What is the Mass of the Late Arrival in a Hot-Air Balloon Ride?

AI Thread Summary
In a hot-air balloon ride scenario, the initial combined weight of the basket and passengers is 1200 kg, making the balloon neutrally buoyant. When a late arrival climbs aboard, the balloon accelerates downward at 0.52 m/s². To determine the mass of the late arrival, the relationship F = ma can be applied, where the unbalanced force is due to the additional weight of the passenger. The forces acting on the system include the downward gravitational force and the upward buoyant force, which balance out before the extra passenger enters. The discussion concludes with the original poster successfully solving the problem.
paste
Messages
7
Reaction score
0

Homework Statement


For a birthday gift, you and some friends take a hot-air balloon ride. One friend is late, so the balloon floats a couple of feet off the ground as you wait. Before this person arrives, the combined weight of the basket and people is 1200kg , and the balloon is neutrally buoyant. When the late arrival climbs up into the basket, the balloon begins to accelerate downward at 0.52m/s^2

What was the mass of the last person to climb aboard?


Homework Equations


F=ma


The Attempt at a Solution


Looked through my notes and I am struggling to find any formula without having a known mass. Any help is appreciated, thanks.
 
Physics news on Phys.org
Welcome to PF, paste!
Try turning it around. If you did know the mass, could you calculate the acceleration?
But you do know the acceleration, so use the same formula to find the mass.
 
Thanks, how is the mass found if the force with the extra mass unknown.
 
There is a downward force of gravity on the whole mass. And an upward buoyant force. These cancel out before the extra passenger gets on. The unbalanced force of the passenger's weight is the F in F = ma. Weight is a force, found using F or W = mg.
 
The new force and mass are both unknowns, how does this help?

I've solved it now, thanks.

(1200+m)*0.52=mg
624+0.52m = mg
=9.81*m
624+0.52m-0.52m=9.81m-0.52m
624=9.29m
624/9.29=9.29m/9.29
m=67.17kg
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top