# A rock weighs 2.25 Newtons when suspended in air. In water, it appears to weigh only 1.8 Newtons. Why?

Jun 19, 2014

There is a buoyant force acting vertically upwards on the rock.

The most likely scenario was that it was hung on a Newton meter or similar whilst being fully/partially submerged in water.

But let's first consider the method of determining the weight of the rock in air with a Newton meter. There would be two vertical forces acting on the rock: tension, $T$, and weight, $w$. Therefore, $T = w$. The tension is the value displayed on the Newton meter and in this case it is equal to the weight.

In the case of full/partial submersion there would be three vertical forces acting on the rock: tension, $T$, weight, $w$, and a buoyant force, ${F}_{b u o y}$. $T$ and ${F}_{b u o y}$ act upwards. Therefore, $T = w - {F}_{b u o y}$. So $T$ is less than the weight, as $T$ is the reading on the Newton meter that is why the rock appears to have less weight (in reality the reading on the Newton meter is the difference between weight and the buoyant force).

If you would like to know how to calculate the buoyant force see the answer to this question: https://socratic.org/questions/how-do-you-calculate-the-buoyant-force-of-an-object