- #1
PremedBeauty
- 25
- 0
The L-shaped tank shown in Figure 14-33 is filled with water and is open at the top.
(a) If d = 5.0 m, what is the force on face A due to the water?
N (up)
(b) What is the force on face B due to the water?
N (right)
The Force of water on the face A
FA = PA*AA
= rwg hA* AA
= rwg ( 2d ) * AA
0r FA = rwg ( 2d ) * d2
putting the values we get FA = 2.5*106 N
The atmospheric force on the face A = FA= ( 1.013*105 )*( 52)
= 2.5*106 N
Therefore Net force FAnet = Fatmp + FA
FAnet = 5*106 N
b) Force on the face B due to the water alone = FB
FB = PB*AB
= rwg ( 5d /2) d2
FB = 3.1*106 N
Thus, net force on face B FBnet = Fatmp + FBnet
FBnet = 5.6*106 N
I have given the problem a try. Please confirm the answer
(a) If d = 5.0 m, what is the force on face A due to the water?
N (up)
(b) What is the force on face B due to the water?
N (right)
The Force of water on the face A
FA = PA*AA
= rwg hA* AA
= rwg ( 2d ) * AA
0r FA = rwg ( 2d ) * d2
putting the values we get FA = 2.5*106 N
The atmospheric force on the face A = FA= ( 1.013*105 )*( 52)
= 2.5*106 N
Therefore Net force FAnet = Fatmp + FA
FAnet = 5*106 N
b) Force on the face B due to the water alone = FB
FB = PB*AB
= rwg ( 5d /2) d2
FB = 3.1*106 N
Thus, net force on face B FBnet = Fatmp + FBnet
FBnet = 5.6*106 N
I have given the problem a try. Please confirm the answer
Last edited: