# CH06.Problems JH.131 (132 updated). Top: N-mg = -mV2/R and N = 0 So, mV2/R = mg Bottom: N-mg = +mV2/R  N = mg +mV2/R= 2 mg =1.37 x10^3 N.

## Presentation on theme: "CH06.Problems JH.131 (132 updated). Top: N-mg = -mV2/R and N = 0 So, mV2/R = mg Bottom: N-mg = +mV2/R  N = mg +mV2/R= 2 mg =1.37 x10^3 N."— Presentation transcript:

CH06.Problems JH.131 (132 updated)

Top: N-mg = -mV2/R and N = 0 So, mV2/R = mg Bottom: N-mg = +mV2/R  N = mg +mV2/R= 2 mg =1.37 x10^3 N

Triangle: theta (=80 deg) Vertical: T sin =mg (T=0.4 N) Horizontal: Tcos = mv2/R V=0.49 m/s, Rev=2pi r, t= 1.9s

Break mg along T T-mgcos = mV2/R T=m(g cos + V2/R)

N = 200 Sin + mg 200 Cos – fk = m a ma = 200 cos(20) – 0.2*(200sin(20)+25*9.8) = 125 a = 5 m/s^2

Vertical: N + F sin = W Horizontal: a = 0: mu N = F Cos So: N = F/mu *cos F/mu Cos + F sin = w  F = w/(cos/mu +sin) F = 400/ (cos30/0.25 + sin30) = 101 N

Vertical: P Sin = N + mg Horizontal: -fk + P Cos = m a a = (P cos – mu * N) /m = (P Cos – mu * (Psin - mg) ) / m = 3.4 m/s^2

Break F to parallel (Fcos) and normal Fsin. Break mg as well: parallel (mgsin) and normal mg cos Find N then friction Find acceleration by parallel to surface forces.

L cos = mg, L sin = m V2/R  R = V2/(g tan)

M g = m V2/R  V =sqrt(R g M/m) = 1.81 m/s

theta = string to horizontal angle :sin(theta) =(d/2) / L, R = sqrt(L^2 – (d/2)^2) Tup cos +Tdn cos = mV2/R Tup sin – Tdn sin – mg = 0  Tdn Find V Fnet = Tupcos +Tdn cos pointing toward center

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