How does the depth of an object in water affect its horizontal force?

  • Thread starter Thread starter jimmy42
  • Start date Start date
  • Tags Tags
    Forces
AI Thread Summary
The discussion centers on how the depth of an object submerged in water affects its horizontal force. When an object is half in the water, the horizontal forces acting on it from the water cancel each other out, resulting in a net horizontal force of zero. The upward buoyant force, however, is influenced by the volume of water displaced, which is half when the object is half-submerged compared to being fully submerged. This aligns with Archimedes' principle, indicating that the buoyant force is directly proportional to the submerged volume. Thus, while horizontal forces remain unchanged, the buoyant force is indeed halved when the object is only partially submerged.
jimmy42
Messages
50
Reaction score
0
If an object is only half in the water, so floating. Will its horizontal force be half of that if it was fully underwater?

Thanks.
 
Physics news on Phys.org
What do you mean by "its horizontal force"?
 
So, the water will be pushing on it in the positive and negative X ( left and right ) direction. If there is a box floating then only the bottom part will feel this force, as they are both equal they will cancel out. That means the box will not move.

I know that the force acting in the Y direction, so up down will be different and can work that out. I just wanted to know if half the box is in the water does it follow that because half as much water is exposed to the box then the force is half as much as when it's fully in the water.

Thanks.
 
jimmy42 said:
So, the water will be pushing on it in the positive and negative X ( left and right ) direction. If there is a box floating then only the bottom part will feel this force, as they are both equal they will cancel out. That means the box will not move.
The net horizontal force exerted by the water on the object will be zero. However the water does exert an upward buoyant force on the object.

I know that the force acting in the Y direction, so up down will be different and can work that out. I just wanted to know if half the box is in the water does it follow that because half as much water is exposed to the box then the force is half as much as when it's fully in the water.
The buoyant force equals the weight of the displaced liquid (this is Archimedes' principle). So the buoyant force when half-submerged is half as much as when fully submerged.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .

Similar threads

Back
Top