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A door has a height of 2.1 m along a y axis that extends vertically upward and a width of 0.975 m along an x axis that extends outward from the hinged edge of the door. A hinge 0.47 m from the top and a hinge 0.47 m from the bottom each support half the door's mass, which is 29 kg. In unit-vector notation, what are the forces on the door at (a) the top hinge and (b) the bottom hinge

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MechEngineer

Answer:

[tex]\vec{F_T} = \vec{F_B} = 142.245 N[/tex] downward

Explanation:

As each hinge supports half the door mass, which is 29kg, then each of them are subjected to 29/2 = 14.5 kg of door mass. Let g = 9.81 m/s2, the weight on each hinge would be:

[tex]W = mg = 14.5*9.81 = 142.245 N[/tex] acting downward

In unit-vector notation, the force vectors on the top and the bottom hinges are:

[tex]\vec{F_T} = \vec{F_B} = 142.245 N[/tex] downward

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