What is the tension in the string connecting a cork and a cube in water?

[college boy19]

One end of a light thin string is attached to the bottom of a 3.7 g, 42 cm3 cork. The other end is attached to a 58 g cube. The arrangement is placed in water and it is found that with 82% of the cork submerged the cube is held in equilibrium. What is the density of the cube in g/cm3?

I need help figuring out this problem.

 

[tiny-tim]

Welcome to PF!

Hi college boy19! Welcome to PF!

(try using the X² tag just above the Reply box )

Show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[college boy19]

okay,
the density of an object is its mass divided by its volume:

D=m/v

so for the cube all i have in the equation is :
D= 58g/v

So i know that in order to get the density i need the volume of the cube.

this is where i am stuck.

 

[tiny-tim]

Hi college boy19!

So i know that in order to get the density i need the volume of the cube.

That's not the only way to get the density …

you can also work it out from the forces on the cork

 

[college boy19]

Well the forces acting on the cork are the buoyant force which is the the weight of the cork which is mg which is:

BF= 3.7*g
but gravity in SI units is 9.8kg/m^3 and I am using CGS units so gravity is 98g/cm^3
So the weight of the cork is

BF= 3.7(98)=362.6

the buoyant force = to the weight of the displaced fluid

 

[tiny-tim]

Well the forces acting on the cork are the buoyant force which is the the weight of the cork

No, the forces acting on the cork are the buoyant force, the weight, and the tension in the string.

 

[college boy19]

the buoyant force:
362.6
the weight:
mg=362.6
how do i figure out the tension in the string?

is it the buoyant force - the weight of the cube

 

[tiny-tim]

the buoyant force:
362.6
the weight:
mg=362.6

How can the buoyant force be equal to the weight?

How is buoyant force defined?

how do i figure out the tension in the string?

You have two equations for it …

one for the cork, and one for the mass …

that should be enough to solve the whole problem.