Calculate Forces on Parachute & Person: Parachute Homework

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The discussion revolves around calculating the forces acting on a parachute and a person during a parachute descent. For a 78 kg person experiencing a downward acceleration of 2.8 m/s² and a parachute weighing 5.4 kg, the upward force exerted by the air on the parachute was calculated as 1050.84 N. The downward force exerted by the person on the parachute, including the tension in the string and the person's weight, was found to be 1747.2 N. However, the calculations were noted to have sign errors, as the net force must align with the direction of acceleration. Correcting these signs is crucial for accurate results in the physics problem.
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Homework Statement


A 78 kg person is parachuting and experiencing a downward acceleration of 2.8 m/s2. The mass of the parachute is 5.4 kg.
(a) What upward force is exerted on the open parachute by the air?
(b) What downward force is exerted by the person on the parachute?



Homework Equations



Fnet = ma

The Attempt at a Solution



Forces on parachute:
Weight of parachute => W1 = mg = 5.4kg * 9.8 m/s^2 = 52.92N down
Force of Air on parachute => A N up
Tension of string => T N down

Forces on Person:
Tension of string => T N up
Weight of person => W2 = 764.4 (found the same way as above) down


Fnet on the person = 78kg * 2.8 m/s^2 = 218.4 N = T - 764.4N
solve for T... T = 982.8N

Fnet on parachute = 5.4kg * 2.8m/s^2 = 15.12N = A - T - W1,
T and W1 known:
15.12N = A - 982.8N - 52.92N
solve for A... A = answer to question (a) air resistance = 1050.84N
and the tension, T, in the string, plus the weight of the person, W2 is the answer to (b) The total downward force on the parachute is 1747.2 N.
This, however, is not the right answer...
Any help is really appreciated!
Thanks
 
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A_lilah said:

Homework Statement


A 78 kg person is parachuting and experiencing a downward acceleration of 2.8 m/s2. The mass of the parachute is 5.4 kg.
(a) What upward force is exerted on the open parachute by the air?
(b) What downward force is exerted by the person on the parachute?



Homework Equations



Fnet = ma

The Attempt at a Solution



Forces on parachute:
Weight of parachute => W1 = mg = 5.4kg * 9.8 m/s^2 = 52.92N down
Force of Air on parachute => A N up
Tension of string => T N down

Forces on Person:
Tension of string => T N up
Weight of person => W2 = 764.4 (found the same way as above) down


Fnet on the person = 78kg * 2.8 m/s^2 = 218.4 N = T - 764.4N
solve for T... T = 982.8N

Fnet on parachute = 5.4kg * 2.8m/s^2 = 15.12N = A - T - W1,
T and W1 known:
15.12N = A - 982.8N - 52.92N
solve for A... A = answer to question (a) air resistance = 1050.84N
and the tension, T, in the string, plus the weight of the person, W2 is the answer to (b) The total downward force on the parachute is 1747.2 N.
This, however, is not the right answer...
Any help is really appreciated!
Thanks
You've got your plus and minus signs mixed up. The net force must be in the direction of the acceleration. For the person, for example, his weight must be greater than T, in order for him to accelerate downwards.
 
Thank you!
 
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