Find Specific Gravity of 8.3 kg Block in Water

[jti3066]

Homework Statement

A 8.3 kg block of some unknown material is suspended from a spring scale and submerged in water. The spring scale reads 46.9 N. What is the specific gravity of the block?

Homework Equations

SG = density of object/density of reference material (water = 1000 kg/m^3)

density = mass/volume

The Attempt at a Solution

How do I find the density of the object if I do not know its volume?

 

[rl.bhat]

Find the weight of the block mg in air.
Weight in water is given. Find loss of weight. That is equal to weight of the displaced liquid. From that you can find the volume of the block.

 

[jti3066]

so...

weight in air = 8.3 * 9.81 = 81.42 N

differnce = 81.42 - 46.9 = 35.523 N (weight of displaced liquid)

Density = m/v so

1000 = 35.523/v ...v = .035523 (volume of block)

density of block = m/v = 8.3/.035523 = 233.65

SG ( of block) = 233.65/1000 = .233

I did this and I get the wrong answer

 

[rl.bhat]

35.523 N is the weight of the displaced water. Find the mass of the displaced water. And then find the density of the block.

 

[jti3066]

ok...

mass of displaced water is 3.21 kg

1000 = 3.21/v...v= 0.00321

Density of block = 8.3/0.00321 = 2585.669

SG = 2585.669/1000 = 2.585

Still getting wrong answer

 

[jti3066]

can someone just show me the step by step solution please...That way I can know exactly know what I am doing wrong instead of doing the wrong steps over and over...TIA

 

[rl.bhat]

35.523n/9.8 =---?

 

[jti3066]

SG = 81.42/(81.42 - 46.9) = 2.36

Thanks for your help though

 

[jti3066]

Now for the density of block...

D = 8.3/.0035 = 2.37 * 10^3 kg/m^3