Recent content by daisy7777

For this problem, I took the moment about point B to 'get rid' of the tension force in the cables BA and BC. By doing this, I figured I could solve for the x and y components of the reaction force at D. From there, I could solve for the tension in BA and BC in the x and y and then solve for Dz...

I've solved for the moment using the cross product where my r = 0i+0.06j+0.03k m and my F = 100i + 150j + 0k. I got my moment to be M= 4.5,3,-6. I know a wrench equivalent force must be parallel to the force (I am assuming this means the resultant force in this case), but how do I apply that...

Ok so I got this to be my equation, however, it yields an answer that doesn't match the one in the answer key which is 65.8cm. I'm getting 0.0213m. Is there something I'm doing wrong? At t = 0s, the mass is 10cm above the origin.

So, would my c value be the Δx between the string and mass then? The 0.122m I calculated earlier? I'm so sorry for the myriad of questions, it's just I haven't had much practice w/ modelling damped SHM functions.

Ok, so then would my Δx just be 10cm as that's the displacement from equilibrium that the spring is pulled and therefore it would be my amplitude?

Wouldn't it be at the unstretched end?

Would it be 0.1m + 0.122m?

So my net force would be equal to Fs = Δx*k in both directions, right? I'm writing it out now and I get a magnitude of 60N in both directions as I'm using 50cm as Δx since that's the length when the spring is upstretched, would this be correct? And if so, would I have to consider the value I...

That's the equation my teacher's given us, so I'm assuming it's the one he wants us to use for problems like this. Thank you though!

It's negative, right?

I first solved for the extension of the spring when its at equilibrium w/ the mass. I got Δx = 0.122m. I thought that if I added this with 0.1m, this would give me my amplitude. I then set 0.5m as my c value and plugged the rest of my values in from there. What am I doing wrong?

I also tried to solve for T when the centripetal force is equal to the tension in the x-dir, but I got 2.174s and that's not the right answer either. I'm not sure what I'm doing wrong exactly.

Would this be correct then? I have the centripetal force as the force of gravity and the force of tension b/c I think they'd both be in the same dir. at the top of the circle.

I calculated the acceleration which is 0.804m/s^2. From there I calculated the centripetal force which is 0.402N. I think my lack of answer is due to my lack of understanding of the concept of what the centripetal force is at the top of the circle. Would it not be Fc = Fg - Ft as the ball...

Wait so the force of friction is gonna be Ff = coeff of friction * (Fstraps + Fgravity)?