Den Hartog’s Mechanics

A web-based solutions manual for statics and dynamics

Problem 10

The drawing in the book shows us everything we need to write the equations for the resultants. The x-component (right positive) is

R_x = \frac{F_2}{\sqrt{2}} + F_3 + \frac{F_4}{\sqrt{2}} - \frac{F_6}{\sqrt{2}} - F_7 - \frac{F_8}{\sqrt{2}}

or

R_x = \frac{1}{\sqrt{2}}(F_2 + F_4 - F_6 - F_8) + (F_3 - F_7)

and the y-component (up positive) is

R_y = F_1 + \frac{F_2}{\sqrt{2}} - \frac{F_4}{\sqrt{2}} -F_5 - \frac{F_6}{\sqrt{2}} + \frac{F_8}{\sqrt{2}}

or

R_y = \frac{1}{\sqrt{2}}(F_2 - F_4 - F_6 + F_8) + (F_1 - F_5)

With these equations in hand, we can substitute in the cylinder forces. For part a, the resultants are: R_x = 12.1, R_y = 5, and therefore R = \sqrt{12.1^2 + 5^2} = 13.1 with the resultant directed at an angle of \tan^{-1}(5/12.1) = 22.5^\circ from horizontal up and to the right.

For part b, the resultants are: R_x = -3, R_y = 19.7, and therefore R = \sqrt{3^2 + 19.7^2} = 19.9 with the resultant directed at an angle of \tan^{-1}(3/19.7) = 8.7^\circ from vertical down and to the left.

Radial engines were fairly common from the early days of flight through WWII and would have been fresh in the minds of students in 1948, when the first edition of Mechanics was published. Here’s a picture of one

taken from a site devoted the history of the Tuskegee Airmen.

Problem 11Problem 9

Last modified: January 22, 2009 at 8:32 PM.