Find the magnitude of the the moment MAB Z-axisX-axisу-аxisI ICIn the accompanying sketch, the following variables apply.Force F = (490i + 100j + 270k ) N passes through point C(2490i – 4560j – 2830k ) Nm about point= (3, 5, – 4) m and-has a moment MA =А — (— 3, — 2, 2) т.Point B = (2, – 6,— 6) т.(You can rotate the figure by holding in the right-click of the mouse on the figure.)

Answered: Image /qna-images/answer/f028e080-cfb8-47f7-a859-daae6667251c.jpg

Answered: Find the magnitude of the the moment…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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3D Moment Find the moment about O if F = {11i + -13j} kN, i = 4m, j = 4m, k = 2m. Upload your work for this 3D Moment problem in the next question. Type 4 as the answer to this question. im jm km F B
MEC107-Lectur... Prashant Kumar Pan... A lever AB = 2 m long is inclined at 35⁰ with horizontal A force of 180 N is acting at B vertically downwards. The moment about 'A' is (a) 29.5 Nm B (b) 295 Nm -2m (c) 2950 Nm 180N (d) 2.950 Nm 6/9/2022 35°
a = 4 mb = 3.2 mc = 2 mF = {-7i+4j+12k} NSpecified axis: AF Find the magnitude of the moment of F about the specified axis.If the calculated answer is not whole number, express it in 4 significant figures.
у- ахis F4 F5 X- axis BC F1 х- ахis y - axis F1 F2 Figure (1) Figure (2) 250 N 36.870 D F2 60° 0.5 m 1 m B 0.2 m 1 m 300 NE F1 1 m Figure (3) Figure (4) Q1] Answer the following questions: For figure (1): The object is subjected to two forces F1 = 25 N, and F2 = 50 N. Set x = 30°, 0 = 45°, and ß = 30°. 1- What is the magnitude of the resultant of these two forces? a) 39.4 N b) 80.1N c) 41.6 N d) 69.8 N e) 71.4 N f) 46.1 N g) None of them 2 - What is the direction of the resultant force measured counterclockwise from the positive x-axis? a) 14. 3° b) 392.6° c) 9.2° d) 322.9° e) 6.8° f) 344.2° g) None of them 500 N
Cartesian coordinat system; A(0,0,0) B(3,4,0) C(3,4,4) F=-20i +10 j+15k N is applied to point C Moment of force F about point A; M, Moment of force F about point B; M, Moment of force F about line AB; M AB Which following one is TRUE? ON MAB(-3i+4j)OM, O A) M = (3i – 4 j)®(-20i +10 j+15k) M, =(j)®(-20i +10 j +15k) O D Mg = 80i – 40j o M, = (3i+4 j+ 4k)® (-20i+10j+15k)
a = 4 mb = 3.2 mc = 2 mF = {-7i+4j+12k} NSpecified axis: AF Find the following: Find the z-component of the moment of F about the specified axis. Find the x-component of the moment of F about the specified axis. Find the magnitude of the moment of F about the specified axis. Find the y-component of the moment of F about the specified axis.
can you pls help me solve this as for my understanding.. I am trying to answer exercises. subject: Statics •PLS ANSWER WITH COMPLETE SOLUTION •3 DEC PLACES •PROVIDE FREE BODY DIAGRAM (if applicable) •Correct UNITS 4. Determine the moment about the origin O of the force F = 2i + 3j -4k that acts at a point A. Assume that the position vector of A is (a) r = 3i -6j +5k, (b) r = i - 4j - 2k, (c) r = 4i + 6j -8k. Thank you. *I'll give your answer a LIKE for helping me.
Cartesian coordinat system; A(0,0,0) B(3,4,0) C(3,4,4) A/ M АВ AB OM, F =-20i+10 j +15k N is applied to point C Moment of force F about point A; M, B/ M = AC ® F C/ M AB = AB O F Moment of force F about point B; M, D/ M, = BC ® F Moment of force F about line AB; MAB Which following one is WRONG? E/ M AB OM, AB
Compare the magnitudes of the equilibrant vectors measured from the experiment with those obtained from the graphical and component methods. Example: A: 200 g 60° above +x axis B: 300 g 45° above -x axis C: 400 g 30°below -x-axis A, A cos a = 1.96 N x cos 60° = 0.98 N B₂B cos b = 2.94 N x cos 135º = -2.08 N C, C cos g = 3.92 N x cos 210°= -3.39 N R₂-A, + B + C₂ = -4.49 N A, A sin a = 1.96 N x sin 60° = 1.70 N By B sin b = 2.94 N x sin 135º = 2.08 N C, C sin g = 3.92 N x sin 210° = -1.96 N Ry= Ay+ By + Cy = 1.82 N Questions: I: (a) A: 200 g along +x axis B: 100 g 45° above -x axis A₂ = A cos a = B₂ =B cos b = R₂-A₂+ B₁₂= A₂ = A sin a = B, = B sin b = R₂ = A + B₂ = R=(R₂. R₂):_ Quadrant R = √ (R₂²+R₂²) = Direction:q=tan [R, /R.] (c) A: 100 g along -y axis B: 200 g along -x axis A = A cos a = B = B cos b = R₂-A₂+ B₂ = A = A sin a = B, B sin b = R=A, +B₂ = R=(R₁, R₂): Quadrant R = √ (R₂²+R₂²) = Direction:q tan¹ [R, /R]
l Z 4.5G+ ? 5:23 PM A docs.google.com Q18 If we use force F = 80N to pull the nail .Find the nail force resistance if the distance d = 0.5m and nail force distance 4cm Pivot إجابتك Q19 Determine the moment of the force about A F = 200 N 45° 100 mm A 100 mm'100 mm Determine the moment of the force about A. 16 إجابتك
Questions Ib and le refer to the figure shown. F=|-201 + 10j + 15k) N lb. What is the moment of F about point A? (a) - 60 i +0j+ 120 k (N-m) 4 m (b) 60 i + 10 j - 120 k (N-m) (c) - 20 i + 125j-110 k (N-m) 4m (d) 20 i - 125 j+ 10 k(N-m) lc. What is the moment of F about line AB? (a) 53.64 N-m (b) 88 N-m (c) - 88 N-m (d) - 53.64 N-m
onthly Exam of Mechanics/Firs x E Monthly Exam of Mechanics/Firs X /e/1FAIPQLSFCZQE_ZSh4aZJdw9EBqR0OfLVx_M7KUGF6 9bgBbd-YGOXA/formResponse For the drawing shown, find the moment at point (A) (Clock + and counter clock 10 kN 8kN A 15 kN-m 2 m 3m MA=90 KN. m O MA--80 KN m O MA=75 KN m O None of them O For the drawing shown, find the moment at point (0) (Clock + and counter clock

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