p170-w5.2-sols

pdf

School

University of British Columbia *

*We aren’t endorsed by this school

Course

170

Subject

Mechanical Engineering

Date

Oct 30, 2023

Type

pdf

Pages

Uploaded by MinisterKookabura3556

PHYS 170 Worksheet 5.2 Student ID Family Name First Name Email _________ ____________ __________ _________________________ _________ ____________ __________ _________________________ _________ ____________ __________ _________________________ _________ ____________ __________ _________________________ _________ ____________ __________ _________________________ _________ ____________ __________ _________________________ _________ ____________ __________ _________________________ _________ ____________ __________ _________________________ 1. Find the tensions in cables A and B, and the components of reaction at ball and socket joint D. A. Draw the free body diagram for the strut by adding force vectors B. Write the coordinates of A, B, C, and D in the xyz coordinate system as column vectors C. Write as a column vector D. Write the force vector at D as a sum of the unit column vectors times magnitudes . ! F B , ! F C , ! F DX , ! F DY , and ! F DZ ! A = 6 0 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ! B = 0 − 3 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ! C = 0 0 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ! D = 0 0 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ! F 1 ! F 1 = 0 500sin30 ° − 500cos30 ° ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = 0 250.0 − 433.0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ! F D ˆ i = 1 0 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ , ˆ j = 0 1 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ , ˆ k = 0 0 1 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ F DX , F DY , and F DZ ! F D = 1 0 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⋅ F DX + 0 1 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⋅ F DY + 0 0 1 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⋅ F DZ 500 N ! F 1 ! F B ! F C ! F DX ! F DY ! F DZ

E. Write the unit vectors for F. Write the force-balance equations using unit column vectors and G. Using D as the reference point, write the moment of as a column vector H. Using D as the reference point, write the moment of as a column vector times ! F B and ! F C Δ ! R BA = ! B − ! A = 0 − 3 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ − 6 0 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = − 6 − 3 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ˆ u B = Δ ! R BA Δ ! R BA = 1 6 2 + 3 2 + 2 2 − 6 − 3 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = 1 7 − 6 − 3 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = − 0.85714 − 0.42857 0.28571 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ Δ ! R CA = ! C − ! A = 0 0 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ − 6 0 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = − 6 0 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ˆ u C = Δ ! R CA Δ ! R CA = 1 6 2 + 0 2 + 2 2 − 6 0 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = 1 40 − 6 0 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = − 0.94868 0 0.31623 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ F B , F C , F DX , F DY , F DZ and ! F 1 ! F B + ! F C + ! F D + ! F 1 = 0 → ! F B + ! F C + ! F D = − ! F 1 → − 0.85714 − 0.42857 0.28571 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⋅ F B + − 0.94868 0 0.31623 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⋅ F C + 1 0 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⋅ F DX + 0 1 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⋅ F DY + 0 0 1 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⋅ F DZ = 0 − 250.0 433.0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ! F 1 ! M 1 = ! R A × ! F 1 = 6 0 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ × 0 250.0 433.0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = 0 2598.1 1500.0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ! F B F B ! M B = ! R B × ! F B = ! R B × ˆ u B ( ) ⋅ F B = 0 − 3 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ × − 0.85714 − 0.42857 0.28571 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⋅ F B = 0 − 1.71429 − 2.57143 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⋅ F B

I. Using D as the reference point, write the moment of as a column vector times J. Write the moment balance equation using column vectors and K. Write the augmented matrix for the system by combining the force-balance equation and the moment-balance equation, with an extra line at the top to show which unknown is associated with each column. You will need to add some zeros since don’t appear in the moment balance equation. L. Solve the system by row-reduction. While we have 5 unknowns and 6 equations, note that one of the equation rows is all zeros. You could drop the one equation and enter it as 5 rows and 6 columns, but row-reduction actually works anyway. M. Do the forces at D make any sense? It seems strange that the y and z forces at D are exactly zero. But if we had taken A as the reference point, then would not make any moments. So must be zero so they don’t either. is non-zero because it must balance the tensions in the cables. ! F C F C ! M C = ! R C × ! F C = ! R C × ˆ u C ( ) ⋅ F C = 0 0 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ × − 0.94868 0 0.31623 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⋅ F C = 0 − 1.89737 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⋅ F C F B and F C ! M B + ! M C + ! M 1 = 0 → ! M B + ! M C = − ! M 1 0 1.71429 − 2.57143 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⋅ F B + 0 − 1.89737 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⋅ F C = 0 − 2598.1 − 1500.0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ! F DX , ! F DY , and ! F DZ F B F C F DX F DY F DZ − 0.85714 − 0.94868 1 0 0 0 − 0.42857 0 0 1 0 − 2500 0.28571 0.31623 0 0 1 433.0 0 0 0 0 0 0 − 1.71429 − 1.89737 0 0 0 − 2598.1 − 2.57143 0 0 0 0 − 1500.0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ F B = 583.33 N F C = 842.26 N F DX = 1299.04 N F DY = 0 N F DZ = 0 N ! F B , ! F C , and ! F 1 ! F DY and ! F DZ ! F DX

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version