# Homework Help: Forces and Vectors Problem?

1. Sep 12, 2004

### pinky2468

First here is the problem: Only two forces act on an object(mass 3.00kg) as in the drawing(40.0N in the x direction and 60.0N north-east and a 45 degree angle between them) Find the magnitude and direction, relative to the x-axis of the acceleration of the object.

I have tried finding the x and y components of each vector and then adding the x components and y components together to get the resultant and then using a=F/M and the tangent to determine the angle(direction) BUT I keep getting the wrong answer!! Please explain!!

Last edited: Sep 12, 2004
2. Sep 12, 2004

### marlon

Did you get this for the components : first force = 40e_x
second force = 60(cos(45)e_x + sin(45)e_y) and cos(45) = sin (45)= 1/sqrt(2)???

e_x and e_y just denote the basis-vectors...

If you did get this, you are right...

Remeber that is this exercise you only need to add the x-components in order to find the answer. We are only looking in the x-direction here..

3. Sep 12, 2004

### marlon

Shouldn't the answer be 27,47546896 m/s² ???

4. Sep 12, 2004

### pinky2468

That is how I did the problem, but the answer in the book is a=30.9(m/s2) and the degree is 27.2 degrees above the x-axis??!!

5. Sep 12, 2004

### marlon

Ahh, got it

They are asking the resulting force when you add the two given forces together

use the cosinus-rule...

6. Sep 12, 2004

### marlon

F² = 40² + 60² -2*40*60*cos(135) Then take the squareroot and devide by three and you will get 30,9 m/s²

7. Sep 12, 2004

### marlon

Or with the components

(total force)² = (60/sqrt(2) + 40)² + (60/sqrt(2))² = 92,7

The angle is calculated using tan(x)=Y/X and Y component = 60/sqrt(2)
and X component = (40 + 60/sqrt(2))

regards
marlon

8. Sep 12, 2004

### Pyrrhus

Do what marlon says or

$$R_{x} = 60cos(45) + 40cos(0)$$

$$R_{y} = 60sin(45) + 40sin(0)$$

$$\vec{F_{R}} = (82.42i + 42.42j) N$$

$$\vec{A} = (\frac{82.42}{3}i + \frac{42.42}{3}j) m/s^2$$

$$\vec{A} = (27.473i + \14.14j) m/s^2$$

$$|\vec{A}| = \sqrt{27.473^2+14.14^2}$$

$$|\vec{A}| = 30.89$$

$$\theta_{A} = arctan(\frac{14.14}{27.473})$$

$$\theta_{A} = 27.23^o$$

9. Sep 12, 2004

### pinky2468

Thanks! I understand I was using cos(45) not cos(0)!