11-3-99
Sections 9.1 - 9.3
Objects in equilibrium
We've talked about equilibrium before, stating that an object is in equilibrium when it has no net force acting on it. This definition is incomplete, and it should be extended to include torque. An object at equilibrium has no net force acting on it, and has no net torque acting on it.
To see how the conditions are applied, let's work through a couple of examples.
Example 1
The first example will make use of the hinged rod supported by a rope, as discussed above. The rod has a mass of 1.4 kg, and there is an angle of 34° between the rope and the rod.
(a) What is the tension in the rope?
(b) What are the two components of the support force exerted by the hinge?
The free-body diagram is shown below, with the support force provided by the hinge split up into x and y components. If you aren't sure which way such forces go, simply guess, and if you guess wrong you'll just get a negative sign for that force.
Something we'll assume in this example is that the rod is uniform, so
the weight acts at the center of the rod (the center is the center of
mass, in other words). As usual, sum the forces in the x and y
directions:
There are too many unknowns here, and this is why summing the torques
can be so useful. To sum torques, choose a point to take torques
around; a sensible point to choose is one that one or two unknown
forces go through, because they will nt appear in the torque equation.
In this case, choosing the hinge as the point to take torques around
eliminated both components of the support force at the hinge. As with
forces, where you choose plus and minus directions, choose a positive
and negative direction for torques. In this case, let's make
counter-clockwise negative and clockwise positive.
This can be solved for T, the tension in the rope. Note that r, which represents the length of the rod, cancels out:
This can be substituted back into the force equations to find the components of the hinge force:
Example 2 - a step-ladder
A step-ladder stands on a frictionless horizontal surface, with just the crossbar keeping the ladder standing. The mass is 20 kg; what is the tension in the crossbar?
This is something of a tricky problem, because you have to draw the free-body diagram of the entire ladder to figure out the normal forces, and then draw the free-body diagram of one half of the ladder to complete the solution. This is also what makes it a good example to look at, however.
Consider first the free-body diagram of the entire ladder. The floor is
frictionless, so there are no horizontal forces exerted by the floor.
The ladder is uniform, so the weight acts at the center of mass, which
is halfway up the ladder and halfway between the two legs. Summing
forces in the y-direction gives:
One way to approach this is to say that the ladder is symmetric, and
there is no reason for the two normal forces to be different; each one
should be equal to half the weight of the ladder. If you don't like
this argument, simply take torques about one of the points where the
ladder touches the floor. This will give you an equation saying that
one normal force is equal to half the ladder's weight, so the other
normal force must be equal to half the weight, too. Either way, you
should be able to show that:
Now consider the free-body diagram of the left-hand side of the ladder.
I'll attach a 1/2 as a subscript to the mass, to remind us that the
mass of half the ladder is half the mass of the entire ladder.
Taking torques around the top of the ladder eliminates the unknown
contact force (F) coming from the other half of the ladder, and gives
(this time taking clockwise to be positive):
This can be solved to find the tension in the crossbar:
