- #1
werson tan
- 183
- 1
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 ?
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?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>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.
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>