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.

CH06.Problems

JH.131 (132 updated)

Top: N-mg = -mV2/R and N = 0 So, mV2/R = mgBottom: 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/RT=m(g cos + V2/R)

N = 200 Sin + mg200 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 cosFind N then frictionFind 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/RTup sin – Tdn sin – mg = 0 TdnFind VFnet = Tupcos +Tdn cos pointing toward center