# 3 Force Vectors

1. Feb 11, 2008

### zcabral

1. The problem statement, all variables and given/known data
Three forces acting on an object are given by F1 = ( 1.55 i - 1.90 j ) N, F2 = ( - 5.20 i - 3.15 j ) N, and F3 = ( - 50.0 i + 44.0 j ) N.
(a) What is the direction of the acceleration?
° (from the positive x axis)
(b) What is the mass of the object?
kg
(c) If the object is initially at rest, what is its speed after 16.0 s?
m/s
(d) What are the velocity components of the object after 16.0 s?
( i + j ) m/s

2. Relevant equations

F=ma

3. The attempt at a solution

the acceleration is 3.70 m/s2. i know to get the direction in part (a) i have to use arctan ay/ax but i cant figure out how to get those values. the velocity is 59.2m/s but i dont know how to get the unit vectors.

Last edited: Feb 11, 2008
2. Feb 11, 2008

### chrsr34

You need to add the 3 vectors vectorally. So add up the i components for your resultant x-value, and add up your j's for the resultant y-value. Then to get the direction, use the arctan. This should get you going.

To get unit vectors you need to divide each coordinate by the magnitude of the resultant vector.

Chris

Last edited: Feb 11, 2008
3. Feb 11, 2008

### blochwave

I'm assuming the acceleration is given?(you stuck it in the solution part so I'm not sure if it's something you figured out but I think it's given, considering the problem)

So what's the resultant force? You just add F1+F2+F3=Fr, those are all vectors so add the x and y components separately. Then that's the y and x component you use in your arctan equation.

You know the magnitude of the acceleration, so if you find the magnitude of that resultant force you can use F=ma to find m

Knowing a I assume you solved c correctly, so...

Edit: That guy's way for d is obviously the better way than what I typed >_>

Last edited: Feb 11, 2008
4. Feb 11, 2008

### zcabral

ok i figured out the direction is 144 degress approximately. however i still dont get how to find the components of velocity (not acceleration)

5. Feb 11, 2008

### blochwave

Actually maybe what I deleted up there was what we needed. Aw what the heck, I'll throw it out here

You know velocity's magnitude, you want the x and y components Vx and Vy

We know sqrt(Vx^2+Vy^2)=|V|(and you know |V|)

We also know the angle(because the velocity will be in the same direction as the force)

so tan(that angle)=Vy/Vx, tan(that angle)=some# so Vy=some#*Vx, and you can plug that back into the other equation and solve for Vy, then find Vx

6. Feb 11, 2008

### zcabral

ok but i still dont get how to find vy and vx, obviously system of equations but wont that make it tan 144=0?

7. Feb 11, 2008

### blochwave

Why would that happen? tan(144)=-.727, so -Vx*.727=Vy

so sqrt[(-Vx*.727)^2+Vx^2]=|V|, solve for Vx, then put that back into -Vx*.727=Vy to solve for Vy