A hot-air balloon consists of a basket, one passenger, and some cargo. Let the total mass be M?. Even though there is an upward lift force on the balloon, the balloon is initially accelerating downward at a rate of ?g?/3. (a) Draw a free-body diagram for the descending balloon. (b) Find the upward lift force in terms of the initial total weight ?Mg?. (c) The passenger notices that he is heading straight for a waterfall and decides he needs to go up. What fraction of the total weight must he drop overboard so that the balloon accelerates upward? at a rate of ?g?/2? Assume that the upward lift force remains the same.
Solution 56P Total mass is M. The initial acceleration of the balloon downwards was g/3. There is an upward lift by the air drag. So, the final equation for this system can be written as, Ma = Mg upward lift M × g/3 = Mg upward lift upward lift = Mg Mg/3 = Mg ---3-------------(1) a) The free body diagram is given below, b) the upward lift we found as, upward lift = Mg2 3 c) Now the balloon should go up at an acceleration of g/2. So the equation will be, 2 ma = Mg3 mg --------------------(2) Here m is the reduced mass after throwing some mass from the total mass M. So, let’s put the values in equation (2), mg/2 = 2Mg/3 mg mg/2 + mg = 2Mg/3 3mg / 2 = 2Mg / 3 2 2 4 m = 3× 3 = M .9 So the mass which has to be thrown is, M m = M M = M . 5 9 9 Conclusion: So, 5/9 th of the total mass has to be thrown out side to move upwards with an acceleration of g/2.