# Buoyant force help

1. Apr 27, 2006

### cscott

70% of a mass is supported by a slab of ice and the ice sinks down so that only half of what was previously exposed now is exposed. What is the mass assuming that the ice has a volume of 10 m^3 and the mass has a specific gravity of 1.0?

Why can't I use the buoyant force of the ice before and after the extra weight is added and subtract to get the weight of the object itself? I get 539.5 kg while the textbook says 790 kg. Only hints please!

2. Apr 27, 2006

### Staff: Mentor

Sounds good to me. Show what you did exactly.

3. Apr 27, 2006

### cscott

Alright, well since $F_B = w$ when an object is floating, the buoyant force is $F_B = \rho_{ice} Vg = (0.917 \times 10^3)(10)(9.8) = 9.0 \times 10^4N$.

$$\frac{0.917 \times 10^3}{1.00 \times 10^3} + \frac{1}{2}(1 - 0.917) = 0.9585$$
for the fraction of the ice submerged once the unkown mass is put on.

With this, the buoyant force is $F_b = (1.00 \times 10^3)(0.9585 \cdot 10)(9.8) = 9.4 \times 10^4N$

$$W_{object} = 9.4 \times 10^4 - 9.0 \times 10^4 = 4.0 \times 10^3N$$

$$\frac{4.0 \times 10^4 \cdot 1.3}{9.8} = 530 kg$$

I rounded the numbers this time.

Last edited: Apr 27, 2006
4. Apr 27, 2006

### Staff: Mentor

Your method looks OK to me (assuming we are interpreting the problem correctly--why do they specify the specific gravity of the mass?) but I would divide by 0.7 instead of multiply by 1.3.

5. Apr 27, 2006

### cscott

The fact that they gave the specific gravity had me starting to work with volume but I could never get any sensible answers. If it makes any difference, this is how the question is worded in the book:

6. Apr 27, 2006

### Staff: Mentor

Since the specific gravity is one, it makes no difference--the buoyant force on the bear equals the weight of her submerged portion. (Were it something else, it would matter.)

7. Apr 27, 2006

### cscott

Ah, ok. I guess the textbook answer is just wrong...?