12.1 Conditions for static equilibrium (Page 6/12)

University physics volume 1 Page 6 / 12

View this demonstration to see three weights that are connected by strings over pulleys and tied together in a knot. You can experiment with the weights to see how they affect the equilibrium position of the knot and, at the same time, see the vector-diagram representation of the first equilibrium condition at work.

A breaking tension

A small pan of mass 42.0 g is supported by two strings, as shown in [link] . The maximum tension that the string can support is 2.80 N. Mass is added gradually to the pan until one of the strings snaps. Which string is it? How much mass must be added for this to occur?

Figure shows small pan of mass supported by two strings intersecting at a 90 degree angle. The length of one string is 5 centimeters, the length of another string is 10 centimeters. — Mass is added gradually to the pan until one of the strings snaps.

Strategy

This mechanical system consisting of strings, masses, and the pan is in static equilibrium. Specifically, the knot that ties the strings to the pan is in static equilibrium. The knot can be treated as a point; therefore, we need only the first equilibrium condition. The three forces pulling at the knot are the tension

{\vec{T}}_{1}

in the 5.0-cm string, the tension

{\vec{T}}_{2}

in the 10.0-cm string, and the weight

\vec{w}

of the pan holding the masses. We adopt a rectangular coordinate system with the y -axis pointing opposite to the direction of gravity and draw the free-body diagram for the knot (see [link] ). To find the tension components, we must identify the direction angles

α_{1}

and

α_{2}

that the strings make with the horizontal direction that is the x -axis. As you can see in [link] , the strings make two sides of a right triangle. We can use the Pythagorean theorem to solve this triangle, shown in [link] , and find the sine and cosine of the angles

α_{1}

and

α_{2} .

Then we can resolve the tensions into their rectangular components, substitute in the first condition for equilibrium ( [link] and [link] ), and solve for the tensions in the strings. The string with a greater tension will break first.

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. — Free-body diagram for the knot in [link] .

Solution

The weight w pulling on the knot is due to the mass M of the pan and mass m added to the pan, or

w = (M + m) g .

With the help of the free-body diagram in [link] , we can set up the equilibrium conditions for the knot:

\begin{matrix} in the x -direction, & − T_{1 x} + T_{2 x} & = & 0 \\ in the y -direction, & + T_{1 y} + T_{2 y} - w & = & 0 . \end{matrix}

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

\begin{matrix} T_{1 x} = T_{1} cos α_{1} = T_{1} / \sqrt{5}, & T_{1 y} = T_{1} sin α_{1} = 2 T_{1} / \sqrt{5} \\ T_{2 x} = T_{2} cos α_{2} = 2 T_{2} / \sqrt{5}, & T_{2 y} = T_{2} sin α_{2} = T_{2} / \sqrt{5} . \end{matrix}

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

\begin{matrix} in x -direction, & T_{1} & = & 2 T_{2} \\ in y -direction, & \frac{2 T_{1}}{\sqrt{5}} + \frac{T_{2}}{\sqrt{5}} & = & (M + m) g . \end{matrix}

The equilibrium equation for the x -direction tells us that the tension $T_{1}$ in the 5.0-cm string is twice the tension $T_{2}$ in the 10.0-cm string. Therefore, the shorter string will snap. When we use the first equation to eliminate $T_{2}$ from the second equation, we obtain the relation between the mass $m$ on the pan and the tension $T_{1}$ in the shorter string:

2.5 T_{1} / \sqrt{5} = (M + m) g .

The string breaks when the tension reaches the critical value of $T_{1} = 2.80 N .$ 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.

Significance

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.

Summary

A body is in equilibrium when it remains either in uniform motion (both translational and rotational) or at rest. When a body in a selected inertial frame of reference neither rotates nor moves in translational motion, we say the body is in static equilibrium in this frame of reference.
Conditions for equilibrium require that the sum of all external forces acting on the body is zero (first condition of equilibrium), and the sum of all external torques from external forces is zero (second condition of equilibrium). These two conditions must be simultaneously satisfied in equilibrium. If one of them is not satisfied, the body is not in equilibrium.
The free-body diagram for a body is a useful tool that allows us to count correctly all contributions from all external forces and torques acting on the body. Free-body diagrams for the equilibrium of an extended rigid body must indicate a pivot point and lever arms of acting forces with respect to the pivot.

Conceptual questions

What can you say about the velocity of a moving body that is in dynamic equilibrium?

constant

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12.1 Conditions for static equilibrium (Page 6/12)

A breaking tension

Strategy

Solution

Significance

Summary

Conceptual questions

Problems

Questions & Answers

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