Formula of forces

werson tan · Feb 26, 2016

1. The problem statement, all variables and given/known data
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 ?
2. Relevant equations

3. The attempt at a solution

BvU · Feb 26, 2016

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.

haruspex · Feb 26, 2016

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.

BvU · Feb 26, 2016

Tell us what the expressions they use for the ##F_1## and ##F_2## stand for ....

haruspex · Feb 26, 2016

haruspex said: ↑

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.
F_x 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.

werson tan · Feb 26, 2016

haruspex said: ↑

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?

werson tan · Feb 26, 2016

haruspex said: ↑

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>

haruspex · Feb 26, 2016

werson tan said: ↑

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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Formula of forces

werson tan

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BvU

haruspex

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werson tan

werson tan

haruspex

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