Torques exerted by Gravity (CONFUSING)

[avenkat0]

Homework Statement

A door with symmetry and of m 43.1 kg is 4.76 m high and 2.68 wide. The hinges are placed all the way on top and all the way on the bottom of the door. Find the absolute value of the horizontal force supplied by the bottom hinge.

Homework Equations

Sum of torques = 0
Torque = R x F

The Attempt at a Solution

I tried finding the torque exerted by gravity... and equaling that to the torque exerted by the hinges...
F(L)=W/2(mg)
now would i divide the force by 2 since there are 2 hinges?
or is my reasoning wrong from the first place?

 

[Doc Al]

Your reasoning is correct. Note that your answer had nothing to do with the top hinge, so don't be dividing by anything.

 

[avenkat0]

Ohh ok i did exactly that and it came out to be wrong...
119.026 came out to be my answer...

 

[Doc Al]

That answer seems about right to me. (Is this one of those on-line homework systems?)

 

[avenkat0]

Yeah its webassign... i asked the professor and this is what he said:

First, draw a diagram showing the forces (including gravity) acting on the object. Next, pick a point of rotation. (Of course, you always pick a point that eliminates one or more of the unknowns.) And then, you apply the torque equation.

But that's exactly what i did... isn't it?

 

[Doc Al]

Yes, that's exactly what you did.

 

[avenkat0]

Doc Al... i contacted him again and i was told, "Its not just a product but a CROSS PRODUCT"...
but in a vertical door arent all the torques exerted by gravity perpendicular??
im confused

 

[asleight]

Doc Al... i contacted him again and i was told, "Its not just a product but a CROSS PRODUCT"...
but in a vertical door arent all the torques exerted by gravity perpendicular??
im confused

No, they're not all perpendicular. With respect to the axis of rotation, there is an angle between the radius and the force applied.

 

[Doc Al]

Doc Al... i contacted him again and i was told, "Its not just a product but a CROSS PRODUCT"...
but in a vertical door arent all the torques exerted by gravity perpendicular??
im confused

It certainly is a cross-product, not just a scalar product, but what you did was fine. To find the torque due to gravity, you took the force (mg) and multiplied it by the perpendicular component of the distance to the axis (W/2). All perfectly correct.

Similarly for the horizontal force supplied by the hinge.

What you did was completely correct.

To express the point more mathematically (for the torque due to gravity):
\tau = \vec{R} \times \vec{F} = RF\sin\theta = F(R\sin\theta) = mg (W/2) = mgW/2