If a torque of M = 300 N⋅m is applied to the flywheel, determine the force that must be developed in the hydraulic cylinder CD to prevent the flywheel from rotating. The coefficient of static friction between the friction pad at B and the flywheel is μs = 0.4.

866.1 N

Explanation:

The torque on the flywheel = 300 N-m

The force from the hydraulic cylinder will generate a moment on CA about point A.

The part of this moment that will be at point B about A must be proportional to the torque on the cylinder which is 300 N-m

we know that moment = F x d

where F is the force, and

d is the perpendicular distance from the turning point = 1 m

Equating, we have

300 = F x 1

F = 300 N   this is the frictional force that stops the flywheel

From F = μN

where F is the frictional force

μ is the coefficient of static friction = 0.4

N is the normal force from the hydraulic cylinder

substituting, we have

300 = 0.4 x N

N = 300/0.4 = 750 N

This normal force calculated is perpendicular to CA. This actual force, is at 30° from the horizontal. To get the force from the hydraulic cylinder R, we use the relationship

N = R sin (90 - 30)

750 = R sin 60°

750 = 0.866R

R = 750/0.866 = 866.1 N

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