gneill said:Are there any vertical placement measurements for the cutouts? How do we know how high the circle is positioned, or rectangles? How long are the rectangles?
vac said:Yes I listed above.
A of circle = pi * 0.5^2
A of rectangle = 2.5*1
A of rectangle = 2.5*0.5
vac said:All I know is this formula: Mcmx = (m1x1 + m2x2 ... mnxn)/(m1+m2...mn) where x can be x,y, or z.
It does not help me that much in this problem.
gneill said:Okay, but how high are these items from the bottom of the triangle? Their exact placement matters.
And it appears to me from the diagram that the two rectangular cutouts are the same width, so why are you giving one as 2.5*1 and the other 2.5*0.5?
Enigman said:Okay, vector format would have been better but we can work with this. Just need to work out x and y coordinates separately.
Note that you will need all the values specifying their placement...like gneill just said.
- Can you find out CM of the figure without any holes (geometrically)?
- Can you find out the individual CM of the shapes that fit into the holes (geometrically)?
- Can you after all this express the actual CM in terms of these? (Think in terms of the formula.)
Where is your origin? It should be same for all figures.vac said:
- Can you find out CM of the figure without any holes (geometrically)?
I think it is (x,y) should be (4,4) right?- Can you find out the individual CM of the shapes that fit into the holes (geometrically)?
- rectangle one is (0.5, 1.25)
- rectangle tow is (0.25, 1.25)
- Can you after all this express the actual CM in terms of these? (Think in terms of the formula.)
- I don't know?
It doesn't seem like this could be the case. The figure implies that the altitude and base of the triangle are both 8 units. Yet, the figure as drawn looks like an equilateral triangle. So the figure could not be to scale.gneill said:A question for vac: Were you given just the figure and a ruler to work with? In other words, are you expected to take your own measurements directly from the figure?
Chestermiller said:It doesn't seem like this could be the case. The figure implies that the altitude and base of the triangle are both 8 units. Yet, the figure as drawn looks like an equilateral triangle. So the figure could not be to scale.
Chet
The center of mass is a point in an object or system where the entire mass of the object or system can be considered to be concentrated. It is the average position of all the mass in the object or system.
The center of mass is calculated by finding the weighted average of the positions of all the individual masses in the object or system. This is done by multiplying the position of each individual mass by its mass and dividing by the total mass.
The center of mass is important because it is the point around which an object or system can be balanced. It also helps in predicting the motion of an object or system, as it is the point where all the external forces act on the object or system.
The higher the center of mass is above the base of support, the less stable an object or system will be. This is because a higher center of mass means that the object or system is more likely to tip over.
Yes, the center of mass can be outside of an object. In cases where the object has an irregular shape or has varying densities, the center of mass may be located outside of the physical boundaries of the object.