CE2 physics question: An automatic drinking water device for livestock currently being designed

Introduction Second Year Physics Question: An automatic drinking water device designed for livestock IV. (13 points) Reference Answers (1) This device uses pressure, buoyancy, connectors, force balance, etc.

CE2 physics question: An automatic drinking water device for livestock currently being designed

IV. (13 points) Reference Answers

(1) This device applies physical knowledge such as pressure, buoyancy, connectors and balance of forces. …………………………(2 points)

(2) Perform a force analysis on ABC as a whole:

ABC as a whole is affected by the gravity, the water reservoir and its The downward force and the upward buoyant force of the water…………………… (1 point)

When the upward buoyant force on the float is greater than that of the lower cover A, vertical rod B and when the combined force of gravity on the float C and the downward pressure of the water in the water tank on the lower cover A, the lower cover A will resist to the water outlet hole of the water tank. No more water flows from the water tank into the drinking water pool……………… (1 point)

Gravity on the bottom cover A, the vertical rod B and the float C is G=mg=( 600 ×10-3kg+400×10-3kg)×10N/kg=10N……(1 point)

When the water tank is full of water, it is obtained from the formulas p=ρgh and p=F/S:

The downward pressure exerted by the water in the water tank on the bottom cover A

F=ρghS=1.0×103kg /m3×10N/kg× 0.6m×30×10-4m2=18N ……………………………… (1 point)

According to Archimedes' principle, the force maximum thrust on the floating ball can be obtained

F float m =ρwater gVdischarge m=1.0×103kg/m3×10N/kg×2×10-3m3=20N……………… ……………………(1 point)

Because when the water tank is filled with water, the maximum buoyant force on the float (20N) is less than the combined force of the gravity on the bottom cover A, the vertical rod B and the float C., and the downward pressure of the water in the water tank on the bottom cover A (10N + l SN = 28N), so. This automatic device cannot work normally when the water tank is full of water. ………………………………………………………………………………………………………… (2 points)

(3) The first situation: it appears from the calculation process above that the condition for the automatic device to operate Normally is to ensure that the upward buoyancy force on the float is not less than the gravity on the bottom cover A, the vertical rod B and the float C and the gravity in the water tank. The resulting force of the downward pressure of the water on the bottom cover A, hence the gravity (mass) on the float; A and the vertical rod B; the size of the cross section of the water outlet, etc. are all affected. Factors that determine whether this automatic device can work properly. (Other correct statements will be scored equally. Solving a factor will earn 1 point, with a maximum of 4 points)

Second case: So that this automatic device always functions normally when the water tank water is full of water. , you cantake the following improvement measures: ① Appropriately reduce the cross section of the water outlet hole; ② Appropriately reduce the water level when the water tank is full of water ③ Appropriately increase the float volume while keeping the water level; mass of the float unchanged ④ While maintaining the When the volume of the float remains unchanged, Appropriately reduce the mass of the floating ball.

Design experiments prove that the buoyancy of water and objects is directly proportional to the volume of water displaced by the object.

ρ water S ( h + d )

< td> When valve C has just been opened, gravity on the valve is equal to the pulling force of lever A on the valve plus the buoyancy of the water, i.e. GC=FA+F float C=FA+ρwater gSd, --①

According to the lever balancing conditions: FA×OA=FB×OB, ∵OB=2OA, ∴FA=1/2FB,

C is pulled up, and the water pressure on C plus The gravity of C is equal to 2 times the buoyancy of float D,

The pressure on C is equal at ρ water g (3h-d) S,

< p>The gravity of C is mc×g,

The buoyant force on D is ρ water g4S (3h-2.5h )

ρ water g (3h-d) S+ mc×g=2ρ water g4S (3h-2.5h)

mc=2ρ water 4S (3h-2.5h) -ρ water (3h-d) S

=ρ water S (4h-3h+d)

< p>=ρwater S(h+d)

So the answer is: ρwater S(h+d).

According to Archimedes' principle, F float = G row = ρ liquid · g · V row, therefore F float / V row = ρ liquid · g

Experimental steps:

1. Fill the measuring cylinder with an appropriate amount of water (not exceeding the measuring range)

2. Use a spring dynamometer to. hang a hook and immerse it in water, calculate the volume displaced, and after immersion, F float = G object (known), so we can calculate if the displacement F float/V is equal to ρ liquid·g

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