What Is the Mass of the Late Arrival in the Hot-Air Balloon?

[pttest]

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 1220 kg , and the balloon is neutrally buoyant. When the late arrival climbs up into the basket, the balloon begins to accelerate downward at 0.59m/s2 .What was the mass of the last person to climb aboard?

Homework Equations

F=ma

The Attempt at a Solution

do I need to find the net force before and after the last person's arrival?then compare those 2 equations to get the last person's mass? could somebody give me a clue how to approach this kind of a problem?

Thanks in advance.

 

[rock.freak667]

On the air balloon there are two forces, weight (downwards) and upthrust (upwards) .

So yes, you need to compare before (which will give you the upthrust) and then after to get the person's mass.

 

[pttest]

so would the 1st equation be??
upthrust (F_u) - Wt of the basket & people (W) = mass (m). acceleration (a)
F_u - 1220 * 9.81 = 1220 * 0.59
Therefore F_u = 12919.8 N

would the 2nd equation be??
F_u - (1220+mass of last person)g = (1220+mass of last person)a

am I in right way ...? if not could you please help me?

Thanks again...

 

[rock.freak667]

so would the 1st equation be??
upthrust (F_u) - Wt of the basket & people (W) = mass (m). acceleration (a)
F_u - 1220 * 9.81 = 1220 * 0.59
Therefore F_u = 12919.8 N

I think F_U -W =0 since before the other person jumps into the basket it is just hovering. The second equation is correct.

 

[rado5]

The first equation is F_u-W_t=0 ,because at the beginning the acceleration is zero, because "the balloon is neutrally buoyant" at the beginning.