How Does Submerging a Sphere Affect Scale Readings?

[mochigirl]

Well, in that case the W of the contents would be 1.0N as before right? But that's not the correct answer.

 

[Doc Al]

Well, in that case the W of the contents would be 1.0N as before right?

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]

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. 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."

So in that case, the weight of the ball should be counterbalanced by the tension and buoyancy forces right? 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?

 

[Doc Al]

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.

Try 1.00 N.

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."

But there are external forces: The tension in the rod pulls up on the ball.

So in that case, the weight of the ball should be counterbalanced by the tension and buoyancy forces right?

True, but having the ball in the fluid changes things. After all, if you then removed the ball, the fluid level will lower.

Realize that whatever force the fluid exerts upward on the ball (the buoyant force), the ball exerts downward on the fluid (Newton's 3rd law). Thus the ball exerts a downward force on the fluid, exactly equal to the weight of the displaced fluid, thus compensating for the spilled fluid.

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?

That "hint" seems a bit misleading, since the scale reading is not simply the weight of the contents.