The force F (in Newtons) required to move a box of mass m kg in motion by pulling on an attached rope is F(x) = f m g/(cos(x))+ f (sin(x))
where x is the angle between the rope and the horizontal, f is the coefficient of static friction and g = 9.8 m/s^2. Find the angle x that minimizes the required force F, assuming f = .4
To find the angle x that minimizes the required force F, we need to differentiate the force equation F(x) with respect to x, and set the derivative equal to zero.
First, let's compute the derivative of F(x) with respect to x:
F'(x) = -[f m g sin(x) / cos^2(x)] + [f cos(x)]
Now, set the derivative equal to zero to find critical points:
0 = -[f m g sin(x) / cos^2(x)] + [f cos(x)]
Now, let's simplify this equation:
[f m g sin(x) / cos^2(x)] = [f cos(x)]
Next, multiply both sides of the equation by cos^2(x) to eliminate the denominator:
f m g sin(x) = f cos^3(x)
Divide both sides by f to isolate the trigonometric expression:
m g sin(x) = cos^3(x)
Now, square both sides of the equation to eliminate the squared trigonometric expression:
(m g sin(x))^2 = (cos^3(x))^2
Simplify further:
m^2 g^2 sin^2(x) = cos^6(x)
Rewrite cos^6(x) as (cos^2(x))^3:
m^2 g^2 sin^2(x) = (cos^2(x))^3
Now, substitute sin^2(x) = 1 - cos^2(x):
m^2 g^2 (1 - cos^2(x)) = (cos^2(x))^3
Expand and rearrange the equation:
m^2 g^2 - m^2 g^2 cos^2(x) = (cos^2(x))^3
Let's define a temporary variable u = cos^2(x):
m^2 g^2 - m^2 g^2 u = u^3
Now, we have a cubic equation in terms of u. To solve for u, we can rearrange the equation:
u^3 + m^2 g^2 u - m^2 g^2 = 0
We can solve this cubic equation to find the value(s) of u. However, finding the exact value of u might be challenging. In this case, you can use numerical methods or graphing tools to approximate the solutions for u.
Once we have the value(s) of u, we can substitute them back into the equation u = cos^2(x) to find the corresponding values of x.