Formula of forces

  • Thread starter werson tan
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Homework Statement


what is the 2.1 and 0.35 refers to in the equation of (9810)(2.1)(4.2)- 9810(0.35)(0.7) -Fx = 1000(5.88)(8.4-1.401) ??
is it the width ? but , the width is not given ?

Homework Equations




The Attempt at a Solution

 

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Answers and Replies

  • #2
BvU
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Same pattern post: not understandable without the pictures. I have to keep open five windows to have a good look at this. It's late and I'm tired.
 
  • #3
haruspex
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Where the depth is 4.2m, the average depth is 2.1m, so the average pressure on the approach is 1000(9.81)(2.1)N/m2.
Multiplying by the height gives the force per unit width on the approach. Similarly on the exit side.
I do have one concern. The barrier is not 4.2m in height, so the calculated force on the approach side seems to much. To get a more accurate result we would need to find the height of the water surface above the highest point of the barrier.
 
  • #4
BvU
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Tell us what the expressions they use for the ##F_1## and ##F_2## stand for ....
 
  • #5
haruspex
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Where the depth is 4.2m, the average depth is 2.1m, so the average pressure on the approach is 1000(9.81)(2.1)N/m2.
Multiplying by the height gives the force per unit width on the approach. Similarly on the exit side.
I do have one concern. The barrier is not 4.2m in height, so the calculated force on the approach side seems to much. To get a more accurate result we would need to find the height of the water surface above the highest point of the barrier.

Edit: I misunderstood the calculation.
Consider the whole of the system covered by the diagram.
F1 is the force acting to the right on that system from the body of water to the left.
Similarly, F2 is the force acting on it to the left from the body of water to the right.
Fx is the horizontal force from the ground to keep the spillway in place.
The sum of these external forces equals the rate of change of momentum.
 
  • #6
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Where the depth is 4.2m, the average depth is 2.1m, so the average pressure on the approach is 1000(9.81)(2.1)N/m2.
Multiplying by the height gives the force per unit width on the approach. Similarly on the exit side.
I do have one concern. The barrier is not 4.2m in height, so the calculated force on the approach side seems to much. To get a more accurate result we would need to find the height of the water surface above the highest point of the barrier.
ok , i can understand the red part of the equation (9810)(2.1)(4.2)- 9810(0.35)(0.7) -Fx = 1000(5.88)(8.4-1.401) , but 1000(5.88)(8.4-1.401) is force , am i right ? it's not force per unit width..... they are not having the same unit...how cvan they be equqal to each other?
 
  • #7
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Where the depth is 4.2m, the average depth is 2.1m, so the average pressure on the approach is 1000(9.81)(2.1)N/m2.
Multiplying by the height gives the force per unit width on the approach. Similarly on the exit side.
I do have one concern. The barrier is not 4.2m in height, so the calculated force on the approach side seems to much. To get a more accurate result we would need to find the height of the water surface above the highest point of the barrier.
do u mean since the width is unknown and constant through out the whole object , so we can take it out and leave the ans (Fx) is Force per unit width>
 
  • #8
haruspex
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do u mean since the width is unknown and constant through out the whole object , so we can take it out and leave the ans (Fx) is Force per unit width>
Yes, we do not know the width, so all forces are considered per unit width.
 

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