Recent content by zaper

Ok, I just subbed in x = 5 and x = 8 and set them equal to each other. I got b = 13 which seems right to me

Wow, can't believe I got those sides mixed around like that... Ok, so then if I can't factor out 5, how would you recommend I go forward? I'm getting stuck at the 110 = 5b

Homework Statement I need to solve 0 = 5x2 - 50x + 125 with solutions x = 5, x = 8. Homework Equations Referencing this thread I arrived at the equation (x - 5)(x - 8) = 5x2 - 50x + 125. Expanding x2 - 13x + 40 = 5x2 - 50x + 125 The Attempt at a Solution I tried to simply...

The first thing that I would recognize here is that the mortar is in the air for a set amount of time and will go a set distance. If you can figure out the time that it takes to hit and the distance it travels in that time then you can find how far Dr. Evil went when it hit him

So Ny+F-mg=ma with no vertical acceleration so it's just simply Ny+F=mg Horizontally then Nx-Fx=ma but where does this acceleration come from?

Ok so with friction for there to be no vertical acceleration then Ny+Fy=mg to stop it from moving down but also mg+Fy=Ny to stop from moving up. For no horizontal acceleration won't it simply be Nx=Fx?

Well I already have the equation for N right? (N=mg/cosθ) And can I just use friction≤μN here?

So then plugging N into mrω2=N*sinθ (again r is the horizontal circle's radius) I get: mrω2=mg*sinθ/cosθ Simplified this is rω2=g*tanθ Now I believe that r is R*sinθ so subbing again: R*sinθ*ω2=g*tanθ which simplifies again to R*ω2=g/cosθ Hopefully I did all that right and...

Ok so since we don't know what N is and the question doesn't ask for it we need to get rid of it. You said earlier that there is no vertical acceleration so that means N*cosθ=mg or N=mg/cosθ

Not sure if this is what you want or not, but subbing Rω2=arad into Fx I get mRω2=N*sinθ. Now if I solve this all out for θ will that be what I'm looking for?

Well a(rad)=Rw^2 so w=(a/R)^1/2 (sorry I'm on my phone now and I can't do the fancy stuff so I'll be using "w" for omega)

Fy=N*cosθ-mg Fx=m*arad=N*sinθ Right?

Ok so N is the force pulling to the center of the hoop. Would it's x component have the force m*arad? Also the weight obviously has no x component so the only x force is from N.

The normal force? That's about all that's left

Ok I think we're getting back to my point of confusion from before. We've established that the only acceleration in the horizontal circle is the radial acceleration so I'm not quite seeing where this vertical force can come from unless you mean the weight