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Vladimir79 [104]
3 years ago
11

If the 1500-lblb boom ABAB, the 220-lblb cage BCDBCD, and the 169-lblb man have centers of gravity located at points G1G1, G2G2

and G3G3, respectively, determine the resultant moment produced by all the weights about point A.
Mathematics
1 answer:
ANEK [815]3 years ago
4 0

Answer:

 (M)_a = -8171 lb-ft

Step-by-step explanation:

Step 1:

- We will first mark each weight from left most to right most.

               Point:                      Weight:                             Moment arm r_a

               G_3                          W_g3 = 169                     30*Cos(75) + 4.25

               G_2                          W_g2 = 220                    30*Cos(75) + 2.5

               G_1                           W_g1 = 1500                    10*cos(75)

Step 2:

- Set up a sum of moments about pivot point A, the expression would be as follows:

(M)_a = -W_g3*(30cos(75) + 4.25) - W_g2*(30*Cos(75) + 2.5) - W_g1*10*cos(75)

Step 3:

- Plug in the values and solve for (M)_b, as follows:

    (M)_a = -169*(30cos(75) + 4.25) - 220*(30*Cos(75) + 2.5) - 1500*10*cos(75)

    (M)_a = -2030.462559 -2258.205698 - 3882.285677

    (M)_a = -8171 lb-ft

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Martin throws a ball straight up in the air. the equation h(t) = -16t^2 + 40t + 5 gives the height of the ball, in feet, t secon
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Step-by-step explanation:

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t ----> the time in seconds

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we know that

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so

-16t^2+40t+5=0

The formula to solve a quadratic equation of the form

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in this problem we have

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t=\frac{-40+\sqrt{1,920}} {-32}=-0.12\ sec

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The solution  is t=2.62 seconds

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