Here's the question: "A camera of mass 240g is mounted on a small tripod of mass 200g. Assuming that the mass of the camera is uniformly distributed and that the line of action of the weight of the tripod passes through D, determine (a) the vertical components of the reactions at A, B, and C when [itex] \theta [/itex] = 0. (b) the maximum value of [itex] \theta [/itex] if the tripod is not to flip over." (Refer to the attachment to better interpret what is being asked.) I've gotten part (a) easily (summation of the forces and moment about D easily gives the answer). However, for part (b), I'm loss. As far as I can tell, the z component of the momentum about D is zero only for mutiples of 180 degrees, and yet the answer indicates that the maximum occurs at [itex] \theta = 54.1^o [/itex]. Any ideas?
Note: Here's the equations and coordinates I'm using (notice that I've moved the coordinate system down to the bottom, so D is at (0,0,0): A: (-0.045 m, 0, 0) B: (0.035 m, 0, 0.038 m) C: (0.035 m, 0, -0.038 m) D: (0, 0, 0) F: ( [itex] \alpha , 0, \beta [/itex] ) where F is where the line of action of the weight of the camera intersects the xz plane, and where: [tex] \alpha = -0.036 cos( \theta ) [/tex] [tex] \beta = -0.036 sin( \theta ) [/tex] If you find the sum of all the moments at D, it should equal zero if the object is static.
The critical condition is how the mass is distributed relative to the line AC. When the line of action of the camera gets too far over that line, the moment of the tripod will be insufficient to maintain equilibrium.