Lesson Explainer: Equilibrium of a Rigid Body under Parallel Forces Mathematics

In this explainer, we will learn how to solve problems about the equilibrium of a body under the effect of parallel coplanar forces.

If a body is rigid, forces acting on the body cannot produce deformation. Forces have only two possible effects on the body. These effects are the linear acceleration of the body and the rotation of the body about a point.

If the forces exerted on a rigid body produce no net linear acceleration of the body, then the body is in translational equilibrium. The sum of the forces on the body must be zero for this to apply.

If the forces exerted on a rigid body produce no net rotation of the body, then the body is in rotational equilibrium. The sum of the moments on the body must be zero for this to apply.

If the sum of the forces and the sum of the moments on a rigid body are both zero, the body is in equilibrium.

Let us define the conditions for the equilibrium of a rigid body.

Definition: Conditions for the Equilibrium of a Rigid Body

A rigid body is in equilibrium if the sum of the forces and the sum of the moments on the body are zero.

Let us also define the moment due to a force.

Definition: Moment due to a Force

The moment of a force about a point 𝑃 is the distance 𝑑 from 𝑃 to the point where the force acts, multiplied by the component of the force perpendicular to the direction of the line intersecting 𝑃 and the point where the force acts. This can be written as 𝜏=𝐹𝑑𝜃,sin where 𝐹 is the force and 𝜃 is the angle between the direction of the force and the direction of the line intersecting 𝑃 and the point where the force acts.

This explainer specifically deals with examples where the forces that act on a rigid body are parallel and coplanar.

For a rigid body to be in equilibrium when parallel forces act on it, one of two conditions must apply.

The first condition is the trivial condition that the lines of action of all forces are parallel to the length of the body and intersecting the center of mass of the body. The line of action of such forces is shown in the following figure.

No moment is produced by any of the forces at any point along the length of the body that act along this line.

The second condition is that the lines of action of all forces must be perpendicular to the length of the body. A possible line of action of such forces is shown in the following figure.

The angle 𝜃 for forces along this line of action or for any line parallel to this line must be 90.

As sin(90)=1, the calculation of moments in these examples requires only the use of the formula 𝜏=𝐹𝑑.

Let us look at such an example.

Example 1: Finding the Reactions of the Supports of a Rod in Equilibrium

A uniform rod having a weight of 35 N is resting horizontally on two supports 𝐴 and 𝐵 at its ends, where the distance between the supports is 48 cm. If a weight of magnitude 24 N is suspended at a point that is 38 cm away from 𝐴, determine the reactions of the two supports 𝑅 and 𝑅.

Answer

There are two unknown reaction forces. The rod is at rest, so the sum of these forces must equal the sum of the weight of the rod and the weight suspended from the rod; hence, 𝑅+𝑅=35+24=59.NNN

The moments acting about either 𝐴, or 𝐵 can be determined. Let us take the moments about 𝐵.

The rod is in equilibrium, so the clockwise and counterclockwise moments about 𝐵 are equal.

The following figure shows the forces acting on the rod. The rod is uniform, so its weight acts at its midpoint. The distances in centimetres are converted to distances in metres.

The only clockwise moment about 𝐵 is due to 𝑅. The counterclockwise moments about 𝐵 are due to the weight of the rod and the weight suspended from the rod.

Equating the moments about 𝐵 gives (0.24×35)+((0.480.38)×24)=0.48×𝑅.

This can be simplified as follows: 𝑅=(0.24×35)+((0.480.38)×24)0.48𝑅=8.4+2.40.48=22.5.N

As previously stated, 𝑅+𝑅=35+24=59,NNN and so, 𝑅 is given by 𝑅=59𝑅𝑅=36.5.N

Let us look at another such example, this time involving a nonuniform body.

Example 2: Finding the Point where the Weight Acts on a Nonuniform Rod by Finding Resultant Forces and Taking Moments

A nonuniform rod 𝐴𝐵 having a weight of 40 N and a length of 80 cm is suspended vertically from its midpoint by a light string, and it becomes in equilibrium in a horizontal position when a weight of magnitude 29 N is suspended from its end 𝐴. Determine the distance 𝑥 between the point at which the weight of the rod is acting and end 𝐴. After removing the weight at 𝐴, determine the magnitude of the vertical force that would be needed to keep the rod in equilibrium in a horizontal position when it acts at end 𝐵.

Answer

In this example, only the vertically upward force exerted by the string is unknown. The string is light, meaning it has negligible weight, so downward force due to the weight of the string is negligible.

The upward force exerted on the rod is equal to the sum of the downward forces. Therefore, the tension in the string is given by 𝑇=40+29=69.NNN

The rod is nonuniform and so, the position of the center of mass of the rod is unknown other than that it is not at the midpoint of the rod.

When a weight is suspended from the rod at 𝐴, the rod is in equilibrium. If the clockwise and counterclockwise moments are taken about a point 𝑃, where the string suspends the rod, the force due to the string produces zero moments about 𝑃. The only moments on the rod about 𝑃 are due to the weight suspended at 𝐴 and the weight of the rod acting at the center of mass of the rod. As the rod is in equilibrium, these moments must be equal.

We are given that the rod is 80 cm long, so the distance from 𝑃 to 𝐴 is 40 cm. Therefore, the moments about 𝑃 are given by 29×40=40𝑑, where 𝑑 is the distance from 𝑃 to the center of mass of the rod. Solving the equation for 𝑑 gives 𝑑=29×4040=29.cm

The center of mass is 29 cm from the midpoint of the rod. The midpoint is 40 cm from 𝐴, so the distance from 𝐴 to the center of mass of the rod is 29+40=69.cm

When the weight suspended at 𝐴 is removed, the rod ceases to be in equilibrium until a vertical force 𝐹 acts at 𝐵.

Taking moments about 𝑃 again, we note that the moment due to the weight of the rod that is to the right of 𝑃 acts clockwise. The force acting at 𝐵 is further to the right of 𝑃, so the force acting at 𝐵 must act in the opposite direction to the weight of the rod.

The moments about 𝑃 are now given by 40×29=𝐹×40.Ncmcm

Solving the equation for 𝐹 gives 𝐹=40×2940=29.N

It is of some interest to note that in the previous example, it was not necessary to consider the translational equilibrium conditions for the rod to determine the unknown values to be found, only the rotational equilibrium conditions for the rod.

Let us look at another example involving the tensions of strings supporting a body. This example includes more forces acting on a rod than in the previous examples but is solvable in the same way.

Example 3: Finding the Tensions in an Equilibrium Problem

A uniform rod 𝐴𝐵 weighs 70 N and has a length of 95 cm. It is suspended from its ends by two vertical strings, where 𝑇 is the tension of the string at 𝐴 and 𝑇 is the tension of the string at 𝐵. A weight of 100 N is suspended from the rod, 30 cm away from 𝐴, and a weight of 93 N is suspended from the rod, 20 cm away from 𝐵. Determine the values of 𝑇 and 𝑇.

Answer

There are two unknown tension forces. The rod is at rest, so the sum of these forces must equal the sum of the weight of the rod and the two weights suspended from the rod; hence, 𝑇+𝑇=70+100+93=263.NNNN

The moments acting about either 𝐴 or 𝐵 can be determined. Let us take the moments about 𝐴.

The rod is in equilibrium, so the clockwise and counterclockwise moments about 𝐴 are equal.

Because 𝑇 acts at 𝐴, it produces zero moments about 𝐴. 𝑇 acts upward and all the other forces on the rod act downward, so the moment due to 𝑇 equals the sum of the remaining moments, 95×𝑇=(30×100)+952×70+((9520)×93)95×𝑇=3000+3325+6975=13300𝑇=1330095=140.N

As previously stated, 𝑇+𝑇=263,N so 𝑇=263140=123.N

Some forces that act on bodies are not usually considered to be variable. The weight of a body is not usually considered to vary, for example. Technically, if a body moved a significant distance away from Earth, then it would vary in weight. Also, if a body fragmented, then the weight of the fragments removed from the body would be subtracted from the weight of the body. These kinds of situations are not considered in this explainer.

When multiple reaction forces or tension forces act on an extended rigid body, however, the magnitudes of these forces can change only because of the position of the point on the body where an applied force acts on it.

Consider the forces acting on the uniform body in the following figure.

Two reaction forces act on the body. For the body to be in translational equilibrium, the sum of the reaction forces must be the sum of the weight of the body 𝑊 and the downward applied force 𝐹. This must be true wherever 𝐹 acts from.

For the body to be in rotational equilibrium, unless 𝐹 acts at the center of mass of the body, 𝑅 must be unequal to 𝑅. If the point at which 𝐹 acts is varied, 𝑅 and 𝑅 will both vary. The greater the distance of 𝐹 from 𝐴, the less the magnitude of 𝑅 must be and the greater the magnitude of 𝑅 must be for the body to be in rotational equilibrium.

By varying the position of the point at which a force acts on a body, and comparing the moments about a point on the body when the force acts on it at different points, it is possible to determine the position of the point at which the weight of the body acts and also the magnitude of the weight.

Let us now look at an example where an applied force on a body acts at different points.

Example 4: Balanced Moments

𝐴𝐵 is a nonuniform rod having a length of 77 cm resting in a horizontal position on one support, which is 26 cm away from the end 𝐴. It is kept in equilibrium by suspending a weight of 16 N at its end 𝐴 and a weight of 2 N at the end 𝐵. If the distance between the support and the end 𝐴 is changed to be 23 cm, the rod will be kept in equilibrium by suspending a weight of 18 N at the end 𝐴 only. Find the magnitude of the weight of the rod 𝑊 and the distance 𝑥 between its line of action and point 𝐴. Round your answers to the nearest two decimal places.

Answer

What is described by the question can be succinctly represented by the following figure, which shows the two arrangements of the rod. The rod is in equilibrium in both arrangements.

At the point where the reaction force acts, the reaction is given by 𝑅=18+𝑊, but 𝑊 is not known, and neither is 𝑥 These two unknowns could not be uniquely determined by considering either of the arrangements of the rod in isolation, only by comparing both arrangements.

For both arrangements, the rod is in equilibrium, and so the net moment about any point on the rod is zero for both arrangements. Let us consider the net moment about 𝐴 for both arrangements.

For the arrangement with forces acting at both 𝐴 and 𝐵, the moments about 𝐴 are given by (26(18+𝑊))𝑥𝑊2(77)=0. This can be rearranged as 468+26𝑊=𝑥𝑊+154314=𝑥𝑊26𝑊.

For the arrangement with a force acting only at 𝐴, the moments about 𝐴 are given by (23(18+𝑊))𝑥𝑊=0.

This can be rearranged as 414=𝑥𝑊23𝑊.

As the net moments about 𝐴 for both arrangements are equal to zero, they must also be equal to each other, so 314+26𝑊𝑥𝑊=414+23𝑊𝑥𝑊.

This can be rearranged to make 𝑊 the subject of the equation as follows.

Subtract 𝑥𝑊 from both sides of the equation to give 314+26𝑊=414+23𝑊.

Subtract 314 from both sides of the equation to give 26𝑊=100+23𝑊.

Subtract 23𝑊 from both sides of the equation to give 3𝑊=100𝑊=1003.N

To two decimal places, 𝑊=33.33.N

The question states that, for the arrangement with a force acting only at 𝐴. The distance between where the weight of the rod acts and end 𝐴 is 𝑥 in both configurations. To determine 𝑥, moments can be taken about the point where the reaction force acts. The distance between this point and the point where the weight of the rod acts is 𝑑, given by (23×18)=𝑑1003414=𝑑10034143100=𝑑=12.42.cm

The distance from 𝐴 to the point where the weight of the rod acts is given by 𝑥=23+12.42=35.42.cm

Key Points

  • Parallel forces acting on a rigid body in equilibrium must either act parallel to the length of the body, intersecting its center of mass, or act perpendicularly to this direction.
  • The net force and net moment due to the parallel forces must be zero for a body to be in equilibrium.
  • The values of multiple tension or reaction forces acting on a rigid body in equilibrium are dependent on their respective distances from the center of mass of the body.

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