Question

A uniform 17 kg door that is 2.2 m high by 1 m wide is hung from two hinges that are 20 cm from the top and 20 cm from the bottom. If each hinge supports half the weight of the door, find the magnitude and direction of the horizontal components of the forces exerted by the two hinges on the door.

Answer 1

46.3 N , Left

46.3, Right

Step-by-step explanation:

`F_(Ax)` = horizontal component of force at upper hinge A

`m` = mass of the door = 17 kg

`d` = distance between two hinges = 1.8 m

`w` = width of the door = 1 m

Using equilibrium of torque about B

`F_(Ax) d = (mg) ((w)/(2) )\\F_(Ax) (1.8) = (17)(9.8) ((1)/(2) )\\F_(Ax) = 46.3 N`

Direction : Towards left

`F_(Bx)` = horizontal component of force at lower hinge B

Using equilibrium of force along vertical direction

`F_(Bx) = F_(Ax)\\F_(Bx) = 46.3 N`

Direction : Towards right