Question

Two parallel 60Nforces are applied to a lever as shown. Determine the moment of the couple formed by the two forces:

a) by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples,
b) by using the perpendicular distance between the two forces &
c) by summing the moments of the two forces about point A.

The answer for a, b and c is 12.39 Nm.

Answer 1

Concepts and reason

The concepts required to solve this problem are the resolution of vector, moment, and the resultant moment.

Resolution of vector: When a force vector ${\bf{F}}$ makes an angle $\theta$ with the $x$ axis, then the cosine component is along $x$ axis and the sine component is along $y$ axis.

Moment: The moment is the turning effect of force and is usually defined with respect to a fixed reference point. The moment is obtained by taking the product of force and the perpendicular distance of force from a fixed point.

Resultant moment: The resultant moment of the body about a certain point gives the same effect as the sum of all the moment of the body about the same point. The resultant moment is obtained by adding all the moment acting on the body.

Initially, in the first case, the forces along $x$ and $y$ direction are calculated by using the resolution of vector concept. Then, by using the moment and the resultant moment concept, find the value of moment of the couple forces.

Similarly, in the second case, the value of moment by the couple forces are calculated by using the concept of moment. Finally, in the third case, the moment about a different point is obtained by calculating the perpendicular distance of forces about that point and then multiplying the perpendicular distance with the force.

Fundamentals

The expression for moment is,

$M = Fd$

Here, $F$ and $d$ are the force and the perpendicular distance of force from a fixed point.

Resolving vector into its components: Consider a vector $\vec V$ acting at a point making an angle $\theta$ with the positive $x$ axis.

The expression for the component of vector $\vec V$ in $x$ direction is,

${V_x} = V\cos \theta$

Similarly, the expression for the component of vector $\vec V$ in $y$ direction is,

${V_y} = V\sin \theta$

Consider the moments acting on the body about a point $A$ as ${M_1}$ and ${M_2}$ . Then, write the expression of resultant moment about point $A$ .

${M_A} = {M_1} + {M_2}$

Sign convention: Take the moment in clockwise direction as positive and moment in anticlockwise as negative. This sign convention is used throughout the solution.

(a)

Consider the forces acting at point $C$ and $B$ as ${F_1}$ and ${F_2}$ . The force ${F_1}$ makes an angle of $20^\circ$ with the $x$ axis. Also, the angle made by the rod $AC$ with the horizontal is $55^\circ$ . Thus, from the simple geometry, the angle made by the force ${F_1}$ with the rod is taken as $35^\circ$ . Similarly, the force ${F_2}$ makes an angle of $35^\circ$ with the rod $BC$ and $20^\circ$ with the horizontal axis.

Draw the diagram showing the force components along $x$ and $y$ axis.

Now, consider the couple forces. Both the forces are equal and opposite in sign. Consider the origin at point $B$ . Resolve the force ${F_1}$ into horizontal and vertical component. Calculate the $x$ component of force ${F_1}$ .

${F_{1x}} = {F_1}\cos 20^\circ$

Substitute $60{\rm{ N}}$ for ${F_1}$ .

$\begin{array}{c}\\{F_{1x}} = \left( {60{\rm{ N}}} \right)\cos 20^\circ \\\\ = 56.38{\rm{ N}}\\\end{array}$

Similarly, calculate the $y$ component of force ${F_1}$ .

${F_{1y}} = {F_1}\sin 20^\circ$

Substitute $60{\rm{ N}}$ for ${F_1}$ .

$\begin{array}{c}\\{F_{1y}} = \left( {60{\rm{ N}}} \right)\sin 20^\circ \\\\ = 20.52{\rm{ N}}\\\end{array}$

From the definition of moment, the distance should be perpendicular to the force ${F_1}$ . The distance between the couple forces is given by $BC$ . Also, from Figure (b), it is clear that the vertical component of length gives the perpendicular distance of the force $F$ .

Since, the angle between the force and the distance is given by $90^\circ$ . Thus, the angle between the rod length $BC$ with the vertical is $35^\circ$ . This gives the vertical component of length as $\left( {0.36{\rm{ m}}} \right)\cos 35^\circ$ and horizontal component as $\left( {0.36{\rm{ m}}} \right)\sin 35^\circ$ . Draw the diagram for the $x$ and $y$ component of force and the perpendicular distance.

Calculate the moment by the $x$ component of force.

${M_x} = {F_x}d$

Substitute $56.38{\rm{ N}}$ for ${F_x}$ and $\left( {0.36{\rm{ m}}} \right)\cos 35^\circ$ for $d$ .

$\begin{array}{c}\\{M_x} = \left( {56.38{\rm{ N}}} \right)\left( {\left( {0.36{\rm{ m}}} \right)\cos 35^\circ } \right)\\\\ = 16.63{\rm{ N}} \cdot {\rm{m}}\\\end{array}$

The moment in $x$ direction is clockwise as shown in Figure (c), and is therefore, taken as positive.

Similarly, calculate the moment by the $y$ component of force.

${M_y} = - {F_y}d$

Substitute $20.52{\rm{ N}}$ for ${F_y}$ and $\left( {0.36{\rm{ m}}} \right)\sin 35^\circ$ for $d$ .

$\begin{array}{c}\\{M_y} = - \left( {20.52{\rm{ N}}} \right)\left( {\left( {0.36{\rm{ m}}} \right)\sin 35^\circ } \right)\\\\ = - 4.24{\rm{ N}} \cdot {\rm{m}}\\\end{array}$

The moment in $y$ direction is anticlockwise as shown in Figure (c), and is therefore, taken as negative.

Calculate the moment of the couple.

$M = {M_x} + {M_y}$

Substitute $16.63{\rm{ N}} \cdot {\rm{m}}$ for ${M_x}$ , and $- 4.24{\rm{ N}} \cdot {\rm{m}}$ for ${M_y}$ .

$\begin{array}{c}\\M = 16.63{\rm{ N}} \cdot {\rm{m}} - 4.24{\rm{ N}} \cdot {\rm{m}}\\\\ = 12.39{\rm{ N}} \cdot {\rm{m}}\\\end{array}$

(b)

Consider the couple forces. The length of the rod is $0.36{\rm{ m}}$ . Also, the rod is at an angle of $35^\circ$ from the force $F$ . Thus, the perpendicular component of distance between the forces is given by $\left( {0.36{\rm{ m}}} \right)\sin 35^\circ$ . Draw the diagram showing the perpendicular force between the couple forces.

Calculate the moment of the couple by using the perpendicular distance between the two forces.

$M = Fd$

Substitute $60{\rm{ N}}$ for $F$ and $\left( {0.36{\rm{ m}}} \right)\sin 35^\circ$ for $d$ .

$\begin{array}{c}\\M = \left( {60{\rm{ N}}} \right)\left( {\left( {0.36{\rm{ m}}} \right)\sin 35^\circ } \right)\\\\ = 12.39{\rm{ N}} \cdot {\rm{m}}\\\end{array}$

(c)

To calculate the moment about point $A$ , both the couple forces need to be considered as both are acting at a different point than point $A$ . Consider the force ${F_1}$ acting at point $C$ . The distance of the point $C$ and $A$ is given as $\left( {0.36{\rm{ m}} + 0.52{\rm{ m}}} \right)$ that is $0.88{\rm{ m}}$ . Draw the diagram showing the perpendicular distance of force ${F_1}$ from a fixed point $A$ .

Thus, the perpendicular distance of force ${F_1}$ is given by $\left( {0.88{\rm{ m}}} \right)\sin 35^\circ$ . Calculate the moment by the force ${F_1}$ .

${M_1} = {F_1}d$

Substitute $60{\rm{ N}}$ for ${F_1}$ and $\left( {0.88{\rm{ m}}} \right)\sin 35^\circ$ for $d$ .

$\begin{array}{c}\\{M_1} = \left( {60{\rm{ N}}} \right)\left( {\left( {0.88{\rm{ m}}} \right)\sin 35^\circ } \right)\\\\ = 30.28{\rm{ N}} \cdot {\rm{m}}\\\end{array}$

Here, the moment by the force ${F_1}$ is clockwise and is therefore, taken as positive.

Now, the force ${F_2}$ acts at point $B$ . The distance between point $B$ and $A$ is given by $0.52{\rm{ m}}$ . Draw the diagram showing the perpendicular distance of force ${F_2}$ from a fixed point $A$ .

From Figure (f), the angle made by the rod $AB$ with the force ${F_2}$ is $35^\circ$ . Thus, the perpendicular distance is given by $\left( {0.52{\rm{ m}}} \right)\sin 35^\circ$ . Calculate the moment by the force ${F_2}$ .

${M_2} = {F_2}d$

Substitute $60{\rm{ N}}$ for ${F_2}$ and $\left( {0.52{\rm{ m}}} \right)\sin 35^\circ$ for $d$ .

$\begin{array}{c}\\{M_2} = - \left( {60{\rm{ N}}} \right)\left( {\left( {0.52{\rm{ m}}} \right)\sin 35^\circ } \right)\\\\ = - 17.89{\rm{ N}} \cdot {\rm{m}}\\\end{array}$

Here, the moment by the force ${F_2}$ is anticlockwise and is therefore, taken as negative.

Calculate the moment by the couple forces about point $A$ .

${M_A} = {M_1} + {M_2}$

Substitute $30.28{\rm{ N}} \cdot {\rm{m}}$ for ${M_1}$ and $- 17.89{\rm{ N}} \cdot {\rm{m}}$ for ${M_2}$ .

$\begin{array}{c}\\{M_A} = 30.28{\rm{ N}} \cdot {\rm{m}} - 17.89{\rm{ N}} \cdot {\rm{m}}\\\\ = 12.39{\rm{ N}} \cdot {\rm{m}}\\\end{array}$

Ans: Part a

The moment of the couple in this case is $12.39{\rm{ N}} \cdot {\rm{m}}$ .

Part b

The moment by the couple force in this case is $12.39{\rm{ N}} \cdot {\rm{m}}$ .

Part c

The moment of the couple forces about point $A$ is $12.39{\rm{ N}} \cdot {\rm{m}}$ .