A baby elephant is stuck in a mud hole. To help pull it out, game keepers use a rope to apply a force F with arrowA, as part a of the drawing shows. By itself, however, force F with arrowA is insufficient. Therefore, two additional forces F with arrowB and F with arrowC 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

Question

A baby elephant is stuck in a mud hole. To help pull it out, game keepers use a rope to apply a force F with arrowA, as part a of the drawing shows. By itself, however, force F with arrowA is insufficient. Therefore, two additional forces F with arrowB and F with arrowC 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 = 1.80. (Take 𝜃 = 15.0°.)

Answer 1

To find the ratio F/FA when k = 1.80, we can use the concept of vector addition and trigonometry.

In this scenario, we have three forces: F with arrowA (FA), F with arrowB (FB), and F with arrowC (FC). We need to determine the relationship between F and FA.

Using vector addition, we can represent the total force acting on the baby elephant as the sum of these three forces:

Resultant Force = FA + FB + FC

From the problem statement, we are given that the magnitude of the resultant force in part b of the drawing is k times larger than that in part a. Mathematically, this can be expressed as:

|Resultant Force in part b| = k * |Resultant Force in part a|

Now, let's break down each force into its x and y components. Since we are given 𝜃 = 15.0°, we can determine the components using trigonometry:

FAx = FA * cos(𝜃)
FAy = FA * sin(𝜃)

Similarly, for FB and FC, we can write:

FBx = FB * cos(𝜃)
FBy = FB * sin(𝜃)

FCx = FC * cos(𝜃)
FCy = FC * sin(𝜃)

The resultant force can be calculated by summing the components:

Resultant Force in x-direction = FAx + FBx + FCx
Resultant Force in y-direction = FAy + FBy + FCy

Now, let's calculate the magnitudes of these forces. In part a of the drawing, the resultant force only consists of FA:

|Resultant Force in part a| = sqrt(FAx^2 + FAy^2)

In part b of the drawing, the magnitude of the resultant force is k times larger. Therefore, we can write:

|Resultant Force in part b| = k * sqrt(FAx^2 + FAy^2)

Equating the expressions for the resultant forces in part a and part b, we get:

k * sqrt(FAx^2 + FAy^2) = sqrt(FAx^2 + FAy^2) + sqrt((FBx + FCx)^2 + (FBy + FCy)^2)

Squaring both sides of the equation and simplifying, we obtain:

k^2 * (FAx^2 + FAy^2) = (FAx^2 + FAy^2) + (FBx + FCx)^2 + (FBy + FCy)^2

Next, substitute the x and y components using trigonometry:

k^2 * (FA^2 * cos^2(𝜃) + FA^2 * sin^2(𝜃)) = (FA * cos(𝜃))^2 + (FA * sin(𝜃))^2 + (FB * cos(𝜃) + FC * cos(𝜃))^2 + (FB * sin(𝜃) + FC * sin(𝜃))^2

Simplify further:

k^2 * FA^2 = FA^2 + (FB + FC)^2 + 2 * (FB + FC) * FA * cos(𝜃)

After canceling out FA^2 terms and rearranging, we have:

(k^2 - 1) * FA^2 = (FB + FC)^2 + 2 * (FB + FC) * FA * cos(𝜃)

Finally, solving for the ratio F/FA:

F/FA = sqrt((FB + FC)^2 + 2 * (FB + FC) * FA * cos(𝜃)) / (k^2 - 1)

By substituting the given values for k = 1.80 and 𝜃 = 15.0°, you can calculate the numeric value of the ratio F/FA.