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Let’s Torque
Equilibrium (again)
• Its rotation is not speeding up or slowing down.
• AT REST.
• or CONSTANT ANGULAR VELOCITY.
ΣFx = 0
ΣFy = 0
Στ = 0
Reminder about rotational equilibrium!
Your child is sitting at the end of a really heavy see-saw
of total length 4m, and wants you to balance him
horizontally. How far should you sit from the fulcrum,
which is at the see-saw’s center?
You
(60kg)
Child
(25kg)
In this problem, the system can rotate! And we want it in Equilibrium. So let’s think about this with torques and equilibrium. I’d like to help you think through rotational
equilibrium problems. You can use this to apply to things like how much will this strain my back if I pick up a box, e.g.; it’s important in physiology.
“Balancing something”
problems
[DRAW FREE BODY DIAGRAM!]
Pick the right axis
(put it at the location of a force you don’t know!).
Sum forces, sum the torques, set them equal to zero.
Solve.
FIRST TRICK: PICK AN AXIS THAT SETS ONE TORQUE TO ZERO!
There’s another one like this on the homework this weekend.
Where would you put your axis?
Mary is about to do a push-up. Her center of gravity lies directly above
a point on the floor which is d1=1.0 m from her feet and d2=0.7 m from
her hands. If her mass is 50 kg, what is the force exerted by the floor
on her hands, assuming that she holds this position?
If an object is in equilibrium (she is), you can take the axis of rotation
to be anywhere. The trick here is to set it somewhere that you don’t
know the force. Then sum the torques which is equal to 0.
A.
B.
C.
D.
At toes
Q88
At hands
At center of gravity
None of the above
Holding this position
ANSWER: A.
I’m going to show you two more examples of a problem like this that look very different but are the SAME THING! THE POINT OF THIS is to understand that axis choice is up to you but really
matters: make it somewhere that you don’t know a force, but not at the thing you’re solving for because that will make it get multiplied by zero! (When we write the torque at her hands it
would be Fperp multiplied by radius zero!)
A metal advertising sign (weight
w) is suspended from the end of a
massless rod of length L. The rod
is supported at one end by a
hinge at point P and at the other
end by a cable at an angle θ from
the horizontal.
What is the tension in the cable?
A. T = w sin θ
B. T = w cos θ
C. T = w/(sin θ)
[DRAW FREE BODY DIAGRAM!]
Pick the right axis.
Sum forces, sum the torques, set them equal to zero.
Solve.
D. T = w/(cos θ)
E. none of the above
This problem looks very different but is the exact same idea.
DRAW THE FBD FIRST.
Answer: C
Q89