Answer Happy

(20%) Problem 5: It is common to hang objects on doorknobs and over-the-door hooks. There is a limit to the amount of weight that a door can hold because of the forces exerted on the hinges. A door of height &-2.2 m and width 6/2 has a mass of M-3 kg. The mass is distributed uniformly so the center of mass is located at the geometric center of the doce. One hinge is located a distance A4 from the top of the door. The second hinge is a distance A4 from the bottom of the door. Refer to (a) in the figure. The door's weight is supported entirely by the two hinges and each hinge supports half of the weight. In other words, the vertical force exerted by each hinge is exactly one half of the total weight, including any additional load. For this problem, take the positive direction to be directly upward and the positive x-direction pointing from the hinge side of the door to the knob side. Mg 20% Part (a) Choose the correct free-body diagram for the door. Me [Me Me h/4 Ⓡ Hinge 11 Hinge 2 (c) +4+h/4- mg Mg Otheexpertta.com Grade Summary Deductions Potenal Sabindo (pr Submit Hist I give Hists: deduction per hit. His g 20% Part (b) Calculate the force, in newtons with its sign, that the upper hinge exerts on the door in the x-direction. 20% Part (c) Calculate the force, in newtons with its sign, that the lower hinge exerts on the door in the x-direction. 20% Part (d) A bag of mass m= 15 kg is hung on the doorknob, which is located a distance h/2 from the bottom of the door. To make the math easier, you may assume that the load is exerted on the right edge of the door. Refer to (b) in the figure. Calculate the force, in newtons with its sign, that the upper hinge exerts on the door in the x-direction. & 20% Part (e) The same bag of mass m = 15 kg is now hung on a hook in the middle of the top of the door. Refer to (c) in the figure. Calculate the force, in newtons with its sign, that the upper hinge exerts on the door in the x-direction. mg h/2 HOS