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a 10,000 pound boulder sits on a mountain side at an incline of 60° fro…

Question

a 10,000 pound boulder sits on a mountain side at an incline of 60° from the horizontal. ignoring the force of friction, what force is required to keep the boulder from rolling down the hill? (round your answer to the nearest whole pound)
select one:
○ a. 17,321 lb
○ b. 5774 lb
○ c. 8660 lb
○ d. 5000 lb

eliminate the parameter from the parametric equations below, and convert the equation into rectangular form.
$f(t) = (x(t), y(t))$
given
$x(t) = t, y(t) = \frac{-t^2}{3}$
select one:
○ a. $y = \frac{3}{x^2}$
○ b. $y = \frac{-x^2}{3}$
○ c. $y = \frac{x}{3}$
○ d. $y = 3x^2$

Explanation:

Response
First Question (Force on Incline)

Step1: Identify the force component

The force required to keep the boulder from rolling down is the component of the weight along the incline. The weight \( W = 10000 \) lb, and the incline angle \( \theta = 60^\circ \). The formula for the component of weight along the incline is \( F = W\sin\theta \).

Step2: Calculate the sine of the angle

\( \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660 \)

Step3: Compute the force

\( F = 10000 \times \sin(60^\circ) = 10000 \times 0.8660 = 8660 \) lb

Second Question (Eliminating Parameter)

Step1: Substitute \( t \) from \( x(t) \) into \( y(t) \)

Given \( x(t) = t \), so \( t = x \). The equation for \( y(t) \) is \( y(t) = -\frac{t^2}{3} \).

Step2: Replace \( t \) with \( x \)

Substitute \( t = x \) into the equation for \( y(t) \): \( y = -\frac{x^2}{3} \)

Answer:

c. 8660 lb