Lesson Explainer: Equivalent System of Forces to a Couple Mathematics

In this explainer, we will learn how to identify the conditions for a system of coplanar forces to be equivalent to a couple and find its moment.

Let us recall the definition of a force couple.

Definition: Force Couple

A pair of force vectors form a force couple if the following conditions are met:

  • The force vectors are parallel and opposite.
  • The force vectors lie on distinct lines of action.
  • The force vectors have equal magnitudes.

A force couple acting on a rigid body with a reference point causes the rigid body to rotate around the reference point, which is also called the axis of rotation. We can observe this effect in the following diagram.

We can see that the forces āƒ‘š¹ļŠ§ and āƒ‘š¹ļŠØ are parallel, opposite with the same magnitude, and lying on distinct lines of action. Hence, this pair of forces form a force couple. We also note that, with respect to a stationary reference point š‘‚, this force couple would tend to cause a clockwise rotation on a rigid body.

When a system of forces acts on a rigid body, its effect may cause the body to move along a direction and also to rotate with respect to a moving axis of revolution. However, if the net force is equal to zero, then the rigid body will only spin without shifting, and the axis of revolution will be stationary. A system of forces which results in a purely rotational motion is equivalent to a force couple.

Definition: Equivalent System of Forces to a Couple

A system of forces is equivalent to a couple if the net force is equal to zero.

In our first example, we will identify unknown constants in a force system which is equivalent to a couple.

Example 1: Finding Unknown Forces Acting on a Square to Produce an Equivalent Couple

š“šµš¶š· is a square, where the five forces, measured in newtons, are acting on it as shown in the figure. If the system of forces is equivalent to a couple, determine š¹ļŠ§ and š¹ļŠØ.

Answer

We recall that a system of forces is equivalent to a couple if the net force is equal to zero. Since the given system is equivalent to a couple, both the horizontal and vertical components of forces in this system must add up to zero.

Before we can sum the components of the forces, we need to decide on the orientation. In line with the common convention of using the Cartesian coordinate system, let us define the positive horizontal direction to be the right and the positive vertical direction to be upward.

The vertical forces š¹ļŠØ and āˆ’20 do not have any horizontal component. On the other hand, we have the purely horizontal forces āˆ’š¹ļŠ§ and 13 and the diagonal force with magnitude 9āˆš2 N which has both horizontal and vertical components. Let us compute the horizontal and vertical components of the diagonal force using right triangle trigonometry.

Since ļƒ š“š¶ is a diagonal of a square, it bisects the right angle āˆ š·š“šµ. Hence, š‘šāˆ šµš“š¶=45āˆ˜. Let us denote the horizontal component of this force by š¹ļ‚ and the vertical component to be š¹ļ“. Then, we can draw the following right triangle.

The horizontal component š¹ļ‚ forms the adjacent side of the right triangle with respect to the angle 45āˆ˜. Using right triangle trigonometry, we can write cosadjacenthypotenuse45==š¹9āˆš2.āˆ˜ļ‚

Hence, |š¹|=9āˆš245=9āˆš2Ɨāˆš22=9.ļ‚āˆ˜cos

Since š¹ļ‚ is oriented to the right, it should be positive. Hence, š¹=9ļ‚N. Furthermore, since a right triangle with 45āˆ˜ as one of its angles is isosceles, the vertical component š¹ļ“ must be equal in magnitude to the horizontal component |š¹|=|š¹|=9ļ“ļ‚N. Since š¹ļ“ is oriented upward, its sign is positive. Therefore, we obtain š¹=9ļ“N.

We are ready to compute the net horizontal and vertical forces. The net horizontal force is given by āˆ’š¹+13+9=āˆ’š¹+22.ļŠ§ļŠ§

Since the horizontal net force equals zero, we obtain āˆ’š¹+22=0ļŠ§, which leads to š¹=22ļŠ§N.

Next, we compute the net vertical force: š¹āˆ’20+9=š¹āˆ’11.ļŠØļŠØ

Since the vertical net force equals zero, we obtain š¹āˆ’11=0ļŠØ, which leads to š¹=11ļŠØN.

Thus, if the given system of forces is equivalent to a couple, then š¹=22ļŠ§N and š¹=11ļŠØN.

In the previous example, we identified unknown forces in a system of forces equivalent to a couple. If a system of forces is equivalent to a couple, then the system of forces generates a rotational moment, or torque. Let us recall how to compute the moment generated by a force with respect to an axis of rotation.

Definition: Moments

Let š‘‚ be a reference point (or an axis of rotation) and āƒ‘š¹ be a force acting on a rigid body with this reference point. If š‘‘ is the perpendicular distance from the reference point š‘‚ to the line of action of the force āƒ‘š¹, then the magnitude of the moment generated by the force āƒ‘š¹ with respect to š‘‚ is given by ||=ā€–ā€–āƒ‘š¹ā€–ā€–š‘‘.moment

The sign of the moment can be determined when we define the positive orientation. By convention, we often take the counterclockwise moment to be positive, meaning the clockwise moment is negative. However, the opposite of this convention may be specified in a problem.

The net moment of a system of forces is obtained by adding up the moment of each force in the system. In this case, it is important to determine the sign of each moment before we find the sum.

In our next example, we will compute the net moment generated by a system of forces equivalent to a couple acting on a light rod, where we take the axis of rotation to be the midpoint of the rod.

Example 2: Computing the Net Moment of a System of Forces Equivalent to a Couple acting on a Light Rod

In the given figure, if the forces that act on a light rod š“šµ are equivalent to a couple, find the moment of this couple.

Answer

We recall that a system of forces is equivalent to a couple if the net force is equal to zero. In the given diagram, we can easily observe that the net vertical force is equal to zero since 2+3āˆ’5=0.N

Since this system is equivalent to a force couple, it will generate a rotational moment. Let us compute the net moment generated by this system. We are not given the axis of rotation for this problem, so let us take the midpoint š¶ of the light rod to be the axis of rotation, or the reference point.

We recall that the magnitude of the moment generated by a force with magnitude š¹ is |š‘€|=š¹š‘‘, where š‘‘ is the perpendicular distance between the reference point and the line of action for the force.

We also remember that a moment is a signed quantity, where the sign of the moment can be determined once we decide on a positive orientation for moments. Unless otherwise specified, we take the counterclockwise moment to be positive in planar motions. So, let us use this convention.

From the diagram, we can see that the line of action for the force of magnitude 5 N contains the reference point š¶. Hence, the perpendicular distance š‘‘ for this force is 0. From the formula of the moment, we see that the moment š‘€ļŒ¢ generated by this force is also equal to zero. Hence, š‘€=0ā‹….ļŒ¢Ncm

For the force of magnitude 2 N, the perpendicular distance from the reference point š¶ is 8 cm. Then, |š‘€|=2Ɨ8=16.ļŒ 

From the diagram, we can see that this force generates a clockwise rotational moment around the axis š¶. By our convention, the sign of this moment must be negative. Thus, we have š‘€=āˆ’16ā‹…ļŒ Ncm.

For the force of magnitude 3 N, the perpendicular distance from the reference point š¶ is also 8 cm. Then, |š‘€|=3Ɨ8=24.ļŒ¢

We can note from the diagram that this force generates a counterclockwise moment around the axis š¶. Hence, the sign of this moment is positive: š‘€=24ā‹…ļŒ¢Ncm.

Adding up all three moments, we obtain the net moment š‘€ of this system: š‘€=š‘€+š‘€+š‘€=āˆ’16+0+24=8.ļŒ ļŒ”ļŒ¢

The moment of this system is 8 Nā‹…cm.

The perpendicular distance between the reference point and the line of action of a force is used to determine the moment. In other words, the location of the reference point with respect to the line of action is important when computing the moment in general force systems. However, in a force couple, or a system equivalent to a force couple, we will see that the location of reference point does not affect the net moment generated by the system.

Let us consider why the location of the axis of rotation does not affect the moment generated by a force couple. First, we consider a force couple and a reference point between the two parallel lines of action.

In the diagram above, š‘‘ļŠ§ and š‘‘ļŠØ are the perpendicular distances from the axis of rotation š‘‚ to the line of action for the forces āƒ‘š¹ļŠ§ and āƒ‘š¹ļŠØ respectively. Let š‘€ļŠ§ and š‘€ļŠØ be the moments of āƒ‘š¹ļŠ§ and āƒ‘š¹ļŠØ respectively. Then, |š‘€|=ā€–ā€–āƒ‘š¹ā€–ā€–š‘‘,|š‘€|=ā€–ā€–āƒ‘š¹ā€–ā€–š‘‘.ļŠ§ļŠ§ļŠ§ļŠØļŠØļŠØ

Since both forces induce a clockwise rotation around the axis of rotation š‘‚, we take the signs of both moments to be negative according to our convention. Also, since this pair of forces form a couple, their magnitudes must be equal. Let us denote š¹ to be the magnitude of both forces. By summing the moments, we obtain the net moment š‘€: š‘€=š‘€+š‘€=āˆ’ā€–ā€–āƒ‘š¹ā€–ā€–š‘‘āˆ’ā€–ā€–āƒ‘š¹ā€–ā€–š‘‘=āˆ’š¹š‘‘āˆ’š¹š‘‘=āˆ’š¹(š‘‘+š‘‘)=āˆ’š¹š‘‘.ļŠ§ļŠØļŠ§ļŠ§ļŠØļŠØļŠ§ļŠØļŠ§ļŠØ

We note that the last equality is possible by observing that the sum of the perpendicular distances š‘‘ļŠ§ and š‘‘ļŠØ is in fact equal to the distance, š‘‘, between two parallel lines of action for the two forces. Since this quantity š‘‘ remains constant no matter where the axis š‘‚ is located, the moment remains unchanged for different reference points as long as it lies between these two parallel lines of action.

Next, let us consider the action of a force couple on an axis of revolution as diagrammed below.

In this case, the magnitudes of the moments induced by the forces āƒ‘š¹ļŠ§ and āƒ‘š¹ļŠØ with respect to the axis of rotation š‘‚ have the same expression as before. Namely, we have |š‘€|=ā€–ā€–āƒ‘š¹ā€–ā€–š‘‘,|š‘€|=ā€–ā€–āƒ‘š¹ā€–ā€–š‘‘.ļŠ§ļŠ§ļŠ§ļŠØļŠØļŠØ

Let us observe the orientation of these moments. We can see from the diagram that the force āƒ‘š¹ļŠ§ would cause a counterclockwise rotation around the axis š‘‚; hence, š‘€ļŠ§ is positive. On the other hand, the force āƒ‘š¹ļŠØ would induce a clockwise moment about š‘‚, which results in a negative moment š‘€ļŠØ. Denoting once again the magnitudes of both forces by š¹, the net moment š‘€ is given by š‘€=š‘€+š‘€=ā€–ā€–āƒ‘š¹ā€–ā€–š‘‘āˆ’ā€–ā€–āƒ‘š¹ā€–ā€–š‘‘=š¹š‘‘āˆ’š¹š‘‘=š¹(š‘‘āˆ’š‘‘)=āˆ’š¹(š‘‘āˆ’š‘‘)=āˆ’š¹š‘‘.ļŠ§ļŠØļŠ§ļŠ§ļŠØļŠØļŠ§ļŠØļŠ§ļŠØļŠØļŠ§

We note that the last equality is because the difference of the perpendicular distances š‘‘ļŠØ and š‘‘ļŠ§ is equal to the distance š‘‘ between the two parallel lines of action. We note that this results in the same expression for the net moment as before. Hence, the net moment does not change in a force couple when we change the location of the axis of revolution. When a system of forces is equivalent to a force couple, the system inherits this property by its equivalence.

Theorem: Axis of Rotation in a System of Forces Equivalent to a Couple

Let š‘‚ and š‘‚ā€² be two distinct reference points. Then the net moment with respect to š‘‚ of a system of forces equivalent to a couple is equal to its net moment with respect to š‘‚ā€².

In our next example, we will use this property of a system of forces equivalent to a couple to compute the magnitude of its net moment.

Example 3: Computing the Net Moment in a System of Forces Equivalent to a Couple

š“šµš¶š· is a rectangle, in which š“šµ=45cm, šµš¶=55cm, and š·šø=28cm. Forces of magnitudes 225, 275, 265, and 135 newtons act along ļƒ š“šµ, ļƒŸšµš¶, ļƒŸš¶šø, and ļƒ šøš“ respectively. If the system of forces is equivalent to a couple, determine the magnitude of the moment of the forces.

Answer

Let us recall the formula for the moment induced by a force with respect to an axis of rotation. If š‘‘ is the perpendicular distance between the axis and the line of action for the force with magnitude š¹, then the magnitude of its moment š‘€ is given by |š‘€|=š¹š‘‘.

We also remember that a moment is a signed quantity and the sign of a moment can be determined once we set a positive orientation for moments. As it is a common convention in a planar motion, let us define counterclockwise moments to be positive in this example.

We recall that, if a system of forces is equivalent to a couple, then the net moment induced by the force system is the same no matter where the reference point is located. By observing the formula for a moment stated above, we can see that a moment of a force is equal to zero when the reference point lies on the line of action, since the perpendicular distance vanishes in this case.

Hence, the best point to place a reference point is at a point of intersection of multiple lines of action. This leaves us with the choices of vertices š“, šµ, or š¶ or the point šø. Point š¶ would be the optimal choice here, since the perpendicular distances for the two forces not passing through the axis are equal to the lengths of the corresponding sides of the rectangle. We place the axis of rotation at the vertex š¶ and label the lengths of the sides of the rectangle.

Since the axis is on the line of action for the forces ļƒŸš¶šø and ļƒŸšµš¶, we know that these forces do not contribute to the net moment about the axis š¶. Hence, š‘€=0,š‘€=0.ļƒ ļŒ¢ļŒ¤ļƒ ļŒ”ļŒ¢

For ļƒ šøš“, the perpendicular distance is 45 cm and the magnitude of the force is 135 N. We can see that the force ļƒ šøš“ induces a clockwise moment about š¶, so the moment is negative by our convention defined earlier. Hence, the moment from this force is š‘€=āˆ’135Ɨ45=āˆ’6075ā‹….ļƒ ļŒ¤ļŒ Ncm

For ļƒ š“šµ, the perpendicular distance is 55 cm and the magnitude of the force is 225 N. We can see that the force ļƒ š“šµ also induces a clockwise moment about š¶, so the moment is negative. Hence, the moment from this force is š‘€=āˆ’225Ɨ55=āˆ’12375ā‹….ļƒ ļŒ ļŒ”Ncm

Adding up these moments, we obtain the net moment š‘€ induced by this force system acting on the axis š¶: š‘€=š‘€+š‘€+š‘€+š‘€=0+0āˆ’6075āˆ’12375=āˆ’18450.ļƒ ļŒ¢ļŒ¤ļƒ ļŒ”ļŒ¢ļƒ ļŒ¤ļŒ ļƒ ļŒ ļŒ”

Then, the magnitude of the moment of these forces is 18ā€Žā€‰ā€Ž450 Nā‹…cm.

Let us consider another example of computing the net moment generated by a system of forces equivalent to a couple.

Example 4: Computing the Net Moment in a System of Forces Equivalent to a Couple

š“šµš¶š· is a square having a side length of 50 cm. Forces of magnitudes 30, 60, 160, and 10 newtons are acting at ļƒ š“šµ, ļƒŸšµš¶, ļƒ š¶š·, and ļƒ š·š“, respectively, while two forces of magnitudes 40āˆš2 and 90āˆš2 newtons are acting at ļƒ š“š¶ and ļƒ š·šµ respectively. If the system is equivalent to a couple, find its moment considering the positive direction is š·š¶šµš“.

Answer

We begin by drawing a diagram of the given force system.

This example defines the positive direction to be š·š¶šµš“, which is counterclockwise. We recall that, if a system of forces is equivalent to a couple, then the net moment is the same regardless of the location of the axis of rotation. Given an axis of rotation, the magnitude of a moment induced by a force is given by |š‘€|=š¹š‘‘, where š¹ is the magnitude of the force and š‘‘ is the perpendicular distance between the axis and the line of action. In particular, a force in the system does not contribute to the net moment if its line of action contains the axis of rotation.

For this reason, we need to begin by finding the ideal location to place the axis of rotation to make the remaining computations simple. We should pick a point of intersections of multiple lines of action so that we can keep these forces from contributing to the net moment. This leads to the vertices š“, šµ, š¶, and š· and the intersection point of the two diagonals. While choosing one of the four vertices of the square has the benefit of removing three forces whose lines of action intersect at that vertex, it is simpler to pick the intersection of the diagonals because this choice simplifies the computation of the perpendicular distances greatly. Let us draw a new diagram with this axis, labeled š‘‚.

Since the axis š‘‚ lies on the lines of action for the diagonal forces ļƒ š“š¶ and ļƒ š·šµ, the moments induced by these forces are equal to zero. That is, š‘€=0,š‘€=0.ļƒ ļŒ ļŒ¢ļƒ ļŒ£ļŒ”

From the diagram, we note that the perpendicular distances from š‘‚ to the lines of action for the forces ļƒ š“šµ, ļƒŸšµš¶, ļƒ š¶š·, and ļƒ š·š“ are equal to the half of the length of the sides of this square. Hence, š‘‘=502=25cm is the perpendicular distance from the axis of rotation to the four forces on the perimeter of the square. Hence, the magnitudes of these moments are calculated as follows: ||š‘€||=30Ɨ25=750,||š‘€||=60Ɨ25=1500,||š‘€||=160Ɨ25=4000,||š‘€||=10Ɨ25=250.ļƒ ļŒ ļŒ”ļƒ ļŒ”ļŒ¢ļƒ ļŒ¢ļŒ£ļƒ ļŒ£ļŒ 

From the diagram, we can see that each of these forces induces a clockwise rotation. Therefore, the signs of these moments are negative. Adding up these moments, we obtain the net moment š‘€: š‘€=š‘€+š‘€+š‘€+š‘€=āˆ’750āˆ’1500āˆ’4000āˆ’250=āˆ’6500.ļƒ ļŒ ļŒ”ļƒ ļŒ”ļŒ¢ļƒ ļŒ¢ļŒ£ļƒ ļŒ£ļŒ 

Hence, the moment of the given system is āˆ’6500 Nā‹…cm.

In our final example, we will identify unknown forces in a system of forces equivalent to a couple acting on a light rod from a given moment.

Example 5: Identifying Unknown Forces Acting from Moment from a System of Forces Equivalent to a Couple

In the given figure, if the forces that act on a light rod š“šµ are equivalent to a couple and the moment of this couple is equal to 17, find āƒ‘š¹ļŠ§ and āƒ‘š¹ļŠØ

Answer

We recall that a system of forces is equivalent to a couple if the net force is equal to zero. From the given diagram, we can see that all forces in this system are vertical. Hence, the vertical net force must equal zero. Defining the upward direction to be positive, we obtain the net force to be (āˆ’5+2āˆ’š¹+š¹)ļŠØļŠ§ N, which simplifies to (āˆ’3āˆ’š¹+š¹)ļŠØļŠ§ N. Setting this expression equal to zero, we obtain

š¹āˆ’š¹=3.ļŠ§ļŠØ(1)

Next, let us consider the moment. We are given that the moment of this system is 17 Nā‹…cm. We recall that if a system of forces is equivalent to a couple, then the net moment remains the same no matter where the axis of rotation is located.

We recall that the magnitude of the moment induced by a force acting on an axis of rotation is given by |š‘€|=š¹š‘‘, where š¹ is the magnitude of the force and š‘‘ is the perpendicular distance between the line of action of the force and the axis of rotation. In particular, if the line of action of a force contains the axis of rotation, then the moment generated by the force is zero.

If we place the axis of rotation on š·, then the contribution of the force āƒ‘š¹ļŠØ to the net moment would vanish. This would leave only š¹ļŠ§ as an unknown quantity, which we can identify using the given moment. Let us start with a new diagram with the axis of rotation set at š·.

We also recall that a moment is a signed quantity, and the sign of a moment can be determined after defining the positive orientation. Since no orientation is specified in this example, we follow the convention that the counterclockwise moment is positive.

Now, we will compute the moment generated by each force in this system. Let us begin with the force acting on point š“. The magnitude of this force is 5 N, and the perpendicular distance from point š· is 1+2=3cm. Then, the moment generated by this force is |š‘€|=5Ɨ3=15.ļŒ 

From the diagram, we see that the force acting on š“ induces a counterclockwise moment around š·; hence, this moment is positive by our convention. This gives us š‘€=15ā‹…ļŒ Ncm.

The magnitude of the force acting on the point š¶ is 2 N, and the perpendicular distance from the point š· is 2 cm. Then, the moment generated by this force is |š‘€|=2Ɨ2=4.ļŒ¢

From the diagram, we see that this force induces a clockwise moment around š·, so this moment is negative; hence, š‘€=āˆ’4ā‹…ļŒ¢Ncm.

As we noted earlier, the force acting on š· does not contribute to the net moment since the line of action contains the axis of rotation š·. Hence, š‘€=0ā‹….ļŒ£Ncm

Finally, the magnitude of the force acting on šµ is š¹ļŠ§ N, and the perpendicular distance from the point š· is 1 cm. Then, the moment generated by this force is |š‘€|=š¹Ć—1=š¹.ļŒ”ļŠ§ļŠ§

From the diagram, we see that this force induces a counterclockwise moment around š·, so this moment is positive; hence, š‘€=š¹ā‹…ļŒ”ļŠ§Ncm.

Adding up all moments generated by the forces in this system, we obtain the net force š‘€: š‘€=š‘€+š‘€+š‘€+š‘€=15+š¹āˆ’4+0=11+š¹.ļŒ ļŒ”ļŒ¢ļŒ£ļŠ§ļŠ§

We are given that the net moment is equal to 17 Nā‹…cm; hence, 17=11+š¹.ļŠ§

This gives us š¹=6ļŠ§N. Substituting this into (1), we obtain 6āˆ’š¹=3,ļŠØ which leads to š¹=3ļŠØN.

Let us recap a few important concepts from this explainer.

Key Points

  • A system of forces is equivalent to a couple if the net force is equal to zero.
  • The magnitude of the rotational moment š‘€ generated by a force with respect to an axis of rotation (or a reference point) is given by |š‘€|=š¹š‘‘, where š¹ is the magnitude of the force and š‘‘ is the perpendicular distance between the axis of rotation and the line of action for the force.
  • If the axis of rotation (or the reference point) lies on the line of action of a force, then no moment is generated by the force.
  • A moment is a signed quantity. After we define the positive orientation for moments, we can determine the sign of a moment by considering the orientation of the rotation induced by the force. Unless specified otherwise, we follow the convention that the counterclockwise moment is positive.
  • In a system of forces, the net moment is computed by adding up the signed moments generated by each force within the system.
  • If a system of forces is equivalent to a couple, then the net moment does not depend on the location of the axis of rotation. This means that we can pick an ideal location for the reference point to simplify our computation.

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