# Newtons Second Law: Finding net forces

1. Sep 18, 2010

### battlebball

1. The problem statement, all variables and given/known data
A spaceship lifts off vertically from the Moon, where g = 1.6 m/s2. If the ship has an upward acceleration of 1.0 m/s2 as it lifts off, what is the magnitude of the force exerted by the ship on its pilot, who weighs 735 N on Earth?
From Halliday, Fundamentals of Physics, 9e

2. Relevant equations
Newton's Second Law: F(net)= ma

3. The attempt at a solution
I'm more looking for an explanation on why the answer is 195 N.
Granted I can see that they added the two forces. (75kg*1.0m/s2 + 75kg*1.6m/s2)

What I do not understand is why I need to add the two positive forces. I was trying to draw a few body diagram.

What I was picturing: two force vectors in opposite direction. So the magnitude would result in 45N if this were the case.

2. Sep 18, 2010

### Sennap

How much force did the ship exert on the pilot before the ship took off?

When the ship takes off from the ground, does the ship exert more force on the pilot or less?

3. Sep 18, 2010

### snshusat161

Take pilot as a center, now

Force acting vertically downward:

1. weight of the pilot

Force acting vertically upward:

1. Normal force

N - W = m.a

N = W + m.a

= 75(1.6) + 75 (1.0)
= 195 N

Hint: Mass = 75 kg because its 735/9.8

4. Sep 18, 2010

### snshusat161

You need to know, why they added two force. Whenever you consider any body in mechanics, forget everything about its surrounding. Don't think that there is an engine which is also exerting force. When we will consider aircraft as our body then there will be two force W and the force exerted by the engine. And since the aircraft is moving upward it means that the force exerted by the engine is more than the weight of aircraft but here in the problem, pilot is our body and only two forces (normal reaction and weight) are acting on it. Since the pilot is moving upward (in the same direction as that of normal reaction) therefore normal reaction is greater than weight.

5. Sep 18, 2010

### battlebball

Thanks for the responses everyone. This pretty much explained it for me. I wondered why the weight wasn't calculated with a negative, but it was then added to find the normal. Thanks