the springs are not extended by $(3,6)$; they are extended by the component of that vector in their direction. So, for each spring you'll have to apply trigonometric functions twice--once to calculate the $x$ used in $kx$, once to calculate the $y$ component of force.
$$-F_{\rm spring} \cos 30^\circ + m g + F_y = 0$$
where positive spring force is tensile and negative is compressive. From this you solve for $F_{\rm spring}$ regardless of what the spring constant $k$ is. You need the spring constant to find out the displacement of the node, but I do not think this is asked here.