12.1 Conditions for static equilibrium  (Page 5/12)

Solution

This condition is trivially satisfied because when we substitute the data, [link] becomes $+0.52w-w+0.48w=0.$ The second equilibrium condition, [link] , reads

where ${\tau }_{\text{F}}$ is the torque of force $F_{\text{F}},{\tau }_w$ is the gravitational torque of force w , and ${\tau }_{\text{R}}$ is the torque of force $F_{\text{R}}.$ When the pivot is located at CM, the gravitational torque is identically zero because the lever arm of the weight with respect to an axis that passes through CM is zero. The lines of action of both normal reaction forces are perpendicular to their lever arms, so in [link] , we have $\left|\text{sin}\theta \right|=1$ for both forces. From the free-body diagram, we read that torque ${\tau }_{\text{F}}$ causes clockwise rotation about the pivot at CM, so its sense is negative; and torque ${\tau }_{\text{R}}$ causes counterclockwise rotation about the pivot at CM, so its sense is positive. With this information, we write the second equilibrium condition as

With the help of the free-body diagram, we identify the force magnitudes $F_{\text{R}}=0.48w$ and $F_{\text{F}}=0.52w,$ and their corresponding lever arms $r_{\text{R}}=x$ and $r_{\text{F}}=d-x.$ We can now write the second equilibrium condition, [link] , explicitly in terms of the unknown distance x :

Here the weight w cancels and we can solve the equation for the unknown position x of the CM. The answer is $x=0.52d=0.52(2.5\text{m})=1.3\text{m}\text{.}$

Solution

When we substitute the quantities indicated in the diagram, we obtain

The answer obtained by solving [link] is, again, $x=0.52d=1.3\text{m}.$

Significance

A special case of static equilibrium occurs when all external forces on an object act at or along the axis of rotation or when the spatial extension of the object can be disregarded. In such a case, the object can be effectively treated like a point mass. In this special case, we need not worry about the second equilibrium condition, [link] , because all torques are identically zero and the first equilibrium condition (for forces) is the only condition to be satisfied. The free-body diagram and problem-solving strategy for this special case were outlined in Newton’s Laws of Motion and Applications of Newton’s Laws . You will see a typical equilibrium situation involving only the first equilibrium condition in the next example.