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Note that setting up a free-body diagram for a rigid-body equilibrium problem is the most important component in the solution process. Without the correct setup and a correct diagram, you will not be able to write down correct conditions for equilibrium. Also note that a free-body diagram for an extended rigid body that may undergo rotational motion is different from a free-body diagram for a body that experiences only translational motion (as you saw in the chapters on Newton’s laws of motion). In translational dynamics, a body is represented as its CM, where all forces on the body are attached and no torques appear. This does not hold true in rotational dynamics, where an extended rigid body cannot be represented by one point alone. The reason for this is that in analyzing rotation, we must identify torques acting on the body, and torque depends both on the acting force and on its lever arm. Here, the free-body diagram for an extended rigid body helps us identify external torques.

The torque balance

Three masses are attached to a uniform meter stick, as shown in [link] . The mass of the meter stick is 150.0 g and the masses to the left of the fulcrum are m 1 = 50.0 g and m 2 = 75.0 g . Find the mass m 3 that balances the system when it is attached at the right end of the stick, and the normal reaction force at the fulcrum when the system is balanced.

Figure is a schematic drawing of a torque balance, a horizontal beam supported at a fulcrum (indicated by S) and three masses are attached to both sides of the fulcrum. Mass 3 is 30 cm to the right of S. Mass 2 is 40 cm to the left of S. Mass 1 is 30 cm to the left of Mass 2.
In a torque balance, a horizontal beam is supported at a fulcrum (indicated by S ) and masses are attached to both sides of the fulcrum. The system is in static equilibrium when the beam does not rotate. It is balanced when the beam remains level.

Strategy

For the arrangement shown in the figure, we identify the following five forces acting on the meter stick:

w 1 = m 1 g is the weight of mass m 1 ; w 2 = m 2 g is the weight of mass m 2 ;

w = m g is the weight of the entire meter stick; w 3 = m 3 g is the weight of unknown mass m 3 ;

F S is the normal reaction force at the support point S .

We choose a frame of reference where the direction of the y -axis is the direction of gravity, the direction of the x -axis is along the meter stick, and the axis of rotation (the z -axis) is perpendicular to the x -axis and passes through the support point S . In other words, we choose the pivot at the point where the meter stick touches the support. This is a natural choice for the pivot because this point does not move as the stick rotates. Now we are ready to set up the free-body diagram for the meter stick. We indicate the pivot and attach five vectors representing the five forces along the line representing the meter stick, locating the forces with respect to the pivot [link] . At this stage, we can identify the lever arms of the five forces given the information provided in the problem. For the three hanging masses, the problem is explicit about their locations along the stick, but the information about the location of the weight w is given implicitly. The key word here is “uniform.” We know from our previous studies that the CM of a uniform stick is located at its midpoint, so this is where we attach the weight w , at the 50-cm mark.

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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