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Rina8888 [55]
3 years ago
6

47. the beam is supported by two rods ab and cd that have cross-sectional areas of 12mm^2 and 8mm^2, respectively. determine the

position d of the 6-kn load so that the average normal stress in each rod is the same.
Physics
1 answer:
Ugo [173]3 years ago
8 0

Let the beam is of length L

Now the stress on both the end is same

now we can say that torque on the beam due to two forces must be zero

N_1* x =  N_2* (L - x)

also we know that stress at both ends are same

\frac{N_1}{12} = \frac{N_2}{8}

2*N_1 = 3*N_2

Now from two equations we have

\frac{3}{2}N_2*x = N_2* (L - x)

solving above equation we have

x = \frac{2}{5}L

<em>so the load is placed at distance 0.4L from the end of 12 mm^2 area</em>

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Explanation:

1) Since the mass of <span>155 g is suspended at 210 degrees, you need to find the components of its weight on the orthogonal coordinate system (0° and 90°).
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<span>2) You do that using the trignometric ratios sine and cosine.
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<span>Weight is mass × g.
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<span>Weight of the object = 155g × g
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<span>Angle, α = 210°
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<span>Horizontal component (0°)
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⇒ horizontal = 155g × g × cos(210°) = - 134.23g  × g

Vertical component
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3) Conclusion:

Therefore, the masses that must be suspended to balance the forces of the 155g mass are 134.23g at 0° (horizontal) and 77.5g at 90° (vertical).




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