Answer :
The tower can lean over the distance, x = 2.5m at the top.
It is assumed that the top is off-centered by distance, d=4.5m
The height of the tower is, h = 55m
Diameter of the tower is, D = 7.0m
Since we are assuming that the tower is uniform; so, its center of gravity (or center of mass) will be at its geometric center i.e.,
at height [tex]\frac{h}{2}[/tex] = 27.5m from the base of support.
The tower can lean until a line projected downward through its center of gravity will fall outside its base of support of diameter. The center of mass can move a total of 3.5 m off of center and still be over the support base.
Corresponding to the center of mass, the top would have to move distance equal to one diameter (D) in order to put the center gravity over the pivot point of one side of the base.
So, further the top can lean over the distance,
x = D - d
x = 7.0 m - 4.5 m
x = 2.5 m
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