# Force and Motion Question

1. Sep 24, 2009

I keep getting the wrong answers for these two questions no matter what way i seem to structure it.
1. Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 9.0 N, acting due east, and the other is 8.0 N, acting 62` north of west.What is the magnitude of the body's acceleration?

If follow the guideline example in my book this is how i try to answer it.
$$F = m a \Rightarrow F_1 - F_2 = m a$$

$$9 - 8 \cos 62 = (3) a \Rightarrow \frac {9 - 3.75}{3} = a$$

$$\frac {5.2}{3} = a \Rightarrow 1.733... = a$$

But my book says the answer is $$2.9$$, what am I missing?

2. Sep 24, 2009

### ehild

The resultant force has a component to north, too, and you need to calculate the magnitude of the total force. Moreover, I do not know what does 62´ north of west mean.

ehild

3. Sep 24, 2009

From the question I assumed you only had to take the horizontal motion;
And I tried (to the best of my ability) using separate $$x_i$$ and $$y_j$$ coordinates and failed. I can't get that elusive $$2.9$$
.
This would mean what? To take $$|a| \sqrt(x^2 i+ y^2 j )$$ accelerations?

62 Degrees North of the west axis. I assume it's like 118 degrees counterclockwise from $$1,0$$ on a unit circle,

4. Sep 24, 2009

### ehild

You either go to west or north, you move horizontally still you stay on the floor and do not rise up, or sink down...

You have got the x component of the acceleration, using the x component of the north-west force. What is the y (north) component of this force?

And yes, you have to use the magnitude (absolute value) of the acceleration at the end, sqrt(ax2+ay2) .

ehild