The discussion centers on the effects of submerging a ball in a fluid within a beaker on scale readings. Initially, the beaker filled with a fluid of density 890 kg/m³ weighs 1.00 N. When a ball of density 5000 kg/m³ and volume 60.0 cm³ is submerged, the scale reading remains unchanged at 1.00 N because the buoyant force acting on the ball is countered by the force from the rod holding the ball in place. The pressure at the bottom of the beaker before and after submersion does not change, confirming that the scale measures the same force.
PREREQUISITESStudents of physics, educators teaching fluid mechanics, and engineers involved in fluid dynamics will benefit from this discussion.
That would be the apparent weight of the submerged ball. Meaning: If you hung the ball from a scale, then dunked it into the fluid, that's what the scale would read.Vanessa23 said:I understand that the scale will not change its reading, so W_2= 1N, but how would you find the weight of the ball? I thought it would be:
m(ball)*g - rho_f*V*g
That's what I would assume.Vanessa23 said:So the question asks "What is the weight W_b of the ball?" and 2.94-(890*0.00006*9.81)=2.42 is incorrect, then they are simply asking for the weight=mg?
Not so. There is definitely a buoyant force acting on the submerged ball, as usual. It equals rho_f*V*g, just like you would expect.Vanessa23 said:Normally buoyancy force comes into play when a mass is suspended by a spring into the fluid. Here I believe that the ball is held in place and the scale does not change, so there is no buoyancy force.
No. The forces on the submerged ball are:The only thing keeping the ball where it is in the fluid is the rigid rod, so the force applied by the rod is equal and opposite to the force applied by the ball since there is no buoyancy force. Therefore, the force applied by the rod would just be negative the weight.
There are two ways to solve this problem. An easy way (if you spot it) and a "hard" way. Let's do the hard way first.So basically in this whole situation you treat the beaker of fluid as one system and the ball+rod as a separate system because inserting the ball has no affect on either system. By that I mean that the scale doesn't change and the fluid does not affect the ball and rod.
It just seems odd that buoyancy wouldn't factor in...
No, the weight of the contents is greater, since the ball is more dense than the fluid it displaced. But the scale reading is the same, which is what the question asks about. (Realize that the ball is being partly supported by the rod.)mochigirl said:Well, in that case the W of the contents would be 1.0N as before right?
Try 1.00 N.mochigirl said:sorry. that is what I meant. the weight that the scale displays should be 1 N. but the answer says that it is not. So I'm still a bit confused.
But there are external forces: The tension in the rod pulls up on the ball.Because the hint says:
"It makes no difference how the ball and fluid are arranged inside the beaker. If there are no external forces, the weight reading would be equal to the total weight of the beaker and its contents."
True, but having the ball in the fluid changes things. After all, if you then removed the ball, the fluid level will lower.So in that case, the weight of the ball should be counterbalanced by the tension and buoyancy forces right?
That "hint" seems a bit misleading, since the scale reading is not simply the weight of the contents.And then I thought that since they had asked that I calculate the weight of the displaced water=0.524N then, the new weight displays would be 0.476. But that is wrong too...what am I doing wrong?