A uniform door (0.86 m wide and 2.2 m high) weighs 140 N and is hung on two hing

Question

A uniform door (0.86 m wide and 2.2 m high) weighs 140 N and is hung on two hinges that fasten the long left side of the door to a vertical wall. The hinges are 2.2 m apart. Assume that the lower hinge bears all the weight of the door. Find the horizontal component of the forces, including sign, applied to the door by (a) the upper hinge and (b) the lower hinge, taking the direction to the right to be the positive direction. Determine the (c) magnitude and (d) direction (as a positive angle counterclockwise from horizontal) of the force applied by the door to the upper hinge. Determine the (e) magnitude and (f) direction (as a positive angle counterclockwise from horizontal) of the force applied by the door to the lower hinge.

This is a qustion of equallibrium.

Explanation / Answer

The 140 N acts at a perpendicular distance of 0.405 m from the hinges.

F1 and F2 are the horizontal forces that act on the door by the hinges.

Taking moment about the lower hinge,

140*0.405 = F1*2.2

F1 = 25.77N

Taking moment about the upper hinge,

140*0.405 = F2*2 .2.2

F2 = 25.77N

Both forces are away from the door.

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