Force and pressure/ upthrust

  • #1

Homework Statement



1)a water tank has a rectangle base of area 5m x 4 m and is 3 m deep. if it is full of water, calculate the thrust on the base and on each side.
2)a model helicopter of mass 5 kg rises with constant acceleration from rest to a height of 60 m in 10 s. find the thrust exerted by the rotor blades during the ascent.

Homework Equations



1)density= mass/ volume
pressure = force / area
2) s= (u+v)/t
v= ut + 1/2 at^2

The Attempt at a Solution


1) weight of water when it is full = 1000 x (3 x4 x 5) = 60000 x 10 = 600000N
upthrust = P= F/A
on sides = 600000/12 = 50000N
base= 600000/ 20 = 30000 N [ ans : 5.9 x 10 ^5, 1.8 x 10 ^5, 2.2 x 10^5 N]

2) thrust force - weight = ma
60=v/2 (10)^2
v= 12 m/s
12= 1/2 (10)^2 a
a= 0.24 m/s^2
thrust force = 5 x 0.24 + 50 = 51.2 N [ans: 55]
 
Last edited:

Answers and Replies

  • #2
Doc Al
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1) weight of water when it is full = 1000 x (3 x4 x 5) = 60000 x 10 = 600000N
upthrust = P= F/A
on sides = 600000/12 = 50000N
base= 600000/ 20 = 30000 N [ ans : 5.9 x 10 ^5, 1.8 x 10 ^5, 2.2 x 10^5 N]
The 'thrust' is the total force on each side, not the pressure. Hint: Be careful when calculating the force on the sides, since the pressure is not constant.

2) thrust force - weight = ma
Good.
60=v/2 (10)^2
That equation should be: d = ½at²; use it to solve for the acceleration directly.
 
  • #3
The 'thrust' is the total force on each side, not the pressure. Hint: Be careful when calculating the force on the sides, since the pressure is not constant.

P=F/A
P acting on the base (y) = h rho g = 3 x 1000 x 9.81 = 29500 Pa
F= 29500 x 20 = 5.9 x 10 ^5 N

F acting on the sides (x) = 5 x 1000 x 9.81 x 12 = 5.9 x 10 ^5N

F on (z) = 4 x 1000 x 9.81 x 15 = 5.9 x 10 ^5 N

* why is my upthrust force is different from the answer given?

* out of topic ( i just learnt it today)
- is there any relationship between young's modulus with density?
 
  • #4
Doc Al
Mentor
45,093
1,398
P=F/A
P acting on the base (y) = h rho g = 3 x 1000 x 9.81 = 29500 Pa
F= 29500 x 20 = 5.9 x 10 ^5 N
This is correct. You calculated the pressure on the bottom, which is uniform. Then used that to calculate the total force.

F acting on the sides (x) = 5 x 1000 x 9.81 x 12 = 5.9 x 10 ^5N

F on (z) = 4 x 1000 x 9.81 x 15 = 5.9 x 10 ^5 N

* why is my upthrust force is different from the answer given?
Not sure what those calculations are doing. What's the average pressure on each side? What's the area of each side?

* out of topic ( i just learnt it today)
- is there any relationship between young's modulus with density?
Not that I know of.
 
  • #5
Not sure what those calculations are doing. What's the average pressure on each side? What's the area of each side?
"a water tank has a rectangle base of area 5m x 4 m and is 3 m deep"

let the length = 5m
width = 4 m
height = 3 m

F acting on (x) = F that acting on the sides (left / right)
F acting on Z= F acting on the wall behind/ front
 
  • #6
Doc Al
Mentor
45,093
1,398
"a water tank has a rectangle base of area 5m x 4 m and is 3 m deep"

let the length = 5m
width = 4 m
height = 3 m

F acting on (x) = F that acting on the sides (left / right)
F acting on Z= F acting on the wall behind/ front
I understand the problem. So answer my questions.
 
  • #7
arent the pressure is equal on all sides?
the area in the bottom = 5 x 4 = 20 m^2
sides= 3 x 4 = 12 m^2
front= 3 x 5 = 15 m^2
 
  • #8
Doc Al
Mentor
45,093
1,398
arent the pressure is equal on all sides?
No. Pressure depends on depth below the surface.
 

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