{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW 5 Prob 2_94 extra

# HW 5 Prob 2_94 extra - PROBLEM 2.94 Knowing that the...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PROBLEM 2.94 Knowing that the tension is 510 1b in cable AB and 425 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables. SOLUTION ﬂ = (40 in.)i — (45 in.) j + (60 in.)k AB = (4011132 + (45 in)? + (60 in.)2 = 85 in. 2—6" = (100 mi — (45 in.) j + (60 in.)k AC = ‘/(100 in.)2 + (45 my + (60 in.)2 =125 in. AB (40 in.)i~—(45 in.)j+(60 in.)k T =2“ x =1" —= 5101b m AB AB AB AB AB C )I: 85111. TAB = (240 lb)i — (270 lb) j + (360 lb)k A? 100 in. i— 45 in. '+ 60 in.)k TAC : TAG )‘AC = TAC IE =(4251b)[£W—(—] TAC = (340 lb)i — (153 lb) j + (204 lb)k R = TAB + TAC = (5 80 lb)i ~ (423 1b) j + (564 lb)k Then: , 12:91292 lb R=913lb 4 and cosﬂx =—580—1b=0.63532 ax =50.6° 4 912.92 lb 0056 =ﬂz—046335 6 =117.6° 4 y 912.92 lb 1” cosQZ=—5——6ﬁ—1b—=O.6l780 6; =51.8° 4 912.92 lb M— ...
View Full Document

{[ snackBarMessage ]}