- #1

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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 ?

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- Thread starter werson tan
- Start date

- #1

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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

BvU

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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

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^{2}.

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.

F

Similarly, F

F

The sum of these external forces equals the rate of change of momentum.

- #6

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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?^{2}.

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.

- #7

- 183

- 1

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>^{2}.

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.

- #8

- 36,987

- 7,212

Yes, we do not know the width, so all forces are considered per unit width.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>

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