# Vector Problem.

1. Oct 8, 2006

### AznBoi

Two people pull on a stubborn mule, as seen from a helicopter. The picture is like this:
_V_
0

Ok, in the middle of V it is the y-axis. In quadrant I and II there are a vector in each. The 0 is the mule that they are pulling on.

In quadrant II, they angle is 75 degrees, the Force is 80 N.

In quadrant I, the angle is 60 degrees, the Force is 120 N.

Find a) the single force that is equivalent to the two forces shown, and b) the force that a third person would have to exert on the mule to make the net force equal to zero.

I got 184 N force at +78 degree north of west.
I used graphical methods(daisy chain, by connected the tail of vector(80 N) to vector (120 N).

How do I find the force that a third person would have to exert on the mule to make the net force equal zero?? I don't even get what a net force is =P

2. Oct 8, 2006

### PhanthomJay

The angle should be 78 degrees north of east (or 12 degrees east of north). The net force is the vector sum of all forces. To make it zero, what should be the magnitude and direction of the third force?

3. Oct 8, 2006

### AznBoi

yeah I meant 78 degrees N of E.. Would the third Force be the opposite of the equivalent vector?

4. Oct 8, 2006

### AznBoi

When it says "exert on the mule" does it mean that it is on the mule's side or on the 2 people's side?? I think that it is 184 N but the opposite direction of the 78 degrees N of E. Therefore it would be -78 degrees S of W??

5. Oct 8, 2006

### AznBoi

Is that right?? 184 N that is -78 degrees S of W?? Like /

6. Oct 8, 2006

### PhanthomJay

Yes, same magnitude, opposite direction. With the third force applied, the net (resultant) force is 0.

7. Oct 8, 2006

Ok thanks

8. Oct 9, 2006

### PhanthomJay

Yes, but get rid of that minus sign.