- Joined
- Jul 24, 2008
- Messages
- 31
- Reaction score
- 0
Im a little confused about the net forces acting on a vertical spring with a mass attached to it at different times. Downward displacement is positive direction
1.@ new equilibrium pt
2.@Lowest point (most stretched)+A
3.@ Highest point (most compressed) -A
I came to the following conclusion, but I ran into a little problem...can someone tell me what I am doing wrong?
y0=equi pt of spring w/o mass, y'=new equilibrium pt with mass
1.@ new equilibrium pt...Fnet=0=Fs-mg
Fs=-ky' and Fg=mg
Since Fs-mg=0
=-ky'-mg
-ky=mg
-y=mg/k or y=mg/-k
2. @+A, lowest pt, Fs is up and (-)
Fnet=Fs-mg
Fnet=-k(y'+A)-mg
Fnet=-ky'-kA-mg
Fnet=-k(mg/-k)-kA-mg
Fnet=mg-KA-mg
Fnet=-kA ok...this makes sense because Fnet is up and that is the oppsite direction of gravity, therefore negative👍
3.@-A, highestt pt, Fs is down and (+)
Fnet=Fs+mg
Fnet=-k(y'-A)+mg
Fnet=-ky'+kA+mg
Fnet=-k(mg/-k)+kA+mg
Fnet=mg+kA+mg
Fnet=+kA+2mg??????!?!?! this does not make sense...what did I do wrong here??!?!👎
1.@ new equilibrium pt
2.@Lowest point (most stretched)+A
3.@ Highest point (most compressed) -A
I came to the following conclusion, but I ran into a little problem...can someone tell me what I am doing wrong?
y0=equi pt of spring w/o mass, y'=new equilibrium pt with mass
1.@ new equilibrium pt...Fnet=0=Fs-mg
Fs=-ky' and Fg=mg
Since Fs-mg=0
=-ky'-mg
-ky=mg
-y=mg/k or y=mg/-k
2. @+A, lowest pt, Fs is up and (-)
Fnet=Fs-mg
Fnet=-k(y'+A)-mg
Fnet=-ky'-kA-mg
Fnet=-k(mg/-k)-kA-mg
Fnet=mg-KA-mg
Fnet=-kA ok...this makes sense because Fnet is up and that is the oppsite direction of gravity, therefore negative👍
3.@-A, highestt pt, Fs is down and (+)
Fnet=Fs+mg
Fnet=-k(y'-A)+mg
Fnet=-ky'+kA+mg
Fnet=-k(mg/-k)+kA+mg
Fnet=mg+kA+mg
Fnet=+kA+2mg??????!?!?! this does not make sense...what did I do wrong here??!?!👎