F_{P} is the force of air resistance on the parachute

[mcg5132]

Homework Statement

An 84 kg person is parachuting and experiencing a downward acceleration of 3.0 m/s2. The mass of the parachute is 5.5 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
Fg=mg
W=mg

The Attempt at a Solution

How many forces are acting on him? Parachute force, air resistance, weight (mg), and acceleration? Is that the only forces? Would the equation look like Fnet = mg-Fp-Fair=mg ? I cannot seem to come up with the correct answer. Would a) and b) be equal?

 

[PhanthomJay]

The parachute force is the force of air resistance (ignoring any small air resistance on the person), so don't count it twice. Also, acceleration is NOT a force. So there are just 2 forces acting on the person, the net sum of which must be in accord with Newton 2. You say 'a' and 'b' are equal...reasoning?

 

[mcg5132]

Sometimes they put more applicational problems instead of actual computing it was just a total guess. But Newtons law says Fnet=mg-Fp=ma correct? Is that the equation for this type of problem?

 

[PhanthomJay]

Sometimes they put more applicational problems instead of actual computing it was just a total guess. But Newtons law says Fnet=mg-Fp=ma correct? Is that the equation for this type of problem?

Yes, but be careful to identify to which object or system of objects you are applying Newton's laws.

 

[mcg5132]

I'm confused, if these numbers are correct it seems like the answer would turn out exactly the same. Fp=mg(823.2)-ma(252)=571 Yes?

 

[mcg5132]

Anybody?

 

[vladimir69]

for part a you seem to be quite close
but I think what you might need is
$$(m_{1}+m_{2})a=-m_{1}g-m_{2}g+F_{p}$$
where
$$m_{1}$$ is the mass of the person
$$m_{2}$$ is the mass of the parachute
g=9.8
a=3.0