# Newton's Laws magnitude of the force

1. Oct 14, 2009

### Mysteek

First post =]
Requesting guidance with this particular problem:

1. The problem statement, all variables and given/known data
A 1.20 kg object is moving in the x direction at 14.4 m/s. Just 3.31 s later, it is moving at 28.8 m/s at 34.0 degrees to the x axis.

a.) What is the magnitude of the force applied during this time?

b.) What is the direction of the force applied during this time?

2. Relevant equations
Fnet=ma?

3. The attempt at a solution
I first organized all my givens, that is:
mObject = 1.20kg
vObject = 14.4m/s
t = 3.31s
v2Object = 28.8m/s 34 degrees to x axis

I established a coordinate system so that positive would be ^ and -> , and negative would be the opposites. I then drew the corresponding vectors (14.4m/s and 28.8m/s at the angle).

Then I broke down the angled vector into its x and y components:

v2x = 28.8cos34 = 23.88m/s
v2y = 28.8sin34 = 16.1m/s

Here's where I get uncomfortable with my work:

I know Fnet = ma

So F = 1.2(v/t)
= 1.2(v2-v1 / t)
= 1.2kg((23.88m/s - 14.4m/s) / 3.31s)
= 3.44

I'm aware I completely neglected the y component, and I've been having trouble finding out where to fit it really. I'd love a push in the right direction if someone wouldn't mind - thanks!

2. Oct 15, 2009

### rl.bhat

Hi mysteek, welcome to PF.
Force is defined rate of change of momentum.
Initial momentum Px is along the x direction. Resolve the final momentum into Px' and Py'.
Find the magnitude of change in momentum and hence the force.

3. Oct 15, 2009

### Mysteek

Ah, momentum, brings back memories.

So then:

Px(initial) = mass of object * initial velocity
= 1.2kg(14.4m/s)
= 17.28

Px(final) = mass of object * x component final velocity
= 1.2kg(23.88m/s)
= 28.656

Py(final) = mass of object * y component final velocity
= 1.2kg(16.1m/s)
= 19.32

Now that I have the components, I can compute the change in momentum, thus providing me with force:

Fnet = Change in P

Change in Px = P(final) - P(initial)
= 28.656 - 17.28
= 11.376 N

Change in Py = 19.32 N

Magnitude of change in momentum would be:

Sqrt(11.376^2 + 19.32^2)
Sqrt(502.676)
= 22.4 N

Am I missing something? I didn't use the 3.31 seconds of time. Was ''time'' in the question just for the sake of throwing me off?

4. Oct 15, 2009

### rl.bhat

To get F divide change in momentum by time.

5. Oct 15, 2009

### Mysteek

22.4 N would be the magnitude of the change of momentum, if I were to divide this by time I would get

22.4 / 3.31s

= 6.77 N

Oh, right! That makes perfect sense all the sudden, a bit more reading had me discover that momentum is in Newton seconds, so if I divided by time, seconds would cancel and I'd be left with newtons, aka force.

At least I think my logic is correct. I'd like to think I understand the concept now.

6. Oct 15, 2009

### Mysteek

Sorry for double posting - for some reason my browser isn't letting me click edit on my last post :(

When finding the direction of the force applied, could I use sine law?

Ex:

6.77N is the magnitude of the change of momentum, thus it would represent the hypotenuse of a right angle triangle, and because this vector is the hypotenuse, I can use:

(19.32 is the change in momentum in the y direction, aka opposite to the angle we need)

Sinx = opposite/hypotenuse
= 19.32 / 6.77

I get a math error when trying to find x. I assume I'm not using the correct numbers?

Crap, attempt number 2:

I think that the magnitude of the change in momentum has nothing to do with this particular section of the problem.

I found the hypotenuse earlier, did I not?

Sqrt(11.376^2 + 19.32^2)
Sqrt(502.676)
= 22.4 N s

This value is needed to keep units consistent, since the x and y components are also in N s.

Thus, I think I would use my same approach with this value?

sinx = o/h
= 19.32 / 22.4
x = 59.6 degrees?

Last edited: Oct 15, 2009
7. Oct 15, 2009

tanθ = Py/Px