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Cerrena [4.2K]
3 years ago
11

If the 1550-lb boom AB, the 190-lb cage BCD, and the 169-lb man have centers of gravity located at points G1, G2 and G3, respect

ively, determine the resultant moment produced by all the weights about point A.

Engineering
1 answer:
Natasha2012 [34]3 years ago
6 0

Answer:

hello the required diagram is missing attached to the answer is the required diagram

7.9954 kip.ft

Explanation:

AB = 1550-Ib ( weight acting on AB )

BCD = 190 - Ib ( weight of cage )

169-Ib = weight of man inside cage

Attached is the free hand diagram of the question

calculate distance x!

= cos 75⁰ = \frac{x^!}{10ft}

    x! = 10 * cos 75^{o} = 2.59 ft

calculate distance x

= cos 75⁰ = \frac{x}{30ft}

x = 30 * cos 75⁰ = 7.765 ft

The resultant moment  produced by all the weights about point A

∑ Ma = 0

Ma = 1550 * x! + 190 ( x + 2.5 ) + 169 ( x + 2.5 + 1.75 )

Ma = 1550 * 2.59 + 190 ( 7.765 + 2.5 ) + 169 ( 7.765 + 2.5 + 1.75 )

      = 4014.5 + 1950.35 + 2030.535

      = 7995.385 ft. Ib ≈ 7.9954 kip.ft

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<h2>Answer:</h2>

24Ω

<h2>Explanation:</h2>

When resistors are connected in parallel, the reciprocal of their combined resistance, when read with a DMM (Digital Multimeter - for measuring various properties of a circuit or circuit element such as resistance...) is the sum of the reciprocals of their individual resistances.

For example if two resistors of resistances R₁ and R₂ are connected together in parallel, the reciprocal of their combined resistance Rₓ is given by;

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Solving for Rₓ gives;

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From the question;

Let

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R₂ =  resistance of second resistor = 60Ω

Now,

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R_{x} = \frac{40 * 60}{40 + 60}

R_{x} = \frac{2400}{100}

R_{x} = 24 Ω

Therefore, the total resistance is 24Ω

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