Recent content by charon_9

Okay, thanks. I have one more question, though. I decided not to open a new thread, but please tell me if I need to. I took this photo from a very old textbook, where it asked us to write the same equation for this system, where the mass of the disc is ##M=2##, and ##k=37.5## for both springs...

Sorry. I was quick to reply so I mistook ##G## for ##F##, the elastic force. This is my guess. Let's say the springs tilt up for a small angle. This means that the elastic force has a vertical component to it (in the negative direction of the y axis in an XY plane). We need the other component...

Sorry for posting a misleading picture, that's what I have been provided with by my school. I'm once again sorry. I did some testing so please tell me if I'm right or wrong. Because the springs are a bit elevated. their force F has a vertical component and a component in the direction of...

Thank you for taking the time to answer. Wasn't it $$M=r \times F = r \cdot F \sin{\theta}$$

Well, $$I\alpha = M_1 + M_2 + M_3$$ where $$M_1 = M_2$$. $$M_1 = M_2 = -k \cdot l\tan{\theta}$$ Now I get stuck figuring out the radius part. I have no clue where they got the cosine from.

Yes, we're always using an approximation that $$cos(\theta) = 1$$ for small angles. Thing is, I can understand that $$M_1 = kx \cdot r$$, where I already figured out that $$x=l\tan{\theta}$$ However, I really can't figure out where they got the $$l\cos{\theta}$$ in $$M_1=-kl\tan{\theta} \cdot...

I need to write an equation for Newton's second law for the above system, where k1=k2 (both springs are the same). The red line represents a bar with m=2kg, l=2m. I know that I*α = M1 + M2 + M3 If we displace the bar by x, we get the angle of displacement theta. M1=M2=-k*x I know that...