Force exerted normal to the plate

Check your arithmetic again.In summary, the conversation discusses the determination of the normal force exerted on a stationary flat plate when a jet of water with specific diameter and velocity strikes it at a 25 degree angle. The equation used to calculate the normal force is F = ρA(v)(v cos theta), and the resulting answer is 268.8N. The person asking for confirmation of their answer is advised to double check their arithmetic.
  • #1
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4

Homework Statement


A jet of water , 50mm diameter , with a velocity of 18m/s , strikes a flat plate inclined at 25 degree to the axis of jet as shown in figure, Determine the normal force , P exerted on the plate when plate is stationary .

Homework Equations




The Attempt at a Solution


6ffCbqU.jpg

in the diagram attached , i found that the angle between the water velocity make an angle of 65 degree to the normal of the plate ,
so my ans is F = ρA(v)(v cos theta) = 1000(pi)( [(50x10^-3) / 2 ] ^2 ) (18^2)(cos 65) = 278.7 N , is my working correct ? i don't have the ans
 

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  • #2
foo9008 said:

Homework Statement


A jet of water , 50mm diameter , with a velocity of 18m/s , strikes a flat plate inclined at 25 degree to the axis of jet as shown in figure, Determine the normal force , P exerted on the plate when plate is stationary .

Homework Equations




The Attempt at a Solution


6ffCbqU.jpg

in the diagram attached , i found that the angle between the water velocity make an angle of 65 degree to the normal of the plate ,
so my ans is F = ρA(v)(v cos theta) = 1000(pi)( [(50x10^-3) / 2 ] ^2 ) (18^2)(cos 65) = 278.7 N , is my working correct ? i don't have the ans
Your algebraic expression looks right but I don't get same answer (about 4% less). Check your arithmetic.
 
  • #3
haruspex said:
Your algebraic expression looks right but I don't get same answer (about 4% less). Check your arithmetic.
my ans is 268.8N
 
  • #4
foo9008 said:
my ans is 268.8N
That is what I get.
 

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