The bar rests against a6-27 The bar shown in Fig.smooth vertical wall at end D. The support at A is a ball-and-socket joint. The cable is not continuous at B.D

Answered: The bar rests against a 6-27 The bar…

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter1: Tension, Compression, And Shear

Section: Chapter Questions

Problem 1.4.10P: A long re Lai nine: wall is braced by wood shores set at an angle of 30° and supported by concrete...

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Question

The bar rests against a
6-27 The bar shown in Fig.
smooth vertical wall at end D. The support at A is a ball-
and-socket joint. The cable is not continuous at B.
D

Expert Solution

Step 1:-Given data

Let

$m = m a s s o f t h e b a r; a = d i s \tan c e b e t w e e n m a s s (m) & p o i n t A;$

Step 2:-Equilibrium condition

Let the system is in equilibrium condition ;

In that condition we can apply equilibrium condition ;

$\sum_{n} F_{Y} = 0;$

$(R_{A} + R_{D}) - (m \times g) + (T) = 0; (R_{A} + R_{D} + T) = (m \times g) . . . . . . . . . . . . . (1)$

$\sum_{n} F_{X} = 0;$

$(R_{D X}) = 0;$

$\sum_{n} F_{Z} = 0; T = 0; . . . . . . . . . . . . . . . . . (2)$

$\sum_{n} M_{A} = 0; (m g) \times (a) - (R_{D} \times (2 a)) = 0; (R_{D} \times (2 a)) = (m g) \times (a); R_{D} = [\frac{1}{2} \times m \times g]; R_{D} = [\frac{1}{2} \times m \times (9.81)]; R_{D} = [4.905 \times m]; . . . . . . . . . . . . . . . . . . . . . . (3)$

Now putting the equation (2) & equation (3) in equation (1) ;

$(R_{A} + R_{D} + T) = (m \times g); [(R_{A} + (4.905 \times m) + (0))] = (m \times g); R_{A} = [(m g) - (4.905 \times m)]; R_{A} = [(g) - (4.905)] \times m; R_{A} = [(9.81) - (4.905)] \times m; R_{A} = [(4.905) \times m]; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (4)$

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The bar rests against a 6-27 The bar shown in Fig. smooth vertical wall at end D. The support at A is a ball- and-socket joint. The cable is not continuous at B. D