Solving Torque Equation: Door with Mass 20kg, 2 Hinges

[BI]

I have this problem with torque and not sure if I approached it correctly or not.

A door is 1m wide and 2m high, has a mass of 20kg, and is supported by two hinges, each bearing 1/2 the weight of the door. The hinges are attached to the left of the door at locations 1/2m above the bottom edge and 1/2m below the top edge. It is in static equilbrium.

So, I write out the two conditions:
vector sum of net force= (Wdoor along -j)+(Fsupport1 along j)+(Fsupport2along j)=0
vector sum of net torque= 0
I'm stuck at this, I don't know if I wrote the equation right or not. Please help. I don't know where to put the pivot point in this example either. Is the direction for the forces right? Please help.

 

[member 553083]

Yes, your equation is correct. The pivot point in this example is the location of the hinges. For each force, you can use either a positive or negative sign depending on the direction of the force. In this case, it looks like the weight of the door is acting downwards (in the -j direction), so you can use a negative sign for that force. The forces from the two hinges will be acting in the +j direction, so you can use a positive sign for those forces.

 

[thepatientmental]

Your approach to the problem is correct. The pivot point for this problem would be at the bottom hinge since it is the point where the door is rotating about.

To solve for the torque equation, you would need to consider the distance between the pivot point and each force acting on the door. In this case, the distance between the pivot point and the weight of the door would be 1m, the distance between the pivot point and the force of support from the bottom hinge would be 0.5m, and the distance between the pivot point and the force of support from the top hinge would be 1.5m.

The equation for torque is T = F x d, where T is torque, F is the force, and d is the distance between the pivot point and the force.

So for the weight of the door, the torque would be -20kg x 9.8m/s^2 x 1m = -196 Nm (negative because it is in the opposite direction of the clockwise rotation).

For the force of support from the bottom hinge, the torque would be (1/2)(20kg x 9.8m/s^2) x 0.5m = 49 Nm (positive because it is in the same direction as the clockwise rotation).

And for the force of support from the top hinge, the torque would be (1/2)(20kg x 9.8m/s^2) x 1.5m = 147 Nm (positive because it is in the same direction as the clockwise rotation).

Now, to solve for the net torque, you would add all of these torques together: -196 Nm + 49 Nm + 147 Nm = 0 Nm. This means that the net torque is 0 Nm, which is what we want since the door is in static equilibrium.

In summary, your approach to the problem is correct and the direction of the forces is also correct. Just remember to consider the distances between the pivot point and each force when solving for the torque equation.