Problem 6/33  Page 440, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: A uniform steel plate of mass M hinged about one edge.  The plate has length L parallel to the hinged edge and length H perpendicular to the hinged edge.  The hinged edge is supported by two bearings, each of distance B from the midline of the plate.  The plate is initially supported in a horizontal plane prior to being released from rest.

Find: The forces exerted by the bearings on the plate an instant after the plate is released.

Parameter Values:
M = 20 kg
L = 0.25 m
B = 0.205 m
H = 0.4 m

1. Mechanical System: Plate at the instant just after release.

2. Free Body Diagram: At the instant of interest the plate is still contained in a horizontal plane.  A weight force acts vertically downward through the mass center of the plate.  The bearings ( A and B) prevent vertical motion so vertically upward forces act at each bearing.  The bearings also prevent horizontal motion in the direction perpendicular to the hinge axis.  Thus horizontal forces in this direction are shown at both bearings.  One bearing must be capable of exerting a force along the hinge axis in order to prevent the plate from sliding along the hinge axis direction.  Thus a force along the hinge axis is applied at one of the bearings (we will assume at bearing A).  The coordinate origin is chosen to be midway between the two bearings.  The z axis is along the hinge axis direction.  The x axis is vertically upward.  The y axis is in the horizontal plane and is perpendicular to the hinge direction.

3. Equations:
a) S F = ( Ax + Bx - M g ) i + ( Ay + By ) j + Az k
S M_G = (-H/2 j - B k ) X ( Ax i + Ay j + Az k ) + ( -H/2 j + B k ) X ( Bx i + By j )
     = (-H/2 Az + B Ay - B By ) i + ( -B Ax + B Bx ) j + (H/2 Ax  + H/2 Bx ) k

b) m a_G = M ( a_O + w X (w X r_G/O) + a X r_G/O
Point O is on the hinge axis and hence does not move (zero acceleration).  The plate is released from rest so at the instant of interest the angular velocity is zero.  The plate rotates about the z axis so the angular acceleration is in the z direction.  The vector from the origin to the mass center is of magnitude H/2 and is along the y axis.
m a_G = M a k X H/2 j
         = -M H/2 a i
I_G a = 1/12 M H² a k
Mass moment of inertia for a plate about an axis parallel to one of its edges and through its mass center.

4. Solve:
Ax + Bx - M g = -M H/2 a
Ay + By = 0
Az = 0
-H/2 Az + B Ay - B By = 0
-B Ax + B Bx = 0
H/2 Ax  + H/2 Bx =  1/12 M H² a
 
From the third equation:
Az=0
Factoring out B from the remaining terms in the fourth equation:
Ay = By
Noting from the second equation that the sum of Ay and By must be zero and noting that Ay must equal By:
Ay = 0
By = 0
Thus the only forces acting at the bearings at this instant are in the x direction (vertical).  Factoring a B out from the fifth equation:
Ax = Bx
Multiplying the first equation by H/6 and adding it to the sixth equation eliminates the term involving the angular acceleration of the plate:
(H/6 + H/2) (Ax + Bx) - H/6 M g = 0
Noting that Ax and Bx are equal:
Ax = Bx = 1/6 M g / ( 4/3) = 1/8 M g
Ax = Bx = 24.5 N