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Dovator [93]
3 years ago
13

Determine the resultant moment produced by the load and the weights of the tower crane jibs about point A and about point B. Exp

ress your answers using three significant figures separated by a comma.
Physics
1 answer:
grigory [225]3 years ago
5 0

Answer:

(M)_A = (M)_b = 76027.5 Nm

Explanation:

Step 1:

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

                                 Point:                       Weight:

                                  G_3                          W_g3 = 6 Mg*g

                                  G_2                          W_g2 = 0.5 Mg*g

                                  G_1                           W_g1 = 1.5 Mg*g

                                 Load                           W_L = 2 Mg*g

Step 2:

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

                   (M)_b = W_g3 * 7.5 + W_g2*4 - W_g1*9.5 - W_L*12.5

Step 3:

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

                   (M)_b = 9.81*10^3( 6* 7.5 + 0.5*4 - 1.5*9.5 - 2*12.5)

                   (M)_b = 9.81*10^3 * (7.75)

                   (M)_b =  76027.5 Nm

Step 4:

- Extend each line of action of force downwards, we can see that value of each force does not change also the moment arms of each individual weight also remains same. Hence. we can say that (M)_A = (M)_b = 76027.5 Nm

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