# Finding the magnitude of forces acting on a body

• Oct 4th 2010, 06:50 PM
Finding the magnitude of forces acting on a body
Two forces Fa and Fb are applied to an object whose mass is 8.0kg. Fa is the larger force. When both forces point due east, the objects acce. has a magnitude of .50m/s.s. However, when Fa points due east and Fb points due west, the acc is 0.40m/s.s. Find a) magnitude of Fa and b) the magnitude of Fb.

I tried it and came up with a) as 4N, but the answer is 3.6N. I did m x a=F
I don't have a clue how to get 0.40 N for the answer of b).

Can someone help me get started on this please.

I looked thru the little notes I have, which do not help and going thru my book, has nothing like this.
Thanks.
Joanne(Itwasntme)
• Oct 4th 2010, 07:05 PM
Jhevon
Quote:

Two forces Fa and Fb are applied to an object whose mass is 8.0kg. Fa is the larger force. When both forces point due east, the objects acce. has a magnitude of .50m/s.s. However, when Fa points due east and Fb points due west, the acc is 0.40m/s.s. Find a) magnitude of Fa and b) the magnitude of Fb.

I tried it and came up with a) as 4N, but the answer is 3.6N. I did m x a=F
I don't have a clue how to get 0.40 N for the answer of b).

Can someone help me get started on this please.

I looked thru the little notes I have, which do not help and going thru my book, has nothing like this.
Thanks.
Joanne(Itwasntme)

Yes, use $\displaystyle F = ma$

Now, we take east to be the positive direction and west to be the negative.

When both forces point east, the force is the sum of the forces, this is equal to mass times acceleration. That is,

$\displaystyle F_a + F_b = 8 (0.5)$

Similarly, when $\displaystyle F_b$ points in the negative direction, we get

$\displaystyle F_a - F_b = 8(0.4)$

Now you have a system of two equations with two unknows ( $\displaystyle F_a$ and $\displaystyle F_b$), solve this system to get your answer.
• Oct 4th 2010, 07:10 PM
OMG, I did not that it included equation. The teacher never said this would be used. oh man.
I got it now.
It's substitution for one of the letters. geeshhhhhhhhhhhh.

Thanks for the help GOT IT. but would have never thought of this at all tho.
Does this happen at times in Physics to use this????
• Oct 4th 2010, 07:17 PM
Jhevon
Quote: