Calculate Forces on Parachute & Person: Parachute Homework

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SUMMARY

The discussion focuses on calculating the forces acting on a parachute and a person during a parachute descent. A 78 kg person experiences a downward acceleration of 2.8 m/s², while the parachute has a mass of 5.4 kg. The upward force exerted by the air on the parachute is calculated to be 1050.84 N, while the total downward force exerted by the person on the parachute is 1747.2 N. The calculations reveal a misunderstanding of force direction, emphasizing the need for accurate sign conventions in physics problems.

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  • Understanding of Newton's Second Law (Fnet = ma)
  • Knowledge of gravitational force calculations (W = mg)
  • Familiarity with free body diagrams
  • Basic algebra for solving equations
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  • Learn about the implications of net force direction on motion.
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  • Review examples of tension in strings and its applications in various physics problems.
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to forces and motion in parachuting scenarios.

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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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