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

Question

The stationary disc, of 300-mm radius b attached to a pin support at the point D. the disk b held in place by the brake ABC in contact with the disc at point C. The hydraulic actuator BE exerts a horizontal force of400 N on the brake at B. Coefficient of static friction between disk and brake b 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

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.