Respuesta :
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
