A mountain-climber friend with a mass of 82 kg ponders the idea of attaching a helium-filled balloon to himself to effectively reduce his weight by 25% when he climbs. He wonders what the approximate size of such a balloon would be. Hearing of your physics skills, he asks you. Share with him your calculations that find the volume of the balloon.
To find the approximate size of the helium-filled balloon, we can start by understanding the concept of buoyancy. Buoyancy is the upward force exerted on an object immersed in a fluid, which in this case is the air.
First, let's determine the reduced weight of the mountain climber when the balloon is attached. The weight reduction is given as 25% of his actual weight, which is 82 kg. Therefore, the reduced weight would be:
Weight reduction = 25% * 82 kg = 0.25 * 82 kg = 20.5 kg
This means the effective weight of the climber with the balloon attached is 82 kg - 20.5 kg = 61.5 kg.
Now, we can calculate the buoyant force acting on the balloon, which is equal to the weight of the air displaced by the balloon. According to Archimedes' principle, this buoyant force is equal to the weight of the fluid displaced.
The buoyant force (F_b) can be calculated using the formula:
F_b = ρ * g * V
where ρ is the density of the air, g is the acceleration due to gravity, and V is the volume of the balloon.
For simplicity, we can assume the density of air is constant at 1.225 kg/m^3, and the acceleration due to gravity is 9.8 m/s^2.
Substituting the values, we have:
F_b = 1.225 kg/m^3 * 9.8 m/s^2 * V
Since the buoyant force is equal to the weight reduction (20.5 kg), we can write the equation as:
20.5 kg = 1.225 kg/m^3 * 9.8 m/s^2 * V
Now, solving for V, we get:
V = 20.5 kg / (1.225 kg/m^3 * 9.8 m/s^2)
V ≈ 1.682 m^3
Therefore, the approximate volume of the helium-filled balloon would be around 1.682 m^3.