Example 2.9. A horizontal line PQRS is 12 m long, where PQ = QR = RS = 4 m. Forces of 1000 N. 1500 N, 1000 N and 500 N act at P. Q. R and S respectively with downward direction. The lines of action of these forces make angles of 90°, 60°, 45° and 30° respectively with PS. Find the magnitude, direction and position of the resultant force.

Example 2.9. A horizontal line PQRS is 12 m long, where PQ = QR = RS = 4 m. Forces of 1000 N. 1500 N, 1000 N and 500 N act at P. Q. R and S respectively with downward direction. The lines of action of these forces make angles of 90°, 60°, 45° and 30° respectively with PS. Find the magnitude, direction and position of the resultant force.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

= 35.10 Qa" The resultant of two concurrent forces is 1500 N and the angle b/w forces is 90°. The resultant makes an angle of 36⁰° with find the magnitude of one of the force find the each force.
Y. 3. The force system shown consist of a counterclockwise couple of 720 N-m plus a 1200 N force through the origin O, directed down to the left as speçified. Determine the (a) distance between the forces and (b) intercepts of the single force at A(a,0) and B(0, b). 13 12 F=1,200 N A(a,0) X C= 720 N-m A B(0,b) F=1,200 N 200 N 150 N 20 cm 4. Given 3 pulleys, the circular sections of which are concentric, and acted upon by three couples as shown. 300 N 40 cm Determine its resultant moment, 300 N e 3 150 NY 200 N+ 30 cm 100 N 150 cm 5. The force system on frame ABCDE is shown below. Replace the system by a force equal to the resultant of the three forces, acting vertically at B 1sd cm plus a couple acting horizontally at D and E. 100 am D 150 cm 30 N
2. Consider two forces F₁ and F₂, both having the same magnitude. F, makes an angle of 60° with both the y and the z axes. F₂ lies in the xy plane and makes an angle of 60° with the x axis. The directions of both forces are as indicated in the sketch. Find the angle between the two forces.[ans: 38.1°] N 60° F₁ 60⁰ 60⁰
A force F having a magnitude of F = 100 N actsalong the diagonal of the parallelepiped. Determine themoment of F about the point A, using MA = rB x F andMA = rC x F.
5. The resultant of the three forces shown is the force R = 200 N acting along z-axis. Determine P1, P2, and P3. B A P₂ Z C 150 mm 200 mm P₁ 120 mm 150 mm P3 320 mm
A. Force System 1. Determine the components of the 500N force directed up to the right as shown in the figure. F = 500N Fy 40° Fx
143 in Newton 30 40° T15° Compute for the forces on the following based on the above figure: a. F along a is [forceA] b. along b is [forceB]
If the force F = 100 N lies in the plane DBEC, which is parallel to the x–z plane, and makes an angle of 10° with the extended line DB as shown, determine the angle that F makes with the diagonal AB of the crate.
(image attached) 1. Find the force in members AB, BC, BD and CD. 2. Find the absolute vertical displacement of joint C (in inches). Use principal of virtual work. Given: EA= 2000 k for members AB, BC and BD. EA= 4000 k for member CD alpha = 5.0 x 10^-5 degrees Fahrenheit (for all trusses). 6. Does the vertical displacement of C move up or down?
QI/ The rigid pole and cross-arm assembly is supported by the three cables shown. A turnbuckle at D is tightened until it induces a tension T in CD of 1.2 kN. Express T as a vector. Does it make any difference in the result which coordinate system is used? Ans. T = 0.321i + 0.641j – 0.962k kN, IC 1.5 m B 1.5 m G T= 1.2 kN 3 m 2 m 1.5m 3 m
Determine the resultant force at A. for the figure below Hint : 1- Find the position of all points A,B,C (x, y, z) 2- Find FAB and FAC 3- Find IrAnl and ITacl 4- Find FAB = F UAB and FAC = F UAC 5- Finally find R and a, ß,y %3D %3D 3m Fc= 500N FB= 500N 3m 5m 3m B 4m 2m 5m
PROBLEMS Introductory Problems 2/1 The force F has a magnitude of 500 N. Express F as a vector in terms of the unit vectors i and j. Identify the x and y scalar components of F. Ans. F = 383i - 321j N, F, = 383 N, F, 321 N %3D 40 F = 500 N