Reaction force between rod AC and trunnion B in crank mechanism

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In the plane mechanism shown,the driving crank OA,which is 50mm between centers,oscillates about the fixed axis O.The rod ZC,which is uniform,100mm long and of tatol mass 0.05kg,slides through the trunnion(swiveling pin)B.The distance OB is 75mm.When the crank OA is at the position shown,it has an anticlockwise angular velocity of 40rad/s and zero angular acceleration.Assume that the masses of all moving links of other than rod AC and the effects of friction and gravity are to be neglected.For the position shown determine the reaction force between the rod AC and the trunnion B.
(this is my mid term question,but my lecturer nevr discuss about it.)
 
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Your posting mentions a rod ZC but the drawing shows a rod AC. I will use AC in this reply.

1. You are looking for force. Force is a function of acceleration. So I would concentrate on the acceleration that the driving crank, OA delivers to the rod, AC.

2. The location where that acceleration is being delivered to the rod is at point A. Since A is currently moving in a circle about O, it only has acceleration in the radial direction OA. That acceleration is the length of the crank times the angular velocity squared (50 * 40 * 40). Consider this acceleration to have two components, one parallel to the rod AC and one perpendicular. You can forget the parallel component. It just pulls the rod out of the trunnion. The perpendicular component is the one involved in forces on the trunnion. I’ll let you do the geometry to calculate the magnitude of the perpendicular component of acceleration at A.

3. The rod cannot translate perpendicular to its length. So the perpendicular acceleration at A must be completely a tangential acceleration about the center of the trunnion, B. Since tangential acceleration is equal to the angular acceleration times the moment arm, AB, you can calculate the angular acceleration of the rod about B (angular accel = tangential accel / AB).

4. Since torque about a point is the moment of inertia times the angular acceleration, you can calculate the torque about B. You need to calculate the moment of inertia first. Make sure you calculate it about the point B.

5. The torque about B is also equal to the force at A times the moment arm, AB. So you can calculate the force at A.

6. The fact that the rod is not translating perpendicular to its length means that the sum of the forces perpendicular to its length are zero.
There can be only two forces, the one at A and the one at B. They must be equal and opposite. You just calculated the force at A. So you also have the force at B. The force at B must be delivered by the trunion. Again, equal and opposite. The force on the trunion is also B.


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