3/26/2023

Statics - Problem 2.24


 Statics, Chapter 2 - Force Vectors


In (Figure 1), the resultant FR of the two forces acting on the jet aircraft should be directed along the positive x axis and have a magnitude of 11 kN,

A top view of a plane labeled 'A', which is pulled with ropes by two trucks labeled B and C. The horizontal x axis points to the right. The plane is moving along the x axis. Trucks C and B are above and below the x axis, respectively. The towing force of truck C is labeled as F subscript C and makes an angle of 20 degrees with the x axis. The towing force of truck B is labeled as F subscript B and makes an angle theta with the x axis.

▼ PART A
Determine the angle θ of the cable attached to the truck at B so that FB is a minimum. Express your answer in degrees to three significant figures.


▼ PART B
What is the magnitude of force in cable  B when this occurs? Express your answer to three significant figures and include the appropriate units.


▼ PART C
What is the magnitude of force in cable  C when this occurs? Express your answer to three significant figures and include the appropriate units.



FR = resultant, which is directed along the x-axis and = 11 kN.

θ = ? to have to be minimum...

since Fy = 0, Fsin(20) = Fsin(θ)

Fx = FR = Fcos(20) + Fcos(θ) = 11 kN

Let's solve for Fusing the equation for Fy



Now simply the 1/sin(20) and plug Fc  in the equation for FR



Now, take the derivative of 2.92sin(θ) + cos(θ) in order to get the maximum value that compliments the equation




Set the right hand side = 0, and solve the equation.





Divide every term by cos(θ).
Using trigonometric identities, evaluate -sin(θ)/cos(θ) = -tan(θ)


Take the inverse tan of both sides; arctan(2.92) = 71.1

Part A answ = 71.1°

Now plug in 71.1° in this equation from above...


... to get our value for For Fb(min)
when doing so, our Part B answer is, F= 11/3.09 or 3.56 kN

Now for Part C, we can plug in what we now know to find Fc re-using the equation below...


F= 9.84 kN



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