# Force problem

1. Mar 10, 2006

### rculley1970

I have a frictionless force problem. A car is traveling north and there is a Y intersection. 480N to the left at 11 degrees, 400N to the right at 26 degrees. Can I find the magnitude and direction of these 2 forces without the mass of the car?

I am coming up with 155.6N but am still working on the direction.

2. Mar 10, 2006

### lightgrav

We can't tell whether the 11 degrees is measured clockwise from the left direction, (ie, 11 degrees toward negative y from the negative-x axis?)
or whether it is 11 degrees clockwise from the left (11 deg toward +y from-x)

Find each component first, of each Force. Then add each component separately.
Finally use Pythagoras to get the magnitude of the total Force, if needed.
use the inverse tan(Fy/Fx) to get the angle.

3. Mar 10, 2006

### rculley1970

Sorry, I meant it going north as in the y direction. 11 degress going negative from the y-axis and 26 degress going positive from the y-axis. I forgot about the inverse tangent function to find the angle. thank you for that. This is what I have:
F(x)=-480(cos11) + 400(cos26)
F(y)=400(sin26) + 480(sin11)

F(x) = -111
F(y) = 267

a = .099 m/s^2

Just let me know if this is wrong. I will rerun the numbers again to verify.

4. Mar 10, 2006

### lightgrav

From your description, I expect each Force to be mostly in the y-direction.
But using -480N cos(11) makes almost all the Force in the (-) x-direction ...
Should you switch the sines with the cosines?
(yes, if the 11 degree angle is measured from the y-axis)

You can't predict an acceleration without knowing the mass, though.

5. Mar 10, 2006

### rculley1970

Not sure if i can put a graphic on here but:

480N \11 deg |26 deg / 400N
\ | /
\ | /
\ | /
\ | /
\ |/
------------------------------------------------------
/|\
|
|

6. Mar 10, 2006

### rculley1970

figured out the magnitude. 914.8. but still working on the angle.

7. Mar 10, 2006

### lightgrav

That's what I thought - these small angles are being measured from the y-axis.
So the small components are the x-components, with large y-components.
Verify by adding the component VECTORS (tail-to-tip) to get the diagonal.
sin(11) = - F1x / F1 ... sin(26) = + F2x / F2y

8. Mar 10, 2006

### lightgrav

umm, did you forget that the x-components tend to cancel?

9. Mar 10, 2006

### rculley1970

thank you for the help, lightgrav. I did mix up the cos and sin. had them backwards. I am using the ThompsonNOW website where the teacher posts the homework. The mag is 914.8 and the a=.288m/s^2. Just trying to figure out the angle though. Give me a few minutes on this one.

10. Mar 10, 2006

### rculley1970

Sorry, mass was given in part (b) of question: 2900Kg. Part (a) just asked for mag and dir. (b) asks for acceleration

11. Mar 10, 2006

### lightgrav

Even if the Forces were perfecly aligned, they would only add up to 880N !
The result of a vector addition is never greater than that ... NOT 914.8 N
(I got 834N).

12. Mar 10, 2006

### rculley1970

Odd, I can input the answers and it tells me if they are correct or incorrect. On the mag and acceleration, i got 914.8 and .288 respectively. It gave me credit for both of them. I keep getting the angle wrong though. will keep working on it.

13. Mar 10, 2006

### rculley1970

F(x)=-480(sin11) + 400(sin26) = 83.8
F(y)=480(cos11) + 400(cos26) = 831

a(x)=(83.8)/(2900) = .029
a(y)=(831)/(2900) = .287

a = .288

14. Mar 10, 2006

### rculley1970

I keep coming up with 84.2 degrees but the website keeps telling me it's incorrect.

15. Mar 11, 2006

### lightgrav

The magnitude of the Force vector is obtained by Pythagoras, not straight addition.

If their angles are measured from +y-axis, yours probably should be, too.

16. Mar 11, 2006

### rculley1970

sqr root( (.029)^2 + (.287)^2) = .288

17. Mar 11, 2006

### rculley1970

Crashing out for the night. Working on it tomorrow along with some friction problems. (ughhh) Too many equations. Mostly memorization of equations and when to use them from what I can tell. Will look into the current problem tomorrow. Thank you for your help.