Question 14.6: Grain falls from a hopper onto a chute CB at the rate of 240......
Grain falls from a hopper onto a chute CB at the rate of 240 lb/s. It hits the chute at A with a velocity of 20 ft/s and leaves at B with a velocity of 15 ft/s, forming an angle of 10° with the horizontal. Knowing that the combined weight of the chute and of the grain it supports is a force W of magnitude 600 lb applied at G, determine the reaction at the roller support B and the components of the reaction at the hinge C.
We apply the principle of impulse and momentum for the time interval Δt to the system consisting of the chute, the grain it supports, and the amount of grain which hits the chute in the interval Δt. Since the chute does not move, it has no momentum. We also note that the sum \sum m_i \text v _i of the momenta of the particles supported by the chute is the same at t and t + Δt and can thus be omitted.
Since the system formed by the momentum (Δm)\text v _A and the impulses is equipollent to the momentum (Δm)\text v _B, we write
\overset{+}{\text y}~x ~components:~~~~~~~~~C_x \Delta t=(\Delta m) v_B \cos 10^{\circ} (1)
{+}{\text x}~y ~components:~~~~~~~~~-(\Delta m) v_A+C_y ~\Delta t-W~ \Delta t+B~ \Delta t \\ \\=-(\Delta m) v_B ~\sin 10^{\circ} (2)
+l moments about C: -3(\Delta m) v_A-7(W ~\Delta t)+12(B~ \Delta t) \\ \\ =6(\Delta m) v_B~ \cos 10^{\circ}-12(\Delta m) v_B~ \sin 10^{\circ} (3)
Using the given data, W = 600 lb, v_A=20 ft / s , v_B=15 ft / s \text {, }and Δm/Δt = 240/32.2 = 7.45 slugs/s, and solving Eq. (3) for B and Eq. (1) for C_x,
12B = 7(600) + 3(7.45)(20) + 6(7.45)(15)(cos 10° – 2 sin 10°)
12B = 5075 B = 423 lb B = 423 lb x
C_x=(7.45)(15) \cos~ 10^{\circ}=110.1 ~lb \quad C _x=110.1~ lb ~ \text ySubstituting for B and solving Eq. (2) for C_y,
C_y=600-423+(7.45)\left(20-15 \sin 10^{\circ}\right)=307 ~lb \\ \\ C _y=307~ lb~ \text x