Question

The force F (in newtons) of the hydraulic cylinder in a press is proportional to the square of sec x where x is the distance (in meters) that the cylinder is extended in its cycle. The domain of F is [0, pi/3], and F(0) = 500.A) find F as a function of x.F(x)=___________B) find the average force exerted by the press over the interval [0, pi/3] (round your answer to 1 decimal place)F= _________? N

Answer 1

Answer:

Part A) F(x) =500 sec(x)

Part B) Average force exerted by the press = 750 N

Step-by-step explanation:

Given:

F(x) ∝ sec(x)

Range of x = [0,Pi/3] or x = [0 , 60°]

F(0) = 500 N i.e. F is 500 N at x = 0.

As we now that:

Force is proportional to sec(x) therefore we may write an equation by introducing Proportionality Constant (k) as under:

F(x) = k * sec(x) - Say it Equation 1

Then using given information of F=500 N at x = 0 we have:

500 = k * sec(0)

As sec(0) = 1, therefore we get:

k = 500

By putting value of k in equation 1 we have:

F(x) = 500 * sec(x)

Now by putting max value of x from the given range that is Pi/3 in the above equation we get:

F(Pi/3) = 500 * sec(Pi/3)

As Pi = 180 there we simplify the above equation as:

F(60°) = 500 * sec(60°)

F(60°) = 500 * 2 ; By putting sec(60°) = 2

F(60°) = 1000 N

Now the avg. of force exerted by the hydraulic press is given by:

Avg. Force = Minimum Force + [(Maximum Force - Minimum Force) / 2] - Say it equation 2

Considering sec(x) is minimum at x = 0 and maximum at x = 60° within the given range [0 , 60°] therefore we have:

The minimum force being put at x = 0 i.e. F = 500 N and max. force at x = 60° which is F = 1000 N.

Finally we get the average force using equation 2 as under:

Average force = 500 +  [(1000 - 500) / 2]

Average Force = 500 + 250

Average Force = 750 N