(c) Two forces are applied on a beam as shown in Figure 2.3. The beam and both forces are in the plane of the page. Given that the weight of the beam is 20.0 N. F2 = 15.0 N F, = 10.0 N 22.0° K 3.50 m →* 4.50 m Figure 2.3 (A) Calculate the net torque and its direction about point O. Repeat Q2. (c) (i) (A) if the pivot point is at the middle of the beam. (B)

(c) Two forces are applied on a beam as shown in Figure 2.3. The beam and both forces are in the plane of the page. Given that the weight of the beam is 20.0 N. F2 = 15.0 N F, = 10.0 N 22.0° K 3.50 m →* 4.50 m Figure 2.3 (A) Calculate the net torque and its direction about point O. Repeat Q2. (c) (i) (A) if the pivot point is at the middle of the beam. (B)

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

3.) see photo for full equation
F1 F2 A F3 F4 E B F5 Consider the following values: - a = 25 n; b = 5 m; c= 25 m; d-5m3; F1 6 kN; F2 - 7 kN; F3 -8 kN; F4-9 kN; F5- 10 kN; a=60° 0= 45° 1] What is the resultant moment of the five forces acting on the rod about point A? a) 15.8 kN.m b) 74 kN.m )88 kN.m d) 59 kN.m e) 241 kN.m 0-47 kN.m
4
Q3: Three forces are acting on a steel plate with 4m length by 4m width shown in figure. These three forces are required to cause a horizontal resultant acting through point A. If F = 504N, 1) The value of horizontal resultant R is: E A D 853.814N + O 796.894N → O 711.512N - O 640.361N → O None of these
revinus (1548-1642 200 N -100 cos sin 017 F 2.sin 02+ F !. 00•-100sin 60 – 90 sin 43 (R.) +(R»} V(192.4) +(60.8) = 202N Ex ( 2): The 200 N force is a resultant of two forces, one of the forces "P" has the direction along the tine AB and the other force "Q" is on the horizontal direction, determine em. 60° ços OF F2.cos 0 2 Fi.cos 0F F2.cos 02 $ 60+Qcos(0) 1------- ------ 02
Q3: Three forces are acting on a steel plate with 4m length by 4m width shown in figure. These three forces are required to cause a horizontal resultant acting through point A. If F = 504N, 1) The value of horizontal resultant R is: F E A D B.
(Q1) (a) The force P= -40 i + 80 jN located at r= 3 i m is the resultant of a couple C1 = -150 k N.m and 50 N, 60 N , 20 N, 40 N and a fifth force F. Find F and its line of action. 50 N 2 m | 2m 60 N 30° 3 m 20 N 40 N
H.W.: ComPute the magnitude and direction the vesultant of the two forces shown in figure. R=? 45 25 200N | 300M 11.5 360N
I need the answer as soon as possible
3.4 In terms of the eccentricity e, the period T, and the angles a and B (in radians), find the time t required to fly from point 1 to point 2 on the ellipse. C is the center of the ellipse. T B-a-2e cos 2л B+a B-a sin 2 Ans. : t= A a P
Force F = - 4.4Ni + 6.4Nj acts on a small rock with a position vector r = 3.1mi + 8.5m j relative to the origin. In unit vector notation, what is the resulting torque on the rock about the point ( 2.4,4.7,9.9)m^ prime ?
Example 11: The man shown in Figure (a) pulls on the cord with a force of 70 Ib. Represent this force action on the support A as Cartesian components and determine its direction. Find the Cartesian nroiections of the r: