A door made of a uniform piece of aluminum measures 2m by 1m and has a mass of 1

Question

A door made of a uniform piece of aluminum measures 2m by 1m and has a mass of 15kg. It is supported by two hinges, one located 0.1m from the bottom of the door and one located 0.1m from the top of the door at the vertical edge of the door. Assume that the vertical force each hinge exerts is the same and find the force (in vector components) that each hinge exerts on the door. Hint: remember gravity is exerted at the center of the door and each hinge exerts both a vertical and horizontal force. You must use torque balance here.

Explanation / Answer

area of the door is A = 2 * 1 m^2

mass of the door is m = 15 kg

the force acting on the door due to upper hinge is

F1x = mg * cosA

and F1y = mg - mg sinA

similarly due to the lower hinge

F2x = mg * cosA

and F2y = mg - mg * sinA

the total force in the x and y directions are

Fx = F1x + F2x = 2mg * cosA

and Fy = 2(mg - mg * sinA)

the net force on the door is

F = (Fx^2 + Fy^2)^1/2

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