A baby elephant is stuck in a mud hole. To help pull it out, game keepers use a rope to apply a force FA, as part a of the drawing shows. By itself, however, force FA is insufficient. Therefore, two additional forces FB and FC are applied, as in part b of the drawing. Each of these additional forces has the same magnitude F. The magnitude of the resultant force acting on the elephant in part b of the drawing is k times larger than that in part a. Find the ratio F/FA when k = 2 cos = 25
Well, well, well. A mud-dwelling baby elephant in need of some serious pulling power! Let's see if we can solve this physics puzzle with a touch of humor.
First things first, my pachyderm predicament ponderer. We need to find the value of F/FA when k = 2cos = 25. Now, I could give you a serious, straightforward answer, but where's the fun in that?
Let's imagine that the baby elephant is a great admirer of circus acts. It dreams of joining the circus and becoming the next weightlifting sensation. The game keepers, fueled by their love for circus-worthy performances, decide to turn this rescue mission into an impromptu elephant-weightlifting experiment.
Now, picture this: the force FA pulls on the baby elephant, but it's struggling, like an elephant trying to squeeze into a mouse hole. So, the game keepers decide to call in the big guns. Forces FB and FC are applied, both with the same magnitude F. Together, they combine to form the resultant force.
And voila! The resultant force in part b is k times larger than in part a. But just how big is this k? Well, according to your question, k = 25. Quite the elephantine boost!
Now, the ratio F/FA can be found by dividing the magnitude of the resultant force (k * FA) by FA. In this case, k = 25. So, F/FA = (k * FA) / FA = k = 25.
There you have it, my mud-caked friend! When k = 25, the ratio F/FA is the same as k, which is also 25. Quite a hefty increase, isn't it? I hope this answer lightened your mood as much as the baby elephant's mood would be once it's out of that sticky situation. Happy rescuing!
To find the ratio F/FA when k = 2 and cosθ = 25, we can use the equation for the magnitude of the resultant force:
Resultant force (FR) = √(FA^2 + FB^2 + FC^2 + 2FAFBcosθ + 2FAFCcosθ + 2FBFCcosθ)
Given that FR is k times larger than FA, we can write:
FR = k * FA
Substituting the values into the equation:
k * FA = √(FA^2 + FB^2 + FC^2 + 2FAFBcosθ + 2FAFCcosθ + 2FBFCcosθ)
Now, let's substitute k = 2 and cosθ = 25 into the equation:
2 * FA = √(FA^2 + FB^2 + FC^2 + 2FAFB(25) + 2FAFC(25) + 2FBFC(25))
Simplifying further:
4 * FA^2 = FA^2 + FB^2 + FC^2 + 50FAFB + 50FAFC + 50FBFC
3 * FA^2 = FB^2 + FC^2 + 50FAFB + 50FAFC + 50FBFC
To find the ratio F/FA, we can divide both sides of the equation by FA:
3 * FA = F^2/FA + (FB^2/FA) + (FC^2/FA) + 50FB + 50FC + 50FBFC/FA
Simplifying:
3 = (F/FA)^2 + (FB/FA)^2 + (FC/FA)^2 + 50(FB/FA) + 50(FC/FA) + 50FBFC/FA^2
Now, let's substitute cosθ = 25:
3 = (F/FA)^2 + (FB/FA)^2 + (FC/FA)^2 + 50(FB/FA) + 50(FC/FA) + 50FBFC/(FA^2 * cosθ)
Since cosθ = 25,
FA^2 * cosθ = FA^2 * 25
3 = (F/FA)^2 + (FB/FA)^2 + (FC/FA)^2 + 50(FB/FA) + 50(FC/FA) + 50FBFC/(FA^2 * 25)
To find the ratio F/FA, we need more information, specifically the magnitudes of FB and FC.
To solve this problem, we need to use vector addition to find the resultant force acting on the baby elephant.
Let's start by considering the forces applied individually in Part a of the drawing. The force applied by the game keepers is denoted as FA.
Next, we need to find the resultant force when forces FB and FC are added to FA. Since both FB and FC have the same magnitude F, we can represent them as vectors in opposite directions. Let's denote FB as F in the positive x direction and FC as F in the negative x direction.
Now, we have three forces acting on the baby elephant:
FA (force applied by the game keepers)
FB (force in the positive x direction)
FC (force in the negative x direction)
To find the resultant force, we will add these vectors together.
Since the forces FB and FC are in opposite directions along the x-axis, we can subtract their magnitudes:
|FR| = |FA| + |FB| - |FC|
Given that the magnitude of the resultant force FR is k times larger than that in part a of the drawing, we can write:
|FR| = k * |FA|
Substituting the magnitudes of the forces:
|FA| + |FB| - | * |FA|
Now, we can substitute the given value of k = 2:
|FA| + |FB| - |FC| = 2 * |FA|
Simplifying the equation:
|FB| - |FC| = |FA|
Dividing both sides of the equation by |FA|:
(|FB| - |FC|) / |FA| = 1
Now, let's substitute the given value of cos = 25.
Since cos = adjacent/hypotenuse, we can determine the value of the ratio |FB|/|FA|:
cos = |FB|/|FA|
25 = |FB|/|FA|
Simplifying the equation:
|FB| = 25 * |FA|
Now, we can substitute this value for |FB| in our previous equation:
(25 * |FA|) - |FC| = |FA|
(25 - |FC|/|FA|) = 1
Substituting the given value of k = 2:
25 - |FC|/|FA| = 2
Simplifying the equation:
25 - |FC|/|FA| = 2
- |FC|/|FA| = 2 - 25
- |FC|/|FA| = -23
Dividing both sides of the equation by -1:
|FC|/|FA| = 23
Therefore, the ratio F/FA when k = 2 and cos = 25 will be:
F/FA = |FC|/|FA| = 23
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