Question

Assignment 3 - Problem 3 Esl Length: 2:00:00 Yuying He: Allempl 3 Page 1: Determine the moment of force F about an axis extending between A and C. Express the result as a Cartesian vector. Use F - [5i + 10 - 4k] KN, L1 - 5 m, L2 - 4 m, and L3 = 3 m for your calculations. Your answer: x-component of the resultant moment in kNm (round the result to 1 decimal place): y-component of the resultant moment in kNm (round the result to 1 decimal place):

Question 1 options:

Determine the moment of force F about an axis extending between A and C. Express the result as a Cartesian vector. Use F = {5i + 10j - 4k} kN, L1 = 5 m, L2 = 4 m, and L3 = 3 m for your calculations.

Your answer:

x-component of the resultant moment in kNm (round the result to 1 decimal place):

y-component of the resultant moment in kNm (round the result to 1 decimal place):

z-component of the resultant moment in kNm (round the result to 1 decimal place):

Answer 1

Consider the free body diagram as follows: 5 m 4 m 3 m F = [5i +10j - 4k] KN Write the position vector AB as follows: AB=B-A AB = (5i + 4j-3k)-(0i + Oj+0k) = (5-0)i +(4-0)j+(-3+0)k = (5i + 4j – 3k) m Write the position vector AC as follows: AC =C-Ā AC = (5i +4j+ Ok)–(0i+Oj+0k) = ( 51 +4j + Ok) m

Calculate the magnitude of the position vector AC as follows: |AC| = Vr? + y2 +:? |AC|= 152 +4 +(-3)? = 6.4 m Calculate the unit vector for û c as follows: üs - AC À (51+4j+0k) 6.4 Calculate the moment of the force about A to C. Mc = ủsc ·(ABXF) 2.[(5i +47 – 3k) mx(5i +10j – 4k) kN] 6.4 M4C = (Si +4j+0k). 5 10 4 = (-16+30) - ** (-20+15)

Calculate the moment of a force about A to C. MAC = M.c ·ûAC (5i + 4j+Ok 26.4 Mac =(14.06 kN - m). (S1+ = 14.06 Sa + OK = (10.981 +8.78j+Ok ) kN.m The x-component of the resultant moment in 10.98 kN.m. The y-component of the resultant moment in 8.78 KN.m. The z-component of the resultant moment in 0.

Answer 2

L = 5m L2=4m L₂=3m fossi +109-4R? KN A = 01 +0j + OP B = 51 +42-3 T = 51 +4 ht or scaler opproach *) Ma- 4x fz - Z fy=4864) - (-3) x 10 = 114 lb-ft | My = -xx fz + Z x fx=1-5*(-4) + (3)x 5 = 15.66-4t M2 = ny fy-yx fu = 5x10 - 445= (30 lb-ft vector approach Mr = Huxli x F ) = 12 s 4 -3 = 14-26+30) 6-0 foris) 70 (10) 15 10 -41 - 14.20-ft partca) My = 4yX (BXF) = 10 lol s 4 -3 - -1-20+15) => 15 16-44 15 10 -4 | past (4) M2 = Uzx (B x 2 =10 IS o 4 d -3] => 1150-20 ] = 30 lb-ft) . pant (c) 10 -4