![]() W = m g w = m g is the weight of the entire meter stick w 3 = m 3 g w 3 = m 3 g is the weight of unknown mass m 3 m 3 į S F S is the normal reaction force at the support point S. W 1 = m 1 g w 1 = m 1 g is the weight of mass m 1 m 1 w 2 = m 2 g w 2 = m 2 g is the weight of mass m 2 m 2 It is balanced when the beam remains level.įor the arrangement shown in the figure, we identify the following five forces acting on the meter stick: The system is in static equilibrium when the beam does not rotate. Here, the free-body diagram for an extended rigid body helps us identify external torques.įigure 12.9 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 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. This does not hold true in rotational dynamics, where an extended rigid body cannot be represented by one point alone. In translational dynamics, a body is represented as its CM, where all forces on the body are attached and no torques appear. 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). Without the correct setup and a correct diagram, you will not be able to write down correct conditions for equilibrium. ![]() Note that setting up a free-body diagram for a rigid-body equilibrium problem is the most important component in the solution process. Also, you may independently check for your numerical answers by shifting the pivot to a different location and solving the problem again, which is what we did in Example 12.1. ![]() If they do not, then use the previous steps to track back a mistake to its origin and correct it. Your final answers should have correct numerical values and correct physical units.
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