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10.2: Conditions for Static Equilibrium

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  • Top figure shows the distribution of forces for the knot that ties the strings to the pan. T1 and T2 forces are pulling at the knot upward. Weight, a sum of M and m multiplied by g is pulling the knot downward. Projections of T1 and T2 at the x and y axes are shown. Bottom figure shows the representation of the knot that ties the strings to the pan as a right triangle. It has legs of the length a and 2a with a being equal 5 centimeters. Hypotenuse is a square root of five. Angle alpha 1 is formed by shorter leg and hypotenuse. Angle alpha 2 is formed by the longer leg and hypotenuse. Cosine of angle alpha 1 is equal to sine of angle alpha 2 and is equal to one divided by square root of five. Cosine of angle alpha 2 is equal to sine of angle alpha 1 and is equal to two divided by square root of five.
    Figure 12.8, we can set up the equilibrium conditions for the knot:

    in the x-direction, $$-T_{1x} + T_{2x} = 0$$in the y-direction, $$+T_{1y} + T_{2y} - w = 0 \ldotp$$

    From the free-body diagram, the magnitudes of components in these equations are

    $$\begin{split} T_{1x} & = T_{1} \cos \alpha_{1} = \frac{T_{1}}{\sqrt{5}},\quad T_{1y} = T_{1} \sin \alpha_{1} = \frac{2T_{1}}{\sqrt{5}} \\ T_{2x} & = T_{2} \cos \alpha_{2} = \frac{2T_{2}}{\sqrt{5}},\quad T_{2y} = T_{2} \sin \alpha_{2} = \frac{T_{2}}{\sqrt{5}} \ldotp \end{split}$$

    We substitute these components into the equilibrium conditions and simplify. We then obtain two equilibrium equations for the tensions:

    in x-direction, $$T_{1} = 2T_{2}$$in y-direction, $$\frac{2T_{1}}{\sqrt{5}} + \frac{T_{2}}{\sqrt{5}} = (M + m)g \ldotp$$

    The equilibrium equation for the x-direction tells us that the tension T1 in the 5.0-cm string is twice the tension T2 in the 10.0-cm string. Therefore, the shorter string will snap. When we use the first equation to eliminate T2 from the second equation, we obtain the relation between the mass m on the pan and the tension T1 in the shorter string:

    $$\frac{2.5T_{1}}{\sqrt{5}} = (M + m)g \ldotp$$

    The string breaks when the tension reaches the critical value of T1 = 2.80 N. The preceding equation can be solved for the critical mass m that breaks the string:

    $$m = \frac{2.5}{\sqrt{5}} \frac{T_{1}}{g} - M = \frac{2.5}{\sqrt{5}} \frac{2.80\; N}{9.8\; m/s^{2}} - 0.042\; kg = 0.277\; kg = 277.0\; g \ldotp$$


    Suppose that the mechanical system considered in this example is attached to a ceiling inside an elevator going up. As long as the elevator moves up at a constant speed, the result stays the same because the weight w does not change. If the elevator moves up with acceleration, the critical mass is smaller because the weight of M + m becomes larger by an apparent weight due to the acceleration of the elevator. Still, in all cases the shorter string breaks first.


    • Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).