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Alenkasestr [34]

2 years ago

13

A light bar AD is suspended from a cable BE and supports a 20-kg block at C. The ends A and D of the bar are in contact with fri

ctionless vertical walls. Determine the tension in cable BE and the reactions at A and D.

1 answer:

babymother [125]2 years ago

7 0

Answer:

Tension in cable BE= 196.2 N

Reactions A and D both are 73.575 N

Explanation:

The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence

$T_{BE}-W=0$ hence

$T_{BE}=W=20*9.81=196.2 N$

Therefore, tension in the cable, $T_{BE}=196.2 N$

Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then

$196.2\times 0.125- 196.2\times 0.2+ D_x\times 0.2=0$

$24.525-39.24+0.2D_x=0$

$D_x=73.575 N$

Similarly,

$A_x-D_y=0$

$A_x=73.575 N$

Therefore, both reactions at A and D are 73.575 N

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So,

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Svetllana [295]

$79 f t^{3}$ is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

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$V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}$

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