45d.200 Ib.F = 200 Ib2. If 0 = 45°, determine the moment produced by the 4-kN force aboutpoint A.3 m7.21 kN.m (clockwise).7.21 kN.m (counter clockwise).8.48 kN.m (clockwise).8.48 kN.m (counter clockwise).a.b.0,45 mC.d.4 kN

Solution; Given Data; F=4 KN MA=-10 KN-m

F= 200 lb 2. If 0 = 45°, determine the moment produced by the 4-kN force about point A. %3D 3 m 7.21 kN.m (clockwise). b. 7.21 kN.m (counter clockwise). 8.48 kN.m (clockwise). d. 8.48 kN.m (counter clockwise). a. 0.45 m C. 4 kN

F= 200 lb 2. If 0 = 45°, determine the moment produced by the 4-kN force about point A. %3D 3 m 7.21 kN.m (clockwise). b. 7.21 kN.m (counter clockwise). 8.48 kN.m (clockwise). d. 8.48 kN.m (counter clockwise). a. 0.45 m C. 4 kN

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Equilibrium and Support Reactions

Civil engineers deal with structures that are composed of several members connected or with individual members such as beams and columns. To analyze such structures, the condition of statics must be applied. Under the static condition, the members are considered to be in equilibrium under the action of forces, that is, the members are assumed to be having negligible velocity.

Question

100%

45
d.
200 Ib.
F = 200 Ib
2. If 0 = 45°, determine the moment produced by the 4-kN force about
point A.
3 m
7.21 kN.m (clockwise).
7.21 kN.m (counter clockwise).
8.48 kN.m (clockwise).
8.48 kN.m (counter clockwise).
a.
b.
0,45 m
C.
d.
4 kN

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