Posted by **Hannah** on Sunday, May 27, 2012 at 4:45pm.

1) A baby elephant is stuck in a mud hole. To help pull it out, game keepers use a rope to apply a force FA. By itself, however, force FA is insufficient. Therefore, two additional forces FB and FC are applied. Each of these additional forces has the same magnitude F. The magnitude of the resultant force acting on the elephant in part b is k times larger than that in part a. Find the ratio F/FA when k = 2.32.

The angles bewteen anad b and b and c are all 22 degrees.

I do not kno

- Physics(thank you for the help) -
**Hannah**, Sunday, May 27, 2012 at 4:45pm
I do not know how to solve for a ratio.

- Physics(thank you for the help) -
**bobpursley**, Sunday, May 27, 2012 at 4:54pm
Physics(thank you for the help) - bobpursley, Sunday, May 27, 2012 at 4:53pm

I am uncertain what part b is.

I am uncertain what F is in the ratio F/Fa

If F is the total force, then F/Fa=k by definition. So frankly, I have no idea what the question is asking.

- Physics(thank you for the help) -
**Elena**, Sunday, May 27, 2012 at 5:05pm
As the forces FB and FC are equal the horizontal component of force is given as FB = FC = F•cos θ,

θ =22º

The net force applied on the elephant is

Fnet = FA + 2 F•cos θ ........ (1) ( there are two forces on either side of FA)

Given that Fnet is k times FA

Fnet = k• FA where k = 2.32•FA

Fnet = 2.32• Fa .............. (2)

Plug in (2) in (1)

2.32 •FA = FA + 2• F• cos θ

2.32• FA - FA = 2• F•cos θ

(2.32 - 1) • FA = 2 F cos θ

1.32• FA= 2• F• cos θ

Ratio

F / FA = 1.32/( 2• cos θ)=

=1.32/2•cos22º = 0.71

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