Tension in string attached to rock half under water.?

Question

Question: A 7.30 kg rock whose density is 4500 kg/m^3 is suspended by a string such that half of the rock's volume is under water. What is the tension in the string? (in Newtons)

I assume that you need to find the volume of the rock first, so

7.30 kg x ((1m^3)/4500kg)= 32850 m^3

Then you need to find what volume of the rock is under water, so

32850 x .5= 16425 m^3

After that im not quite sure where to go. If any one can help it would be greatly appreciated! Thanks

Answer 1

Here's a hint:

There are 3 forces acting on the rock:

1. Gravity.  This exerts a force equal to mg, downward.

2. Buoyancy.  This exerts an upward force, equal to the weight of the displaced water (figure out the weight of the displaced water by multiplying (volume of displaced water)×(density of water)×(g).)

3. Tension.  This exerts an upward force.

Now, notice that since the rock is not accelerating, all the upward forces exactly balance all the downward forces.  That is:

Upward forces = downward forces

Buoyancy + Tension = Weight of rock

Now just figure out some expressions for "Buoyancy" and "Weight of rock"; and solve for "Tension".