# What did I do wrong? (Newton's Second Law)

1. Nov 7, 2005

### th77

Only two horizontal forces act on a 3.0 kg body. One force is 9.0N, acting due east, and the othr is 8N, acting 62 degrees north of west. What is the magnitude of the body's acceleration?
Fnet= ma ............so I took that to mean F1x + F2x = ma
F1x = 9 cos 0 = 9
f2x = 8 cos 62 = 3.76
a = (9 + 3.76)/3 = 4.25
The solution says 2.9

2. Nov 7, 2005

### arildno

East and west are opposite directions. What does that tell you?

3. Nov 7, 2005

### th77

I think it tells me that it should be F1 - F2 instead. If yes then

a = (9 - 3.76)/3 = 1.75

It still doesn't come to 2.9

4. Nov 7, 2005

### arildno

Oops, sorry to mislead you a bit.
The MAGNITUDE of the acceleration evidently includes the contribution from the VERTICAL acceleration the body also expriences.
That is, you are to find $a=\sqrt{a_{x}^{2}+a_{y}^{2}}$

5. Nov 7, 2005

### th77

You're right because I got the answer using it, but I don't know why. The problem says that only 2 'horizontal' forces act on the object. Why is the vertical involved? I know F2 is making an angle, but then what do they mean by horizontal forces? Shouldn't it just say 'forces'? I have a sample problem in the book that specifies there is one-dimensional motion yet 1 of the forces is making an angle.

6. Nov 7, 2005

### arildno

The phrasing in the book is just dumb, that's all there is to it.

7. Nov 7, 2005