Calculate the moment at point B of the forces shown in figure (clockwise +* and counter clockwise -)La 7120 N60 N1 m100 N3m270 N- m150 N120 N13301 m0.5 m0.5 m5 mMB=280 N.mMB=290 N.mMB=250 N.mMB=260 N.mرجوععدم إرسال كلمات المرور عبر لماذج Go ogle مطلثًا.elasyi selued uc 8 Al-Mansour University College Jagal lia plGoogle L

It is required to determine the moment about point B

Calculate the moment at point B of the forces shown in figure (clockwise + * and counter clockwise -) 120 N 60 N 1 m 100 N 3m 270 N-m 150 N 120 N 30 13 4 0.5 m 1 m 0.5 m 5 m MB=280 N.m O MB=290 N.m O MB=250 N.m O MB=260 N.m O

Calculate the moment at point B of the forces shown in figure (clockwise + * and counter clockwise -) 120 N 60 N 1 m 100 N 3m 270 N-m 150 N 120 N 30 13 4 0.5 m 1 m 0.5 m 5 m MB=280 N.m O MB=290 N.m O MB=250 N.m O MB=260 N.m O

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

See similar textbooks

Similar questions

A tee section is made up of a 30 mm x 150 mm flange and a 30 mm x 160 mm web.4. Which of the following most nearly gives the area of the section in square mm.a. 9000 b. 9300 c. 9500 d. 8700 5. Which of the following most nearly gives the location of the neutral axis from the bottom of top of the flange.a. 64 mm b. 68 mm c. 51 mm d. 84 mm 6. Which of the following most nearly gives the moment of inertia with respect to the centroidal x-axis in million (1 x 10⁶).a. 21.84 mm⁴ b. 27.45 mm⁴ c. 54.21 mm⁴ d. 31.54 mm⁴
Determine the internal normal force at point C. Take F1F1F_1 = 460 lblb , F2F2F_2 = 250 lblb , F3F3F_3 = 346 lblb . Determine the shear force at point C. Determine the moment at point C. Determine the internal normal force at point D. Determine the moment at point D.
F1 = 250 lbx = 6 fty = 2 fta = 3b = 2Total Moment = 400 lb-ft (counterclockwise) Find the perpendicular moment arm of F (in ft). Find F (in lb).
Bending moment equation for segment BC. * a. 7.35x(KN)-0.05x3(KN/m2)-34.3KN.m b. 7.26x(KN)-0.07x3(KN/m2)-35.3KN.m c. 7.28x(KN)-0.06x3(KN/m2)-33.4KN.m d. 7.6x(KN)-0.08x3(KN/m2)-32.4KN.m
Completely determine the resultant of the forces shown in the figure. Option: R= 957lb down to the right at 32.2°, (0, 1.38) (2.19,0) R= 759lb down to the right at 32.2°, (0, 1.38) (2.19,0) R= 957lb down to the left at 32.2°, (0, 1.38) (2.19,0) R= 759lb down to the left at 32.2°, (0, 1.38) (2.19,0)
Solve for the Following Bending moment equation for segment BC. a. 2000 lb(X) - 100 lb/ft (X squared) - 9600 lb-ft b. 2100 lb(X) - 120 lb/ft (X squared) - 9700 lb-ft c. 21500 lb(X) - 200 lb/ft (X squared) - 9800 lb-ft d. 3000 lb(X) - 200 lb/ft (X squared) - 9850 lb-ft The distance between point A and the shear force VAB=0 a. 6 feet b. 8 feet c. 10 feet d. 12 feet The degree of bending moment equations for segments AB and BC. a. 1st degree and 2rd degree b. 3rd degree and 2nd degree c. 2nd degree and 3rd degree d. None of the above.
Determine the magnitude of moment (Nm) of the 383-N force about point A if x = 0.55 m , y = 0.69 m, and θ = 41.4ᵒ. Round off only on the final answer expressed in 3 decimals. Instead of units, type either POSITIVE or NEGATIVE to indicate the direction of moment which is consistent with the sign convention discussed on lectures. Sample answers are 12.908 POSITIVE or 12.908 NEGATIVE
5a. If lengths a=1.2m, b=3.7m, c=1.6m, d=4, e=1.4m, and f=4.3m, determine the angle (in degree) between the forces F_A and F_B. Answer must include 1 place after the decimal point. 5b. If lengths a=1.1m, b=4.8m, c=1.1m, d=3.2, e=1.3m, and f=3.2, determine the moment (in N*m) caused by force F_A about axis OD. Answer must include 1 place after the decimal point. Negative sign must be included if the rotational tendenct about the OD axis (from O to D) is clockwise. 5c. Find out all support reaction components at the fixed support O after being given all forces and lengths. ( use the 2nd image named practice problem 5-2 for this question)
Determine the normal force, shear force and moment at point E and F. Tabulate the final anwer in rounding upto 3 significant figures E F Normal Force (lb) Shear Force (lb) Moment (lb ft) Weight of Crate = 200 lb
Determine the bending moment acting at each of the following locations:(a) x = 13.5- ft (i.e., just to the left of point B)(b) x = 13.5+ ft(i.e., just to the right of point B)(c) x = 17.0 ft (i.e. at point C)(d) x = 23.0 ftNote that x = 0 at support A. When entering your answers, use the shear-force sign convention detailed in Section 7.2.Answers: (a) M = kips-ft (b) M = kips-ft (c) M = kips-ft (d) M = kips-ft Use your shear-force and bending-moment diagrams to determine the maximum positive bending moment, Mmax, pos, the maximum negative bending moment, Mmax, neg, and their respective locations, xmax, pos and xmax, neg. When entering your answers for the maximum bending moments, use the shear-force and bending-moment sign conventions detailed in Section 7.2. The maximum negative bending moment is the negative moment with the largest absolute value. Enter the maximum negative bending moment as a negative value.Answers: Mmax, pos = kips-ft xmax, pos = ft Mmax,…
A member is loaded with the given system of forces as shown in figure. Find the moment about A & B. Where AB = AD = 80 m, F1=35 N F2=5N, F3=39N, F4=32N, theta1=35degree , theta2= 50degree, theta3=55degree & theta4=30degree. The Moment about "A" (in Nm) = The Moment about "B" (in Nm) = The Moment about "C" (in Nm) = The Moment about "D" (in Nm) = Short Answer
1. Determine the resultant of the three concurrent forces. y F=50N F=10N 14 60° X F=60N 2. [Example] A hand crankFi is in the Axy plane. A force F is applied at point D and in a plane perpendicular to the y axis. The angle between the force and the vertical line is a. CD=a, BCllx, CD/ly, AB-BC=1. Determine the moments of the force about x, y and z axes. E A y