2)it is known that the connecting rods exerts on the crankbca 2.5-kn force directed down andto the left along the centerline ofab. determine the moment of this force about for the two casesshown at below.

M_c = 100.8 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two cases shown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

                        M_c = F_a*( 42 / 150 ) *144

                        M_c = 2.5*( 42 / 150 ) *144

                        M_c = 100.8 Nm

- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

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