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 = 1.64.
To find the ratio F/FA when k = 1.64, we need to understand how forces add up when applied at different angles.
In this scenario, force FA is the original force applied to pull the baby elephant out of the mud hole. However, force FA alone is not enough to move the elephant. In order to provide enough force, two additional forces FB and FC are applied at different angles.
To calculate the resultant force acting on the elephant in part b of the drawing, we need to use the concept of vector addition. The magnitude of the resultant force is given by the Pythagorean theorem:
Resultant force (FR) = √(FB^2 + FC^2)
The given information states that the magnitude of the resultant force in part b is k times larger than the magnitude of the resultant force in part a. Mathematically, this can be expressed as:
FR(b) = k * FR(a)
Substituting the expressions for the resultant forces:
√(FB^2 + FC^2)(b) = k * √(FB^2 + FC^2)(a)
Now, let's express FB and FC in terms of F (the magnitude of the additional forces) and the angle between them:
FB = F * cosθ
FC = F * sinθ
Substituting these values into the equation:
√((F * cosθ)^2 + (F * sinθ)^2)(b) = k * √((F * cosθ)^2 + (F * sinθ)^2)(a)
Since sin²θ + cos²θ = 1, we can simplify the equation:
√(F^2 * (cos²θ + sin²θ))(b) = k * √(F^2 * (cos²θ + sin²θ))(a)
Simplifying further:
√(F^2)(b) = k * √(F^2)(a)
Simplifying the square root of F^2 to F:
F(b) = k * F(a)
Finally, since we're looking for the ratio F/FA, we can divide both sides of the equation by FA:
F(b)/FA = k * F(a)/FA
F/FA = k
Substituting the given value of k = 1.64, we find:
F/FA = 1.64
Therefore, when k = 1.64, the ratio F/FA is 1.64.
We are given that the magnitude of the resultant force in part b is k times larger than that in part a. Let's denote the magnitude of force FA as FA, and the magnitude of additional forces FB and FC as F.
In part a, only force FA is applied. Therefore, the magnitude of the resultant force in part a is simply FA.
In part b, we have two additional forces FB and FC applied along with force FA. The magnitude of the resultant force in part b is given as k times larger than that in part a. Mathematically, we can represent this as:
FA + FB + * FA
Substituting k = 1.64, we have:
FA + F + F = (1.64) * FA
2F = (1.64) * FA - FA
2F = 0.64 * FA
Divide both sides of the equation by FA:
2F/FA = 0.64
Therefore, the ratio F/FA when k = 1.64 is:
F/FA = 0.64/2 = 0.32