Is there a trick to finding these 3D vectors in Cartesian coordinates?

[lc99]

Homework Statement

Homework Equations

The Attempt at a Solution

I am having a bit of trouble visualizing the vectors for these type of problems. The angles they give are very ambiguous and so I am not sure why they are there. For the 45 degree angle, how do i know that this is used for finding the x and z components of the weight force?

For the 60 degree angle, i am not sure why that is there and is useful for.

Do you think that it was be useful to analyze these parts individually by drawing the out separately?

 

[andrewkirk]

45 degrees is the angle that the rod makes with the vertical. That is needed to find the x,y,z coordinates of the centre of mass of the rod excluding its handle, and how gravity acting on that will transmit force to the points C and D.

60 degrees is the angle between the handle and the plane containing the vertical line and the rod. It is needed in order to find the coordinates of the CoM of the handle and hence the torque that the handle's weight applies to the rod, about the rod's axis, as well as what the handle adds to the force the rod applies to points C and D.

The rod will, unless opposed, rotate so that the handle goes to its lowest point. The moment M is what must be applied to stop that rotation from commencing.

 

[lc99]

45 degrees is the angle that the rod makes with the vertical. That is needed to find the x,y,z coordinates of the centre of mass of the rod excluding its handle, and how gravity acting on that will transmit force to the points C and D.

60 degrees is the angle between the handle and the plane containing the vertical line and the rod. It is needed in order to find the coordinates of the CoM of the handle and hence the torque that the handle's weight applies to the rod, about the rod's axis, as well as what the handle adds to the force the rod applies to points C and D.

The rod will, unless opposed, rotate so that the handle goes to its lowest point. The moment M is what must be applied to stop that rotation from commencing.

I'm not sure if I am seeing the moment of the weight force about the x axis. Where is the perpendicular force?

 

[andrewkirk]

I'm not sure if I am seeing the moment of the weight force about the x axis. Where is the perpendicular force?

Taking C as the origin, the CoM (call it point M) of the rod+handle combination will have nonzero x, y and z coordinates.

Assuming the y-axis is supposed to be horizontal (it looks so, although the diagram does not explicitly state that), the weight force vector will be $-\frac{W}{\sqrt2}(1,0,1)$. To get the moment of that about C, take the cross product of that weight vector with the vector CM. We can then observe that only one of the moment's coordinates can possibly be zero and it is not the x coordinate.

 

[lc99]

Taking C as the origin, the CoM (call it point M) of the rod+handle combination will have nonzero x, y and z coordinates.

Assuming the y-axis is supposed to be horizontal (it looks so, although the diagram does not explicitly state that), the weight force vector will be $-\frac{W}{\sqrt2}(1,0,1)$. To get the moment of that about C, take the cross product of that weight vector with the vector CM. We can then observe that only one of the moment's coordinates can possibly be zero and it is not the x coordinate.

Hi, I am still not getting it. cause the answer key i think says that it would have moment about x::