Question

A couple acts on a beam as shown below. If x=1.60 m, h=0.45 m, and F=950 N, what is the moment of the couple? Express your answer to three significant figures and include the appropriate units.

Answer 1

Concepts and reason

Moment:

“The magnitude of moment of force acting about a point or axis is directly proportional to the distance of the force from the point or axis”.

Fundamentals

Resolution of forces:

The resultant forces can be resolved into two components as shown in Figure (1).

The horizontal force $\left( {{F_H}} \right)$ is calculated as shown in Figure (1).

${F_H} = F\cos \theta$

Here, force is F and angle of force with the horizontal is cos $\theta$ .

The vertical force $\left( {{F_V}} \right)$ is calculated as shown in Figure (1).

${F_V} = F\sin \theta$

Here, the angle of force with the vertical is $\sin \theta$ .

Write the Basic trigonometry formula from the Figure (2).

From the Figure (2),

Write the law of tangent.

“The tangent of an angle is the ratio of the length of the opposite side to the length of the adjacent.”

Tangent of an angle is mathematically expressed as follows:

$\begin{array}{c}\\\tan \theta = \frac{{{\rm{opposite side}}}}{{{\rm{adjacent side}}}}\\\\ = \frac{{AC}}{{BC}}\\\end{array}$

From the Figure (2),

Write the law of sine.

“The sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse.”

Mathematically expressed as,

$\begin{array}{c}\\\sin \theta = \frac{{{\rm{opposite side}}}}{{{\rm{hypotenuse side}}}}\\\\ = \frac{{AC}}{{AB}}\\\end{array}$

Write the law of cosine.

“The cosine of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse.”

Mathematically expressed as,

$\begin{array}{c}\\\cos \theta = \frac{{{\rm{adjacent side}}}}{{{\rm{hypotenuse side}}}}\\\\ = \frac{{BC}}{{AB}}\\\end{array}$

The moment can be calculated as follows.

It is calculated by the product of the magnitude of force with the perpendicular distance between the points of application of force.

${\rm{Moment}}\left( M \right){\rm{ = Force \times perpendicular distance}}$

General sign conventions for Moment:

Moment is considered as negative in clockwise direction and positive in counter clockwise direction.

Draw the diagram as shown in figure (3).

Resolve the horizontal and vertical forces in respective direction.

${F_H} = 950{\rm{N}}\sin \theta$

${F_V} = 950{\rm{N}}\cos \theta$

Take moment about $z$ axis.

$\begin{array}{c}\\{M_Z} = \left( {{F_H} \times 2h} \right) - \left( {{F_V} \times x} \right)\\\\ = \left( {950{\rm{ N}} \times {\rm{cos}}\theta \times 2h} \right) - \left( {950{\rm{ N}} \times {\rm{sin}}\theta \times x} \right)\\\\ = \left( {950{\rm{ N}} \times \frac{{3{\rm{ m}}}}{{5\,{\rm{m}}}} \times 2 \times 0.45{\rm{ m}}} \right) - \left( {950{\rm{ N}} \times \frac{{4{\rm{ m}}}}{{5\,{\rm{m}}}} \times 1.6{\rm{ m}}} \right)\\\\ = - 703{\rm{ N}} \cdot {\rm{m}}\\\end{array}$

Ans:

The resultant couple moment ${M_Z}$ is $703{\rm{ N}} \cdot {\rm{m}}\left( {{\rm{Clockwise}}} \right)$ .