- #1

JJBladester

Gold Member

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## Homework Statement

Belt A conveys sand with a velocity of 6.0 ft/s (constant). The velocity of belt B is 8 ft/s (constant). Determine the velocity of the sand relative to B as it lands on belt B.

*Answer: V*=20.1 ft/s at 85.1 degrees

_{S/B}## Homework Equations

V

_{S/B}= V

_{S}-V

_{B}

(Velocity of sand relative to belt B = velocity of sand as it hits belt B - velocity of belt B)

## The Attempt at a Solution

V

_{B}=8cos(15°)+8sin(15°) = -7.727[tex]\hat{i}[/tex]+2.071[tex]\hat{j}[/tex](ft/s)

My trouble is with finding V

_{S}. I know that on the horizontal section of belt A, the sand's velocity (V

_{S}) is -6[tex]\hat{i}[/tex]+0[tex]\hat{j}[/tex](ft/s).

When the sand hits the bend where belt A is not horizontal anymore, but starts to bend into belt B, the velocity of the sand changes.

I think I'm stuck on the trig of what the angle is and how the 5ft contributes to finding the correct angle that the sand runs into belt B.