Question

1. Estimate the buoyant force that air exerts on a man. (To do this, you can estimate his volume by knowing his weight and by assuming that his weight density is about equal to that of water. Assume his weight is 940 N.) answer in N
2.On a perfect fall day, you are hovering at low altitude in a hot-air balloon, accelerated neither upward nor downward. The total weight of the balloon, including its load and the hot air in it, is 17000 N.
(a) What is the weight of the displaced air?
answer in N
(b) What is the volume of the displaced air?
answer in m^3

Answer 1

Solution :

1. We know that : Buoyant force = weight of the liquid displace

                                                  = volume displaced x density of the fluid

Now volume of the man = $\frac{\text{mass}}{\text{density}}$

Mass = weight / g

         $=\frac{940}{9.8}$

         = 95.92 kg

And density = 1000 $kg/m^3$

Therefore,

$\text{volume} = \frac{\text{mass}}{\text{density}}$

           $=\frac{95.92}{1000}$

           = 0.0959 $m^3$

We know density of air = 1.225 $kg/m^3$

∴ Mass of air displaced = 0.0959 x 1.225

                                       = 0.1175 kg

Weight of the air displaced = 1.1515 N

Therefore, the buoyant force = 1.1515 N

2). As the balloon is not accelerated, the net force acting on it is zero.

Thus the weight that acts downwards = buoyant force upwards

So, the weight of the air displaced = weight of the balloon

                                                          = 17000 N

Therefore, the mass of the air displaced = volume of the air displaced (volume of the balloon) x density of air

$\frac{17000}{9.8} = \text{volume of air} \times 1.225$

$\text{Volume of air displaced} = \frac{1700}{9.8 \times 1.225}$

                                     = 1416.0766 $m^3$