Dynamics HW #4 typical rock will fall in the bed? 1 3) It is proposed to put the

Question

Dynamics HW #4

typical rock will fall in the bed? 1 3) It is proposed to put the barge dock under cover (roof) to be able to load in inclement weather. The end I of the conveyor shall be under the same roof. If the conveyor is at a 10° angle, what is the minimum roof height (above the conveyor end) that will not allow the stone to impact the roof? roof Bed of the barge, note for prob 3-4 the bed is angled 60m 4) A new concept for limestone shipping suggests that the barge decks should be angled so the cargo will automatically move to the front and rear while loading (see angled bed in Figure). Given the slope of the bed is -.1, how far up from the rear of the barge will a typical rock land (use horizontal convey-I or 5) A robotic spray gun paints the rear end of an automobile that has a slope modeled by the equation y- -5x'. The paint spray exits the nozzle at 2 m/sec, which sits 1.5 m over the car bod The spray angle goes from 0 to 90°. What is the furthest point on the vehicle (x) that paint will reach?

Explanation / Answer

4. let velocity of the conveyor belt be v

given that horizontal conveyor is being used

so let the time the blocks stay in air after leaving the conveyor be t

also, let the horizontal distance travelled by the block in this time be d

then given slope = 0.1

or vertical deprerssion of the barge bed be t'

0.1 = t'/d

t' = 0.1d

hence the blocks have to travel a vertical distance of h = 60 + 0.1d

so from initial 0 vertical velocity

h = 0.5gt^2

60 + 0.1d = 0.5gt^2

but v = d/t

t = d/v

so

60 + 0.1d = 4.9*d^2/v^2

(4.9/v^2)*d^2 - 0.1d - 60 = 0

solving for d

d = v^2(0.1 + sqroot(0.01 + 4*60*4.9/v^2))/9.81

d = v^2(0.1 + sqroot(0.01 + 1176/v^2))/9.81

where v is horizontal velocity of the conveyor

so distance along the conveyor from the back of the barge wherfe the block get dropped = l

lcos(theta) = d

l = d/cos(theta)

but tan(theta) = 0.1

so theta = 5.71 deg

so l = v^2(0.1 + sqroot(0.01 + 1176/v^2))/9.81*cos(5.71) = 0.1024*v^2(0.1 + sqroot(0.01 + 1176/v^2))

where v is the horizontal speed of the conveyor

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