Horizontal Forces exerted by door hinges

[sdcmma]

Homework Statement

A door of width 1.00 m and height 2.30 m weighs 272 N and is supported by two hinges, one 0.40 m from the top and the other 0.40 m from the bottom. Each hinge supports half the total weight of the door.

Homework Equations

Torque = rFsin(theta)

The Attempt at a Solution

I know that since the door is in equilibrium the net torque from either hinge must be zero.

I also know (I think) that the vertical component of the force exerted by each hinge must be equal to half the weight of the door in order for the net force to be zero; thus, at each hinge there is a vertical force of 136 N.

If we take one of the hinges as a pivot point, then the two forces that would produce torque are the horizontal force on the other hinge and the weight of the door.

The torque from the other hinge is rFsin(theta) = 1.5 F
The torque from the weight of the door is rFsin(theta) = sqrt(0.8125)(136)sin(33.7)
... sqrt(0.8125) is from the Pythagorean theorem; the linear distance from the hinge to the center of mass of the door is the hypotenuse of the right triangle with legs equal to 0.75 (half the distance between the hinges) and 0.5 (half the width of the door).
...136 is half the weight of the door, since the problem says that each hinge supports half the door's weight.
...33.7 degrees is tan(0.5/0.75) from the right triangle formed by legs having length half the distance between the hinges and half the width of the door.

Setting these equal to each other yields a force of 45.3 N, which is not the right answer... what am I doing wrong?

Thank you so much!

 

[PhanthomJay]

Homework Statement

A door of width 1.00 m and height 2.30 m weighs 272 N and is supported by two hinges, one 0.40 m from the top and the other 0.40 m from the bottom. Each hinge supports half the total weight of the door.

Homework Equations

Torque = rFsin(theta)

The Attempt at a Solution

I know that since the door is in equilibrium the net torque from either hinge must be zero.

you meant to say the net torque about one hinge form the other hinge and door's weight must be zero, right?

I also know (I think) that the vertical component of the force exerted by each hinge must be equal to half the weight of the door in order for the net force to be zero; thus, at each hinge there is a vertical force of 136 N.

this is given

If we take one of the hinges as a pivot point, then the two forces that would produce torque are the horizontal force on the other hinge and the weight of the door.

The torque from the other hinge is rFsin(theta) = 1.5 F

yes.

The torque from the weight of the door is rFsin(theta) = sqrt(0.8125)(136)sin(33.7)
... sqrt(0.8125) is from the Pythagorean theorem; the linear distance from the hinge to the center of mass of the door is the hypotenuse of the right triangle with legs equal to 0.75 (half the distance between the hinges) and 0.5 (half the width of the door).
...136 is half the weight of the door, since the problem says that each hinge supports half the door's weight.
...33.7 degrees is tan(0.5/0.75) from the right triangle formed by legs having length half the distance between the hinges and half the width of the door.

The weight is 272 N. But rather than use rFsin theta in computing the torque, it is much simpler to use the alternate equivalent definition of torque: Torque equals the force times the perpendicular distance from the line of action of the force to the pivot point.

 

[sdcmma]

Got it, thanks for your help!