a A bent member fixed at at C with, a = 4.2 ft, = b 6.0 ft, M 138.0 ft-lb CW, 0=29.0°, and, F = 73.0 و lb. What is the resultant moment from M and F acting at point C? G Ꮎ F B b M C
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- Find the moment about A & B where AB = 8m, BC = CD = 6m & all forces makes 40° to the C3 %3D horizontal axis. 7kN 8 kN 10kN 15kN 6kN 5kN 4kN 3kNFind the moment of the forces shown * in fig w.r.t. point A? If P= 250 N Q = 150 N 0.2 m A P 35° -0.18 m -0.12 m-Find the moment about A & B where AB = 8m, BC = CD = 6m & all forces makes 40° to the %3D %3D horizontal axis. 7kN 8 kN 10KN 15kN 6kN 5kN B 3kN 4kN
- (2) Jakas The resultant of the force system shown in fig. is 750 N.cm clockwise ?moment, Find P, F, and the couple C 50 N 12 Tcm F 100 N upl- P= 200 N and F= 260 N and C=1440 N.cm O P= 345 N and F= 407 N and C=3095 N.cm O P= 407 N and F= 260 N and C=1440 N.cm O P= 407 N and F= 345 N and C= 3095 N.cm OPoint A is at (1.2, 0, 0.8) meters from 0. Force P is in negative y-dir only. Force Qis in negative z-dir only. A bent pipe is welded to a wall at point O. Two forces Q=213 N and P=154 N are applied at point A as shown. Compute the total moment about point O. What is the x-component of this moment in Newton-meters rounded to the nearest whole number?For the drawing shown, find the moment at * (- point (A) (Clock + and counter clock ור 1m F=7 kN 4m A MA=20 kN.m MA=-21 kN.m MA=-30 kN.m None of them
- (a) Calculate the moment of the 90-N force about point O for the condition @ 15. Also, determine the value of 6 for which the moment about O is (b) zero and (e) a maximum. Ans. (a) Mo-33.5 N m CCw (b) @-36.9 (or 217") (e) 8 126,9 (or 307") F- 90 N 800 mm 600 mm2. Four forces are applied to the machine element ABDE as shown to the left. Find the resultant 200 mm force and the resultant moment at point A. (Assume that points A, B, D, and E are located at the centroid of the cross-section of the bar, and that the forces act at these points.) 50 N 40 mm 20 mm B. 300 N R =-420î – 50ĵ– 250k N Ans.: 160 mm M, =0î +30.8ĵ –22k N - m 250 N D A 100 mm E 120 NFind the moment about point A due to the forces shown in the Figure below. HINT: Choose clockwise as negative and counterclockwise as positive. a. C. 10 N b. M M e. d. M 30 N 30 N = 1 m = -10 N. 1.5 m - 30 N. 1 = 2 m = -10 N 1.5 m + 30 N. 1 M +10 N 1.5 m - 30 N.3 1.5 m = -10 N. 1.5 m - 30 N. 2 M 10 N 2 m + 30 N 1.5 n O O O O
- Example The 30-N force P is applied perpendicular to the portion BC of the bent bar. Determine the moment of P about point B and about point A. P = 30 N %3D 1.6 m 45° 1.6 mFind Resultant (magnitude, Direction, Angle inclination), Moment @ Pt. A, Moment Arm, and x and 300 N y intercept 1.5 m 405 N 4 m G50 N 120 N 30° 60° A 400 NCalculate the moment M about point A, that is created by force F, located at point B (units are in meters). vectors and the cross-product method. determine the correct answer by using F = -2î + 4ĵ – 1k (N) y pt B (6,2,0) X pt A (5,0,2)