Question

A meter stick is at a 23 degree angle with the wall. It is 40cm high. How do I figure out the torque around the point it meets the wall?

Answer 1

How is the mass distributed on the meter stick ? Or Is there a force applied on the stick ?

Answer 2

In answer to the first question:
I am only assuming, so please tell me if there is ANYTHING more to the question. The EASIEST way to do it was to draw a diagram. The diagram is based on the question, but from the question we can get a few things:

*There is a wall. Walls are vertical. correct? lol
*There is an angle measured 23 degrees.
*We are told that the meter stick "meets" the wall
*Wherever it meets the wall, there is an angle, so there is an angle of 23 degrees between the wall and the stick
*We always see a wall with a base, right? or the ground.
*The point where the stick meets the wal is the corner between the wall and the ground THIS IS VERY IMPORTANT
*So we can have the stick at an angle with the wall AND raised above the ground.
*This allows us to have a height of 40cm. FROM the ground TO the top of the stick
@(Fil Rouge) It is good to know if there is a force applied on the stick
*There is a rule that ALL objects are affected by GRAVITY once they are raised above the ground. THEY HAVE WEIGHT and weight is a FORCE

Another IMPORTANT point is this. Torque is defined as Force multiplied by distance
Normally The force act at 90 degrees, like turning a door. I don't want to explain that too much, because this is a SPECIAL situation but remember that TORQUE NEEDS both a Force AND a distance in meters
\[Torque = Force \times Distance\]

The diagram uses the Formula:
\[Torque = Fd \cos \theta\]This is because there is an ANGLE (in this case, between the wall and the stick). The value of the angle is really important.

In the diagram it is the angle closest to the ground

Weight = 10N
Distance = 40 cm = 0.4m
Angle = 90-23 = 67(degrees)

\[Torque = Fd \cos \theta\]\[Torque = 10\times0.4\times \cos 67\]
Answer = 1.563 Nm or 1.6Nm