Q.5) A moment and four forces are applied to a two-dimensional frame as shown in figure below. Replace the given force system by a resultant force system acting at point A. Then replace the resultant force system by the simplest resultant and find x and y intercepts of its line of action. 50 lb 10 in. 80 lb 18 in. 14 in.- 75 lb B E 12 in. 100 in.-lb 90 lb -16 in. 16 in.-

Q.5) A moment and four forces are applied to a two-dimensional frame as shown in figure below. Replace the given force system by a resultant force system acting at point A. Then replace the resultant force system by the simplest resultant and find x and y intercepts of its line of action. 50 lb 10 in. 80 lb 18 in. 14 in.- 75 lb B E 12 in. 100 in.-lb 90 lb -16 in. 16 in.-

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

Elements of the lower arm are shown in the figure. The mass of the forearm is 3.7 kg with the center of mass at G. Determine the combined moment about the elbow pivot O of the weights of the forearm and the sphere. What must the biceps tension force be so that the overall moment about O is zero? T 50 mm 55 G 150 mm 2.3(9.81) N 3.6(9.81) N 325 mm
i need the answer quickly i need all the steps, i need the answer in Handwritten.
A system of Forces is shown in the Figure. All length units are in meters. Determine the moment.
Mechanics #11
Q2 Determine the moment caused by the forces subjected to the frame around the point O, if you know that the traffic light signs have the weight shown in the figure, also the frame weight is 100 kg with a center of gravity showing at point G В G A 36 kg 36 kg 5 m 4 m 7 m
Please help solve
Calculate the value of F and the value of theta θ so that the resultant of the forces is 1200 lb downward
The moment of the force F=360-lb about the point O in Cartesian vector form is. F 1 ft 4 ft' 2 ft 4 ft Select one: O a. Mo = 600j – 1200k b. Mo = –200j + 400k O c. Mo = -200j – 400k O d. Mo = 200 + 400k