Question

The figure below has an n = 120 revolutions per minute and each circle has a weight of 16.1 lb. Determine the counterweight W that will maintain in the position shown.

.

20 w Axis of Rotation

Answer 1

Sol: Here radius of circular path of circle ia given as-

1" 30 45 r r w Axis of Rotation

r=1" + (6 + 4) sin30 = 6' = 0.1524m

300 105) F w Axis of Rotation W

Now centripital force F is given as-

F = 47?mrf

here m = 16.1 lb = 7.30kg r = 0.1524m and f = 120/60 = 2 rps

Hence

$F=4\pi ^{2}\times 7.30\times 0.1524\times 2^{2}=175.68N$

For equilibrium condition-

$(6+4)Fcos60^{\circ}=6Tcos(105^{\circ}-90^{\circ})$

or

$T=\frac{10}{6}\times\frac{Fcos60^{\circ}}{cos15^{\circ}} =0.8627F$

Again for equlibrium condition of W

$W=2Tcos45^{\circ}=1.414T$

now by substituting value of T

$W=1.414\times 0.8627F=1.22F$

or

$W=1.22\times 175.68=214.33N$

or

$W=214.33/9.81=21.85kg=48.17lb$