cure https://ef.engr.utk.edu/ef152-2018-01/sys.php?f-assess/main EF152 jpan8- On
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cure https://ef.engr.utk.edu/ef152-2018-01/sys.php?f-assess/main EF152 jpan8- On-Line HW 1-0-8 Previous Next Close Current time: Wed Jan 10, 09:18:42 Next due date: i 1090 credit: Mon Jan 15, 2018 23:59 Adue date(s) 8. F1 = 160 lb @ 29°Ccw from +x, applied at [ 26, 45 ] ft F2 128 lb @ 15°Cw from-y, applied at [ 70.90 ] ft Point A: [ 5, 9] ft NOTES IMAGES DISCUSS UNITS STATS HELP Part Description Answer Chk What is F2 in a 3D vector component form? (enter as a vector: [x,v,z]) (include units with answer) A: [-33.13, -123.64, 0]lb C: -33.13,-123.6,0.000]ints: 0.0 lb A. What is the torque from F2 about point A in a 3D vector corpponent form? Array (enter as a vector: [x,y,z]) (include units with answer) Format Check 10.78 pts. 108% ok - using first value as 2150 ft-b 2% try penalty Hints:0.0 ft-lb What is the net torque of F1+F2 about point A? (+CCW) (include units with answer) C. Format Check 10.78 pts. 108% oK- using first value as 7360 ft-lb Hints: 0.0 | Previous |Next ivati pdf ^ Epicinstaller-7.00-...msi ^Explanation / Answer
B. torque = r x F
r = (70 - 5)i + (90 - 9)j
r = 65i + 81j
F2 = -33.13i - 123.6j
torque = r x F2 = - 8034 k + 2683.6 k
torque = -5350.5 k lb-ft
Ans: (0 , 0 , - 5350.5 ) lb ft
(C) due to F1,
Torque1 = (21i + 36j) x (-33.13i - 123.6j)
= - 2595.6k + 1192.7k
= -1403 k
net torque = - 5350 - 1403 = - 6753 lb ft .........Ans
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