(II) A spherical balloon has a radius of 7.15 m and is filled ...

Question

(II) A spherical balloon has a radius of 7.15 m and is filled with helium. How large a cargo can it lift, assuming that the skin and structure of the balloon have a mass of 890 kg? Neglect the buoyant force on the cargo volume itself.

Accepted Answer

Hello, fellow physicists today we're gonna solve following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem, determine what the maximum cargo weight value is that a spherical air balloon filled with hydrogen, which has a density of 0.08375 kilometers per meters. Cube can lift up at zero °C and one atmospheric pressure. The balloon has a radius of 8.15 m and the combined mass of the balloon structure and the skin is 880 kg. Assume that the buoyant force on the cargo itself is negligible. Awesome. So that's our end goal. Our end goals were asked to determine what the maximum cargo weight value is that this specific spherical balloon filled with hydrogen can support. So basically, we're trying to see how much and what's the maximum cargo weight. So since this isn't a spherical balloon that's floating, we need to figure out what the maximum weight it it can have in order for it to lift up off the ground. So it can't be too heavy or also be stuck on the ground. Awesome. So now that we have an idea of what we're solving for, let's look at our multiple choice answers to see what our final answer might be. And let us note that they're all in the same units of newtons. So A is 1.9 multiplied by 10 to the power of three, B is 2.3 multiplied by 10 to the power of three C is 3.1 multiplied by 10 to the power of three and D is 4.3 multiplied by 10 to the power of three. Ok. So first off, let us note that the balloon buoyancy must equal the weight of the balloon plus the weight of the hydrogen located inside of the balloon plus the weight of the cargo load. Let us assume that the hydrogen is at a temperature of zero °C and one atmosphere when calculating its weight. So let us note that in this case that the buoyancy is determined by the weight of the air displaced by the balloon's volume. Thus, we should be able to recall and use the following equation that F buoyancy is equal to ro air multiplied by V balloon multiplied by G is equal to MH, multiplied by G plus MB multiplied by G plus M. See multiplied by G where FB is equal to the buoyant force row. A is the density of air VB is the volume of the balloon G is the acceleration due to gravity. M is the mass of hydrogen. MC is the mass of the cargo MB is the mass of the balloon and row H, which we're gonna be discussing row H in just a second. But let us note that row H is the density of hydrogen. So now, at this point, let us recall it that we can also write that MC is equal to row a multiplied by VB minus MH minus MB is equal to row a multiplied by VB minus row H which this is where row H is important, this is the density of hydrogen multiplied by VB minus MB is equal to parenthesis row A minus row H multiplied by VB minus MB. Awesome. So now at this point, we can plug in all of our known variables to solve for MC, which is our final answer that we're trying to ultimately solve for. So MC is equal to bracket RC. 1.29 means now we need to take the density of air which is 1.29 M, 1.29 and its units are kilometer. So kilograms kilograms per meters cubed minus the density of hydrogen, which is given to us as 0.0 8375. And its units are also kilograms per meters cubed. And we need to multiply this by and then don't forget your other parenthesis in your bracket multiplied by the volume which was recalled that the volume of a sphere is four thirds multiplied by pi multiplied by R cubed, which R in this case is given to us as 8.15 m. And don't forget it's to the power of three. So is the power of three. And then we need to subtract MB which the mass of the balloon is given to us as 880 kg. Fantastic. So now let me plug that into our calculator. We will find that MC is equal to 1855.26 kg, which is approximately equal to 1.9 multiplied by 10 to the power of 3 kg when we round to one decimal place and write it in scientific notation and that's it we've solved for this problem. Hooray, we did it. So the final answer has to be the letter A 1.9 multiplied by 10 to the power of three newtons. Thank you so much for watching. Hopefully that helped and I can't wait to see in the next video. Bye.

Struggling with Physics?

(II) A spherical balloon has a radius of 7.15 m and is filled with helium. How large a cargo can it lift, assuming that the skin and structure of the balloon have a mass of 890 kg? Neglect the buoyant force on the cargo volume itself.

Watch next

Buoyancy & Buoyant Force practice set