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Chapter 15: Problem 2

Prove that a couple, together with a force in the same plane, is equivalent to a single force. Describe completely the possible resultants of a force of \(10 \mathrm{~N}\) acting in the same plane as a couple of magnitude \(20 \mathrm{Nm}\).

Short Answer

Expert verified
The resultant is a 10 N force moved 2 meters.

Step by step solution

01

Understanding the Terms

A 'couple' refers to a pair of equal and opposite forces whose lines of action do not coincide. This creates a rotational effect (moment) but no net force. A 'single force' is a force with a net effect.
02

Express the Force and Couple

Let's denote the force as \(\textbf{F} = 10 \text{ N}\), and the couple as a moment \(\textbf{M} = 20 \text{ Nm}\). The force's line of action is in the same plane as the couple.
03

Combine the Force and Couple

A force \(\textbf{F}\) and a couple \(\textbf{M}\) can be combined into a single resultant force by moving the point of application of \(\textbf{F}\) to produce the moment \(\textbf{M}\). The distance \(d\) from the original application point of \(\textbf{F}\) to the new application point can be found from \(\textbf{M} = \textbf{F} \times d\).
04

Calculate Distance d

Since \(\textbf{M} = \textbf{F} \times d\), substituting the given values: \(\textbf{20 Nm} = 10 \text{ N} \times d\). Solve for \(d\): \(\textbf{d} = \frac{20 \text{ Nm}}{10 \text{ N}} = 2 \text{ m}\).
05

State the Resultant Force

The couple \(20 \text{ Nm}\) combined with the force \(10 \text{ N}\) is equivalent to the same force \(10 \text{ N}\) moving 2 meters along its line of action to produce the same rotational effect.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

force and couple system
In mechanics, a 'force and couple system' can be visualized as a force acting at a certain point, coupled with a moment created by a pair of equal and opposite forces. A couple generates a pure rotational effect without producing translational motion.
For example:
  • A couple consists of two equal and opposite forces
  • The lines of action do not coincide
  • The result is a pure rotation with no net force
This means if we have a single force and a moment in the same plane, it can be represented as just a force after proper adjustments to its point of action. The force and the couple together act as a single resultant force, hence simplifying the analysis.
moment of a force
The 'moment of a force', also referred to as torque, is the measure of the tendency of the force to rotate an object about an axis or pivot. The formula for the moment is: \[ \text{Moment} (M) = \text{Force} (F) \times \text{Perpendicular Distance} (d) \]
In our exercise, we were given:
  • A force, \[ F = 10 \text{ N} \]
  • A couple creating a moment, \[ M = 20 \text{ Nm} \]
To find the distance where the force creates the same moment:
  • Use the formula: \[ M = F \times d \]
  • Substitute the given values: \[ 20 \text{ Nm} = 10 \text{ N} \times d \]
  • Solve for \[ d = 2 \text{ m} \]
This calculation confirms the force needs to be moved 2 meters to create the same 20 Nm moment.
equilibrium in two dimensions
Equilibrium in two dimensions refers to a state where all the forces and moments acting on a body result in no motion. For an object to be in equilibrium, two conditions should be met:
  • The sum of all horizontal and vertical forces must be zero: \[\textstyle \text{鈭慒} = 0 \]
  • The sum of all moments about any point must be zero: \[ \textstyle \text{鈭慚} = 0 \]
This means for our force and couple system example:
  • Both the force \[ 10 \text{ N} \] and moment \[ 20 \text{ Nm} \] must balance each other
  • This results in no translational or rotational motion
Thus, in our solution, adjusting the point of application of the force ensures equilibrium by matching the moment produced by the couple.
rotational effect
Rotational effect describes how forces create rotation around a point or axis. A couple's main characteristic is to produce this rotational effect without causing linear displacement. Key points include:
  • The couple's opposing forces create a turning effect
  • The magnitude of this effect is determined by the moment formula: \[ M = F \times d \]
In the exercise:
  • A force of \[ 10 \text{ N} \] and a couple of \[ 20 \text{ Nm} \]
  • By moving the force to a point \[ 2 \text{ m} \] from its original position
  • The same rotational effect (moment) is produced
This demonstrates the concept that a combination of forces can be replaced by an equivalent single force acting at a different location, simplifying analysis and maintaining the same rotational effect on the body.

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Most popular questions from this chapter

Find the magnitude and direction of the resultant of each of the following two systems of forces which act in the plane of a rectangle \(A B C D\). in which \(A B=4 a\), and \(\mathrm{BC}=3 a\) (a) Forces \(4 P\) along \(A B, P\) along \(A D\) and \(10 P\) along \(D B\). (b) Forces which have total moment \(-P a,+15 P a\) and \(-5 P a\) about A. B and D respectively ( \(+\) indicates the sense \(\mathrm{ABC}\) ). Find also, in each case, the point of intersection of the line of action of the resultant with the line \(\mathrm{AB}\) (produced if necessary).

A system of forces acting in the plane of perpendicular axes \(O x\) and \(O y\) consists of: a force \(10 P\) along \(\mathrm{Ox}\), a force \(-9 P\) along \(\mathrm{O} y\), a force \(13 P\) along \(\mathrm{OA}\), where \(\mathrm{A}\) is the point \((12 a, 5 a)\). a force \(20 P\) along \(A B\), where \(B\) is the point \((8 a, 8 a)\). Find the magnitude, direction and equation of the line of action of the resultant of this system. A clockwise coplanar couple of magnitude \(240 P a\) is added to the system. Find the magnitude, direction and equation of the line of action of the resultant of the new system.

Six forces act round the sides of a hexagon. Find the equation of the line of action of their resultant. (a) Their magnitudes are \(P, 2 P, 4 P, 3 P, P, 2 P\) along \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{CD}}, \overrightarrow{\mathrm{ED}}, \overrightarrow{\mathrm{FE}}, \overrightarrow{\mathrm{AF}}\) respectively. (b) The hexagon is regular. (c) The co-ordinates of vertex \(\mathbf{B}\) are \((1,1)\).

A system of coplanar forces consists of forces of \(4,3,2,5\) and 6 newton acting along the sides \(\mathrm{AB}, \mathrm{BC}, \mathrm{CD}, \mathrm{DE}\) and \(\mathrm{EF}\) respectively of a regular hexagon \(\mathrm{ABCDEF}\) of side \(2 \mathrm{~m}\), the forces acting in the directions indicated by the order of the letters. Find, in magnitude and direction, the force \(P\) newtons, acting at \(\mathrm{F}\), which will reduce the system to a couple and find the magnitude and sense of this couple. If the force \(P\) newtons is replaced by a force of 7 newton along \(A F\), show that the system now reduces to a single force and find the magnitude of this resultant and the point of intersection of its line of action with \(\mathrm{AB}\), produced if necessary.
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