FIGURE 1 shows a 10 kg uniform metal rod with 2.0 m long pivoted at its centre C. A 20 N force is exerted on the rod, 0.5 m from point C. If the rod is in equilibrium, determine the a) force, F needed.
b) magnitude and the direction of the force exerted at point C.

Explanation:

Assuming we are looking parallel to a horizontal pivot pin with positive UP to the top of the picture, and the 20 N force acts perpendicular to the rod.

Sum moments about C to zero assume CW is the positive moment.

The rod mass causes no moment because there is no moment arm about C.

Fsin50[1.0} - 20[0.5] = 0

F = 10/sin50

F = 13 N

Summing vertical forces to zero With R being the pin reaction at C

Ry - F - 20sin50 - mg = 0

Ry = 13 + 20sin50 + 10(9.8)

Ry = 126.32 N

Summing horizontal forces to zero assuming a right directed force is positive.

Rx + 20cos50 = 0

Rx = -12.855...

R = √(-12.855² + 126.32²) = 126.972

R = 127 N

θ =arctan(126.32/-12.855) = 95.8107... = 5.8° CCW from UP

v=ir = 3x15 =45 v

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