How can the weight force of a rigid body be applied in a moments equation?

[greg_rack]

Homework Statement

A uniform square trap door of side 0.80 m and mass 14kg has a smooth hinge at one edge and
is held open at an angle of 30° to the horizontal. It is supported by a single rigid rod placed so
that it meets the surface of the trap door at 90° at a distance 0.10m from the top edge of the
trap door.
What is the normal contact force exerted on the trap door by the rod?

Relevant Equations

P=mg

I'm having a hard time solving this problem since I don't really know how to apply the weight force of the trap door.
Ideally, in order to find the contact force exerted by the rod, I would find out the weight force of the trap in the point of contact and then find its component radial to the rod... but I indeed don't know how to apply weight.

 

[berkeman]

Homework Statement:: A uniform square trap door of side 0.80 m and mass 14kg has a smooth hinge at one edge and
is held open at an angle of 30° to the horizontal. It is supported by a single rigid rod placed so
that it meets the surface of the trap door at 90° at a distance 0.10m from the top edge of the
trap door.
What is the normal contact force exerted on the trap door by the rod?
Relevant Equations:: P=mg

View attachment 270070
I'm having a hard time solving this problem since I don't really know how to apply the weight force of the trap door.
Ideally, in order to find the contact force exerted by the rod, I would find out the weight force of the trap in the point of contact and then find its component radial to the rod... but I indeed don't know how to apply weight.

The best way to start most of these types of problems is by drawing free body diagrams (FBDs) of each mechanical piece. Can you show us your FBD of the trap door, with the forces at the hinge and where the support connects? And if the trap door is not moving, what is true about the sums of the forces and moments that are applied to it?

Thanks.

 

[haruspex]

In a 2D statics problem, there are three standard equations available. What are they?

 

[Lnewqban]

... I'm having a hard time solving this problem since I don't really know how to apply the weight force of the trap door.
...

Consider that the weight of the trap is uniformly distributed across its square shape, since its thickness and density are constant.

In that particular case, we can assume that an imaginary vector weight (mass x gravity acceleration) is located at the centroid of that square and that is always pointing vertically down, regardless the angle the trap makes with the ground.

 

[greg_rack]

In a 2D statics problem, there are three standard equations available. What are they?

Aren't those Ftot=0(and thus a=0) and Mtot=0?

 

[greg_rack]

Consider that the weight of the trap is uniformly distributed across its square shape, since its thickness and density are constant.

In that particular case, we can assume that an imaginary vector weight (mass x gravity acceleration) is located at the centroid of that square and that is always pointing vertically down, regardless the angle the trap makes with the ground.

View attachment 270074

View attachment 270073

The images you've attached are really eloquent... thank you!

 

[haruspex]

Aren't those Ftot=0(and thus a=0) and Mtot=0?

Right, except that F_tot = 0 is available in each of two dimensions, so that makes three equations altogether.
Sometimes you do not need all three. The moments equation leaves you the choice of which point to take as the axis. If there are forces that you do not know and do not need to find then picking an axis on the line of action of such a force is a smart move, since that force will not feature in the moments equation.
Can you see such a axis in this case?

In regard to the weight of the trap door, for a rigid body in a uniform gravitational field, it is always ok to consider its weight as acting wholly at its mass centre.

 

[greg_rack]

Right, except that F_tot = 0 is available in each of two dimensions, so that makes three equations altogether.
Sometimes you do not need all three. The moments equation leaves you the choice of which point to take as the axis. If there are forces that you do not know and do not need to find then picking an axis on the line of action of such a force is a smart move, since that force will not feature in the moments equation.
Can you see such a axis in this case?

In regard to the weight of the trap door, for a rigid body in a uniform gravitational field, it is always ok to consider its weight as acting wholly at its mass centre.

Got it! I did it by equalling momentums of weight and contact force... the thing that was confusing me was where to locate the weight force on the trap :)
Thank you very much!