Lesson Explainer: Equilibrium of a Rigid Body Mathematics

In this explainer, we will learn how to solve problems about the equilibrium of rigid bodies in 2D where the sum of forces and the sum of moments equal zero.

If a body is rigid, forces acting on it 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, then the body is in equilibrium.

Definition: 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.

Consider a uniform rod of length 2๐‘™, negligible thickness, and weight ๐‘Š. The rod is suspended at one end by a rope that applies a tension ๐‘‡ vertically upward. The magnitudes of ๐‘‡ and ๐‘Š are equal.

The rod is in translational equilibrium as the magnitudes of the vertical forces are equal to each other and no horizontal forces act on the rod.

The rod cannot be in rotational equilibrium, as can be seen by taking moments about the rodโ€™s center of mass and either of its ends.

Definition: The Moment of a Force About a Point

The moment of a force about a point ๐‘ƒ is the distance ๐‘‘ from ๐‘ƒ to the point where the force acts, multiplied by the component of the force perpendicular to the direction of the line intersecting ๐‘ƒ and the point where the force acts. This can be written as ๐œ=๐นโ‹…๐‘‘๐œƒ,sin where ๐น is the force and ๐œƒ is the angle between the direction of the force and the direction of the line intersecting ๐‘ƒ and the point where the force acts.

Consider points ๐ด, ๐ต, and ๐ถ on the rod.

Taking counterclockwise moments as positive, if moments are taken about ๐ต, then we find that the moments about ๐ต are given by netmoment=โˆ’๐‘‡๐‘™.

If moments are taken about ๐ด, then we find that the moments about ๐ด are given by netmoment=โˆ’๐‘Š๐‘™.

If moments are taken about ๐ถ, then we find that the moments about ๐ถ are given by netmoment=๐‘Š๐‘™โˆ’2๐‘‡๐‘™.

As ๐‘‡=๐‘Š, the net moment is the same whatever point moments are taken about.

Consider an identical rod suspended at both ends by identical ropes.

For the body to be in translational equilibrium, it must be the case that ๐‘‡=๐‘Š2.

Taking counterclockwise moments as positive, if moments are taken about ๐ต, then the moments about ๐ต are netmoment=๐‘‡๐‘™โˆ’๐‘‡๐‘™=0.

If moments are taken about ๐ด, then the moments about ๐ด are netmoment=2๏€ผ๐‘Š2๏ˆ๐‘™โˆ’๐‘Š๐‘™=0.

If moments are taken about ๐ถ, then the moments about ๐ถ are netmoment=๐‘Š๐‘™โˆ’2๏€ผ๐‘Š2๏ˆ๐‘™=0.

The net moment is zero whatever point moments are taken about. The rod is, therefore, in rotational equilibrium. It is important to note that we only need to take the moments about one point, and if the moments about this point sum to zero, then the moments about any point will sum to zero.

Forces acting on a body do not necessarily act horizontally or vertically, but the component of a force acting at an angle ๐œƒ from the direction of the force can be determined, as shown in the following figure.

Let us look at a series of examples of using the conditions for equilibrium of a rigid body to find the magnitudes and directions of forces that act on some rigid bodies.

Example 1: Solving a Static Equilibrium Problem by Equating Moments

In the given figure, determine the magnitude of the force ๐น that makes the rod in equilibrium, given that the magnitude of the given force is 7 N and cos๐œƒ=45.

Answer

It would initially seem that the rod cannot be in translational equilibrium, as both forces acting on it have downward vertical components. The figure does not actually show all the forces acting on the rod. The weight of the rod is not shown and neither is the reaction force from the surface at ๐ด. Since the rod is in translational equilibrium, the net vertical force acting on it must be zero, so the reaction force at ๐ด must be equal to the sum of the weight and the vertically downward components of the applied forces.

It would also initially seem that the rod cannot be in rotational equilibrium, because if moments were taken about either of the points where an applied force acts, there would only be one nonzero moment acting on the rod. This shows that another force must exist that is not included in the original figure. The missing force is the force due to friction between the base of the rod and the surface that the rod rests on. The force due to friction for an object in equilibrium is equal in magnitude to the net horizontal force acting on the rod, and it acts in the opposite direction to the net horizontal force on the rod.

The following figure shows the forces that should be included, where ๐‘Š is the weight of the rod, ๐‘… is the reaction at ๐ด, and ๐‘ is the force due to friction. Because these forces pass through point ๐ด, they produce zero moment about ๐ด.

Each of the applied forces has a horizontal component equal to the cosine of the angle from the horizontal at which the force acts. For the unknown force, this angle is not stated, but it is the co-angle of angle ๐œƒ and so it is equal to ๐œƒ.

Since the rod is in rotational equilibrium, the net moment on it must be zero. The counterclockwise moments on the rod must therefore be equal to the clockwise moments on it.

We notice that ๐‘Š, ๐‘…, and ๐‘ pass through point ๐ด. Therefore, if we take moments about ๐ด, we can eliminate those forces. The counterclockwise moments about ๐ด are 2.1๐น๐œƒcos and the clockwise moments about ๐ด on the rod are (2.1+4.7)ร—730cos. Equating the counterclockwise and clockwise moments gives 2.1๐น๐œƒ=(2.1+4.7)ร—730,2.1ร—๐นร—45=6.8ร—7ร—โˆš32,coscos which determines the magnitude of force ๐น, which is given by ๐น=85โˆš36.N

The magnitudes and directions of reaction forces on rigid bodies can be determined from the type of contact the body has with the source of the reaction forces. When an object is on an inclined surface, the direction of the reaction force on the object is perpendicular to the direction of the surface. This force is called the normal reaction. If an extended object that is itself inclined at some angle is supported at a point, then the direction of the reaction force on the object is perpendicular to the inclination of the object.

Example 2: Solving a Static Equilibrium Problem Involving Friction

A uniform rod ๐ด๐ต of weight 10 N and length 12.5 m is resting with its end ๐ด on a rough horizontal plane and point ๐ถ (between ๐ด and ๐ต) resting against a smooth horizontal nail, which is 5.7 m above the horizontal plane. If the rod is about to slide when it is inclined to the horizontal at an angle whose tangent is 34, determine the coefficient of friction between the rod and the horizontal plane.

Answer

The directions of the forces acting on the rod are shown in the following figure. Force ๐น is the force of friction.

The maximum possible value of frictional force is given by ๐น=๐œ‡๐‘…,max๏Œ  where ๐œ‡ is the coefficient of friction. Because the rod is about to slip, ๐น has its maximum value, so ๐œ‡=๐น๐‘….๏Œ 

The reaction force at ๐ถ acts perpendicularly to ๐ด๐ต, so the angle between ๐‘…๏Œข and the vertical is equal to the angle of ๐ด๐ต with the horizontal, as shown in the following figure.

๐‘…๏Œ  and ๐น cannot be directly determined, but knowing ๐‘…๏Œข allows ๐‘…๏Œ  to be determined, as for the rod to be in translational equilibrium vertically, the magnitude of the weight of the rod must equal the sum of the magnitudes of ๐‘…๏Œ  and the vertical component of ๐‘…๏Œข, so ๐‘…+๐‘…๐œƒ=10.๏Œ ๏Œขcos

Looking at our system of forces, we can see that if we take moments about point ๐ด, then we will eliminate two of our unknowns, ๐‘…๏Œ  and ๐น, since they both act through point ๐ด. This will enable us to find the other unknown, ๐‘…๏Œข.

It is worth noting that the distance to the center of mass of the rod along ๐ด๐ต is 12.52 m and the vertical distance of ๐ถ above ๐ด is 5.7 m.

The weight, ๐‘Š, acts vertically, so the perpendicular distance from ๐ด to the center of mass of the rod is ๐‘‘, the horizontal distance from ๐ด to the point vertically below the center of mass, as shown in the following figure.

We have that ๐‘‘=6.25๐œƒ.cos

The reaction ๐‘…๏Œข acts perpendicularly to ๐ด๐ต, so the perpendicular distance from ๐ด to ๐ถ is ๐ด๐ถ, as shown in the following figure.

So ๐ด๐ถ=5.7๐œƒ.sin

The rod is in equilibrium, so the counterclockwise moments can be equated to the clockwise moments: 6.25๐œƒร—10=๐‘…5.7๐œƒ.cossin๏Œข

Because tan๐œƒ=34, sin๐œƒ=35, and cos๐œƒ=45, this can be rewritten as 6.25ร—45ร—10=5.7ร—53๐‘…,๏Œข which shows that the magnitude of ๐‘…๏Œข is given by ๐‘…=10019.๏ŒขN

For the rod to be in vertical translational equilibrium, the magnitude of the weight must equal the sum of ๐‘…๏Œ  and the vertical component of ๐‘…๏Œข. The vertical component of ๐‘…๏Œข is ๐‘…๐œƒ๏Œขcos, so ๐‘…+45๐‘…=10,๏Œ ๏Œข and it follows that the magnitude of ๐‘…๏Œ  is given by ๐‘…=10โˆ’4510019=11019.๏Œ N

For the rod to be in translational equilibrium horizontally, the horizontal component of ๐‘…๏Œข must have the same magnitude as the frictional force, so ๐น=๐‘…๐œƒ๏Œขsin: ๐น=1001935=6019.N

As previously stated, the coefficient of friction ๐œ‡=๐น๐‘…๏Œ . The value of ๐œ‡ can then be found from ๐œ‡=๏€ป๏‡๏€ป๏‡,๏Šฌ๏Šฆ๏Šง๏Šฏ๏Šง๏Šง๏Šฆ๏Šง๏Šฏ which simplifies to ๐œ‡=611.

The reaction force on a rigid body in contact with a surface is easily determinable. If a rigid body is attached to a surface, however, the direction of the reaction force acting on the body is not necessarily perpendicular to the surface.

Example 3: Solving a Problem Involving a Hinged Rod in Static Equilibrium

A uniform rod that has a length of 128 cm and a weight of 10 N is attached from one of its ends to a hinge that is fixed on a vertical wall. A weight of 10 N is suspended from the rod at a point located 96 cm away from the hinge. The rod is kept in a horizontal position by a string which is attached to the opposite end of the rod from the hinge and fixed to a point on the wall directly above the hinge. Given that the string is inclined to the horizontal at an angle of 60โˆ˜, determine the tension ๐‘‡ in the string, the reaction of the hinge ๐‘…, and the angle ๐œƒ between the reactionโ€™s line of action and the horizontal ground rounded to the nearest minute.

Answer

The tension ๐‘‡ acts along the string. The tension has both horizontal and vertical components, so we can resolve it as follows: ๐‘‡=๐‘‡60=๐‘‡โˆš32verticalsin and ๐‘‡=๐‘‡60=๐‘‡2,horizontalcos as shown in the following figure.

In determining ๐‘‡, the reaction force can be ignored by taking moments about the point where it acts. The moments on the rod about the hinge can be determined from the following figure. Because the rod is uniform, the weight acts at the midpoint of the rod.

The horizontal component of ๐‘‡ acts along the line that passes through the hinge, which is where we are taking the moment, hence producing zero moment. Since the rod is in equilibrium, the counterclockwise and clockwise moments can be equated: 10(0.96)+10(0.64)=๐‘‡1.28๏€ฟโˆš32๏‹.

The left-hand-side terms simplify to 9.6+6.4=16. Dividing both sides of the equation by 1.28 gives 252=๐‘‡๏€ฟโˆš32๏‹.

The factor of โˆš32 can be removed by multiplying both sides of the equation by 2โˆš3 to give ๐‘‡=25โˆš3.

This can be simplified further to ๐‘‡=25โˆš3โˆš3โˆš3=25๏€ฟโˆš33๏‹, so the magnitude of ๐‘‡ is given by ๐‘‡=25๏€ฟโˆš33๏‹.N

When the magnitude of ๐‘‡ is determined, for the rod to be in equilibrium, the magnitudes of the vertical and horizontal components of the reaction force at the hinge must be such that the net horizontal and vertical forces on the rod are zero, as shown in the following figure.

The figure shows that ๐‘…=20โˆ’๐‘‡โˆš32vertical and ๐‘…=๐‘‡2horizontal.

The magnitude of the reaction and the angle from the horizontal at which it acts are determined by finding the length and angle ๐œƒ of the hypotenuse of a right triangle with opposite and adjacent side lengths equal to the vertical and horizontal components of the reaction, as shown in the following figure.

The length of the hypotenuse, โ„Ž, is found using the Pythagorean theorem: โ„Ž=๏€ฟ20โˆ’25โˆš3โˆš32๏‹+๏€ฟ252โˆš3๏‹โ„Ž=๏€ผ20โˆ’252๏ˆ+๏€ฟ252โˆš3๏‹โ„Ž=๏€ผ152๏ˆ+๏€ฟ252โˆš3๏‹โ„Ž=2254+62512=3253.๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ

Taking square roots of both sides of the equation gives โ„Ž=5โˆš393, meaning that the reaction force magnitude is ๐‘…=5โˆš393.N

The tangent of ๐œƒ is the fraction of ๐‘…v over ๐‘…h, so tan๐œƒ=๏€ป๏‡๏€ผ๏ˆ=15โˆš325,๏Šง๏Šซ๏Šจ๏Šจ๏Šซ๏Šจโˆš๏Šฉ and ๐œƒ=466โ€ฒ.โˆ˜

In our final example, let us examine a ladder resting against a smooth wall with its base on rough horizontal ground.

Example 4: Solving a Problem Involving a Ladder in Static Equilibrium

A uniform ladder is resting in a vertical plane with its upper end against a smooth vertical wall and its lower one on a rough horizontal floor, where the coefficient of friction between the ladder and the floor is 23. The ladder is inclined to the horizontal at an angle measuring 48โˆ˜. Given that the ladder weighs 295 N and has a length ๐ฟ, find, in terms of ๐ฟ, the maximum distance a man weighing 610 N can climb up the ladder without it slipping, rounding your answer to two decimal places.

Answer

With static equilibrium problems involving ladders, it is often sensible to start by drawing a force diagram using the information given in the question to help us visualize the situation.

In our diagram, ๐‘…๏Œถ is the reaction force perpendicular to the wall, ๐‘…๏Œฅ is the reaction force perpendicular to the floor, and ๐น๐‘Ÿ is the friction between the base of the ladder and the rough ground.

Now, recall that for an object to be in equilibrium, the sum of the horizontal forces and the sum of the vertical forces must be equal to zero and the sum of the clockwise and counterclockwise moments must be zero.

If we start by looking at the horizontal forces, we can see that ๐‘…=๐น๐‘Ÿ.๏Œถ

Recalling that ๐น๐‘Ÿ=๐œ‡๐‘…๏Œฅ, where ๐œ‡ is the coefficient of friction, we have that

๐‘…=๐œ‡๐‘….๏Œถ๏Œฅ(1)

Now, if we look at the vertical forces, we can see that ๐‘…=610+296=905.๏ŒฅN

Substituting this into equation (1), we have that ๐‘…=๐œ‡(905)=23ร—905=18103.๏ŒถN

As we noted earlier, for a body to be in equilibrium, the sum of any moments must be equal to zero, or equivalently, the clockwise moments must be equal to the counterclockwise moments. We can take moments about any point on the ladder, but it is often easiest to choose a point where multiple forces act, as this will simplify the calculation. At the base of the ladder, we have two forces acting, friction and the normal reaction force from the floor, so we will take moments from this point. Doing so allows us to form the following equation: ๐ฟ48โ‹…๐‘…=12๐ฟ48โ‹…295+๐‘ฅ48โ‹…610.sincoscos๏Œถ

Substituting for ๐‘…๏Œถ gives ๐ฟ48โ‹…18103=12๐ฟ48โ‹…295+๐‘ฅ48โ‹…610.sincoscos

Finally, we need to rearrange to make ๐‘ฅ the subject: ๐‘ฅ48โ‹…610=๐ฟ48โ‹…18103โˆ’12๐ฟ48โ‹…295๐‘ฅ=๐ฟ๏€ป48โ‹…โˆ’48โ‹…295๏‡48โ‹…610๐‘ฅ=๐ฟโ‹…0.85667โ€ฆ.cossincossincoscos๏Šง๏Šฎ๏Šง๏Šฆ๏Šฉ๏Šง๏Šจ

Rounding to two decimal places gives an answer of 0.86๐ฟ.

Let us finish by recalling some key points.

Key Points

  • A rigid body is in equilibrium if the sum of the forces and the sum of the moments on the body are zero.
  • Forces acting on rigid bodies other than applied forces are the weight of the body, reaction to the weight of the body, and friction between the body and rough surfaces that it is in contact with.
  • Taking the moments on a rigid body about a point can be useful in determining the magnitudes and directions of forces acting on the body that cannot be determined directly.

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