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A jet of water 10 mm in diameter strikes a flat plate at an angle of 0º and 30º to the normal of the plate with a velocity of 10 m/s. What is the force exerted by the jet on the plate if the plate is:

A) Stationary
B) Moving in the direction of the jet with a velocity of 2 m/s
C) Parallel to the direction of the jet
D) Rotating at a velocity of 5 rad/s

User Hipny
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1 Answer

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Final answer:

The force exerted by the jet of water on the plate is zero when the plate is stationary or parallel to the direction of the jet. When the plate is moving with a velocity of 2 m/s in the direction of the jet, the force exerted is equal to the mass flow rate multiplied by the change in velocity.

Step-by-step explanation:

To calculate the force exerted by the water jet on the plate, we can use the principle of momentum. The force exerted is equal to the rate of change of momentum. Since the water jet is striking the plate at an angle of 0º or 30º, the component of velocity perpendicular to the plate is zero. Therefore, the force exerted by the water jet on the plate is zero in both cases.

Therefore, for both cases A) stationary plate and C) parallel plate, the force exerted by the jet on the plate is zero.

For case B) the plate moving in the direction of the jet with a velocity of 2 m/s, we need to consider the change in momentum of the water jet. The rate of change of momentum is given by:

Force = (mass flow rate) x (change in velocity)

Here, the mass flow rate remains the same at 10 mm in diameter. The change in velocity is given by (10 m/s - 2 m/s) = 8 m/s.

Therefore, the force exerted by the jet on the plate is Force = (10 mm) x (8 m/s).

User Kristianp
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