- #1

A_lilah

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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