Member AB is supported by a cable BC and at A by a square rod which fits loosely through the square hole in the collar fixed to the member as shown. Determine the components of reaction at A and the tension in the cable needed to hold the rod in equilibrium. 3 m A) 1/7 6 m 400 N в) 37 C) 4/7 300 N x'2 m D) 5/7 F:Applied force F:Reaction force at A=F(A);components are A,, A,, A, E) 2/7 Tyc: Tension force in cable BC ignore weight of the member AB. EF. = 0→-400+kTgc =0→k = ? (components of applied force are 300 N and 400 N)

Member AB is supported by a cable BC and at A by a square rod which fits loosely through the square hole in the collar fixed to the member as shown. Determine the components of reaction at A and the tension in the cable needed to hold the rod in equilibrium. 3 m A) 1/7 6 m 400 N в) 37 C) 4/7 300 N x'2 m D) 5/7 F:Applied force F:Reaction force at A=F(A);components are A,, A,, A, E) 2/7 Tyc: Tension force in cable BC ignore weight of the member AB. EF. = 0→-400+kTgc =0→k = ? (components of applied force are 300 N and 400 N)

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

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Concept explainers

Free Body Diagram

The system of forces acting on a body tends to either rotate it or make the body undergo motion. In general, when an external agent exerts a force on a body, the body undergoes translational motion and rotational motion. The body has six degrees of freedom under the influence of that force. Here, three degrees of freedom are from the translational motion while the other three are from the rotation motion. However, sometimes the bodies are not allowed to undergo any motion and are fixed, under such conditions the body is said to be constrained.

Question

Member AB is supported by a cable BC and at A by a
square rod which fits loosely through the square hole in the
collar fixed to the member as shown. Determine the
components of reaction at A and the tension in the cable
needed to hold the rod in equilibrium.
3 m
A) 1/7
6 m
400 N
B) 3/7
C) 4/7
300 N
2 m
D) 5/7
F:Applied force
E) 2/7
F:Reaction force at A=F(A),components are A,, A,, A,
Tạc : Tension force in cable BC
ignore weight of the member AB.
EF, = 0→-400+kT3c =0→k = ?
(components of applied force are 300 N and 400 N)

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