# Magnitude of forces

1. Jan 4, 2006

### hoseA

A 2.7 kg mass accelerates at 7 m/s^2 in a
direction 28 degrees north of east. One of the two
forces acting on the mass has a magnitude of
12.6 N and is directed north.
Determine the magnitude of the second
force. Answer in units of N.

I did F2= sqr rt (ma^2+ F1^2)

= 22.715 N

This is apparently incorrect. This seems too easy and yet I cannot figure it out. I'm assuming F2 is in the positive "x" direction w/o a "y" component.

Any help appreciated.

2. Jan 4, 2006

### cscott

I think your formula should be $F_2 = \sqrt{(ma)^2 - F_1^2}$ (remember $$c^2 = a^2 + b^2$$) to get 14.09 N.

3. Jan 9, 2006

### hoseA

This is not the answer... tried it and got it wrong :(

I also tried splitting up the the acceleration into x and y components and trying to solve it F1x+F2x = MAx; F1y+F2y=MAy F1x=0 F1y=12.6N

I solved it and then did sqr rt(F2x^2+F2y^2)

= 16.2662N

I only have 1 try left so it would be great if someone can help me out, please.

4. Jan 9, 2006

### Mindscrape

How come the force acting north has a magnitude of 12.6N? It should be $$masin(\theta)=F_N$$ which is 8.87N. Was 12.6 given?

5. Jan 9, 2006

### Little_Rascal

The force you're trying to find must also have an y-component, otherwise cscotts formula would've worked.

6. Jan 9, 2006

### hoseA

The question states (as found in the original post):

"One of the two forces acting on the mass has a magnitude of 12.6 N and is directed north."

So assume the sole direction of F1 is North, or in the positive y direction.

Any ideas as to an equation or problem-solving technique to solve this?

7. Jan 9, 2006

### Mindscrape

To make sure I have this right:

-The net force (magnitude) of the two components is the $$7 \frac{m}{s^2} * 2.7kg = 18.9N$$

-This magnitude has an angle of 28º north of east

-The north acting component of the net force is 12.6N

Unless physics isn't allowed to use vectors anymore, this is wrong.

8. Jan 9, 2006

### hoseA

hmm... i don't know. I cut and pasted the question -- maybe the question is wrong?

9. Jan 9, 2006

### mukundpa

ma cos28 = (F2) cos@
ma sin28 = (F1) + (F2) sin@

@ is the angle F2 makes north of east

eliminate @ to get F2

M.P.

10. Jan 9, 2006

### hoseA

ma cos28 / cos@ = ma sin28 - F1 / sin@

tan@ = ma sin28 - F1/ ma cos28

@ = tan -1 (ma sin28 - F1/ ma cos28)

F2x = 17.09883283
F2y = -17.09883283

F2 = sqr rt ( 17.09883283^2 * 2)

= 24.18140129 N?

This comes up as wrong. I dont know -- maybe I made a math mistake.

Thanks for the help though.

11. Jan 9, 2006

### mukundpa

solve in this way

F2 cos@ = macos28 = 16.688N
F2 sin@ = masin28 - F1 = -3.727 N

Who told you that the two forces are perpendicular or F2 is towards east?

M.P.

12. Jan 9, 2006

### hoseA

I think that's it. Thanks for the help!