The stationary disc, of 300-mm radius is attached to a pin support at the point

Question

The stationary disc, of 300-mm radius is attached to a pin support at the point D. the disk is held in place by the brake ABC in contact with the disc at point C. The hydraulic actuator BE exerts a horizontal force of 400 N on the brake at B. Coefficient of static friction between disk and brake is 0.6. Determine the smallest couple moment to be applied to the stationary disc in the clockwise direction, to cause motion of the disc (27.7 N-m). What would be the couple moment's magnitude, if it were to turn the disc counterclockwise instead?

Explanation / Answer

Total moment acting on disk about the pin D (taking clockwise as positive and anticlockwise as negative)

- Fr*300*10^-3 + M , where M is the moment applied in clockwise direction externally

here Fr is frictional force at point C

Fr = N (where N is contact force at C)

Now for the equilbirum of brake about point A (when rotating clockwise)

+400*.2 - N*.4 - Fr*.2 = 0

Putting Fr = .6*N and solving we get

+80 - .4N - .12N = 0

so

N = 80/.52 = 153.84 N

Fr = .6*153.84 = 92.307 N

So for total moment

M = Fr*.3 = 92.307*.3 = 27.7 N-m

Now for the equilbirum of brake about point A (when rotating anti clockwise)

400*.2 - N*.4 + Fr*.2 = 0

So we get

N = 80/.28 = 285.714 N

so Fr = .6*285.714 = 171.428 N

External Moment in the anticlockwise direction will be

Fr*.3 = 171.428 * .3 = 51.428 N-m

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