Solving for Equilibrium

When a force is applied on a body, it may change the state of motion of the body. Motions created by the applied force can be translational or rotational, but not all forces cause translational or rotational motion. You may like understanding the concept of equilibrium.

Balanced Force:

Two or more forces are said to be balanced if the net force of the all the forces is equal to zero. Balanced forces do not cause any motion in the object they are acting on. \(_\square\)

Unbalanced Force:

Two or more forces are said to be unbalanced if the net force of the all the forces is equal to not equal to zero. Unbalanced forces always cause motion in the object they are acting on. \(_\square\)

A rigid body is one that does not deform under the application of an external force. For a rigid body to be in a state of equilibrium, the net force on the object must be zero. That is, if an object remains motionless, then Newton's second law, \(F=ma\), tells us that the total external force acting on the object is zero. The total external force on an object is also zero, if the body moves with a constant nonzero velocity.

The whole goal when solving for equilibrium is to find out what the various forces have to do so that there is zero net force on each body \(\left(\sum { F=0 }\right). \) Since force is a vector, we have to take into consideration the direction in which the force acts.

Consider an object that is on top of a frictionless table. From the figure below what single force should be acted so that the object is in equilibrium? What will be the direction of the force?

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For convention let the right direction be positive, and left direction negative. Also, let the force required be \(X.\) Now balance all the forces:

\[\begin{align} \sum {F} &=0 \\ X+4\text{ N}+3.5\text{ N}-3\text{ N}-1.5\text{ N}&=0\\ \Rightarrow X&=4.5\text{ N}-7.5\text{ N}\\ &=-3\text{ N}. \end{align}\]

Since the negative sign means left direction, a force of \(3N\) must act to the left so that the object is in equilibrium. \(_\square\)

A common force when solving for equilibrium is weight. Weight is the direct application of Newton's law of gravitation and for an object near the surface of the earth it is given by

\[W=mg,\]

where \(m\) is the mass of the body, and \(g\) is the gravitational acceleration and is equal to approximately \(9.8\text{ m/s}^2\). Note that weight always acts downwards toward the center of the earth.

Another common force when solving for equilibrium is tension. Tension is the general name for the pulling force of a rope, stick, cable, etc. Tension has a unit of force and can be measured in newtons.

If a block of mass \(3\text{ kg}\) is hanging from a ceiling by a massless rope, what will be the tension on the rope?

Take \(g\) to be \(9.8\text{ m/s}^{2}\).

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\[\] We know that the block is acted upon by a gravitational force equal to \(mg\), and since the object is motionless there must be another force to balance it. That force is tension, which we can get as follows:

\[\begin{align} \sum F&=0 \\ T-mg&=0\\ \Rightarrow T&=(3\text{ kg})\left(9.8\text{ m/s}^{2}\right)\\ &=29.4\text{ N}. \ _\square \end{align}\]

A stick-man who has a mere mass of \(3\text{ kg}\) is sitting on a swing and is at rest, as shown in the figure below. If the platform he is sitting upon has a mass of \(5\text{ kg},\) what will be the tension in each individual rope that holds the swing?

Assume that the weight is equally divided by the two ropes.

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Consider the system of the stick-man and the platform. Since the mass is equally distributed, let the tension in each rope be \(T.\)

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Then since the system is in equilibrium, we have

\[\begin{align} \sum F&=0 \\ T+T-(5\text{ kg})(g)-(3\text{ kg})(g)&=0\\ 2T&=8g\cdot\text{ kg}\\ \Rightarrow T&=4g\cdot \text{ kg}\\ &\approx4\times 9.8 \text{ m/s}^2 \cdot \text{ kg}\\ &= 39.2\text{ N}. \ _\square \end{align}\]

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