Hello,

 

First check the appendix. This is a photo of the given problem and my FBD The description is in dutch.

The translation:

If the springs are coupled to eachother without the mass they are at resting position. There is no tension in the springs. When the mass is applied the springs will extend. Thereafter a punch (i.e. krachtstoot) is applied.

The only value given is 60 cm

The mass and spring constant and punch i declared myself. 

Mass, m = 2kg

Spring constant, k = 10 kg/m

Punch, Fp = 100 N

Now where do i begin? Im not sure what my second step should be? I cannot calculate the Y position etc. (First step was draw fbd)

Do you guys have an hint?

 

Given the x-y position of the mass you can calculate the lengths of the two springs and so the force they apply to the mass.  Take the point at which the left spring is attached to the platform as (0, 0) so that the other spring is at (0.6, 0) (in meters).  You say that "If the springs are coupled to each other without the mass they are at resting position".  That position would be along the platform at (0.3, 0) so I will interpret that as saying that the natural length of the springs is 0.3 m.   With the mass at (x, y) at rest, the length of each spring will be $\sqrt{x^2+ y^2}$ so the force due to each spring will be [tex]10(\sqrt{x^2+ y^2}- 0,3)[/tex].  At rest, the vertical component of force of the two springs must be mg= 2(9.81)= 19.62 N.  The vertical component is proportional to the vertical component of the position vector, x.  That is [tex]\frac{F_v}{10(\sqrt{x^2+ y^2}- 0.3}= \frac{x}{\sqrt{x^2+ y^2}}[/tex] so [tex]F_v= \frac{10x\sqrt{x^2+ y^2}- 0.3}{\sqrt{x^2+ y^2}}= 19.62[/tex].  It is clear from symmetry that the x component of the rest position is 0.3 m so that becomes [tex]\frac{3\sqrt{y^2+ 0.09}- 0.3}{\sqrt{y^2+ 0.09}}= 19.62}[/tex].  Solve that to find the y component of the initial position.

That should be enough to get you started.  

Edited March 9, 20196 yr by HallsofIvy