14. Moments and equilibrium

From Mechanics

Current revision

Key Points

For a body in equilibrium:

• The resultant force on the body must be zero.

and

• The resultant moment of the forces on the body about all points must be zero.

Sometimes it is more convenient to solve a problem just using moments.

Example 14.1

A uniform beam has length 8 m and mass 60 kg. It is suspended by two ropes, as shown in the diagram below.

Find the tension in each rope.

Solution

The diagram shows the forces acting on the beam.

Take moments about the point where T1 acts to give:

\displaystyle \begin{align} & 5\times {{T}_{2}}=3\times 588 \\ & {{T}_{2}}=\frac{3\times 588}{5}=352\textrm{.}8=353\text{ N (to 3sf)} \\ \end{align}

Take moments about the point where T2 acts to give:

\displaystyle \begin{align} & 5\times {{T}_{1}}=2\times 588 \\ & {{T}_{1}}=\frac{2\times 588}{5}=235\textrm{.}2=235\text{ N (to 3sf)} \\ \end{align}

Finally for vertical equilibrium we require \displaystyle {{T}_{1}}+{{T}_{2}}=588 , which can be used to check the tensions. In is the case we have:

\displaystyle 352\textrm{.}8+235\textrm{.}2=588

Example 14.2

A beam, of mass 50 kg and length 5 m, rests on two supports as shown in the diagram. Find the magnitude of the reaction force exerted by each support.

Find the maximum mass that could be placed at either end of the beam if it is to remain in equilibrium.

Solution

The diagram shows the forces acting on the beam.

Taking moments about the point where R1 acts gives:

\displaystyle \begin{align} & 2\times {{R}_{2}}=1\textrm{.}5\times 490 \\ & {{R}_{2}}=\frac{1\textrm{.}5\times 490}{2}=367\textrm{.}5=368\text{ N (to 3sf)} \\ \end{align}

Taking moments about the point where R2 acts gives:

\displaystyle \begin{align} & 2\times {{R}_{1}}=0\textrm{.}5\times 490 \\ & {{R}_{1}}=\frac{0\textrm{.}5\times 490}{2}=122\textrm{.}5=123\text{ N (to 3sf)} \\ \end{align}

For vertical equilibrium we require \displaystyle {{R}_{1}}+{{R}_{2}}=490 , which can be used to check the tensions. In is the case we have:

\displaystyle 367\textrm{.}5+122\textrm{.}5=490

First consider the greatest mass that can be placed at the left hand end of the beam. The diagram below shows the extra force that must now be considered. When the maximum possible mass is used, \displaystyle {{R}_{2}}=0 .

Taking moments about the point where R1 acts gives:

\displaystyle \begin{align} & 1\times mg=1\textrm{.}5\times 490 \\ & m=\frac{1\textrm{.}5\times 490}{9\textrm{.}8}=75\text{ kg} \\ \end{align}

Similarly for a mass placed at the right hand end of the beam:

\displaystyle \begin{align} & 2\times mg=0\textrm{.}5\times 490 \\ & m=\frac{0\textrm{.}5\times 490}{2g}=12\textrm{.}5\text{ kg} \\ \end{align}

Hence the greatest mass that can be placed at either end of the beam is 12.5 kg.

Example 14.3

A ladder, of length 3 m and mass 20 kg, leans against a smooth, vertical wall so that the angle between the horizontal ground and the ladder is 60\displaystyle {}^\circ .

a) Find the magnitude of the friction and normal reaction forces that act on the ladder, if it is in equilibrium.

b) Find the minimum value of the coefficient of friction between the ladder and the ground.

Solution

The diagram shows the forces acting on the ladder

a) Considering the horizontal forces gives:

\displaystyle F=S

Considering the vertical forces gives:

\displaystyle R=196

Taking moment about the base of the ladder gives:

\displaystyle \begin{align} & 196\times 1\textrm{.}5\cos 60{}^\circ =S\times 3\sin 60{}^\circ \\ & S=\frac{196\times 1\textrm{.}5\cos 60{}^\circ }{3\sin 60{}^\circ }=\frac{196\times \cos 60{}^\circ }{2\sin 60{}^\circ }=\frac{196}{2\tan 60{}^\circ }=56\textrm{.}6\text{ N (to 3 sf)} \\ \end{align}

But since

\displaystyle F=S , the friction force has magnitude \displaystyle 56\textrm{.}6 \text{ N} .

b) Using the friction inequality,

\displaystyle F\le \mu R gives:

\displaystyle \begin{align} & \frac{196}{2\tan 60{}^\circ }\le \mu \times 196 \\ & \mu \ge \frac{1}{2\tan 60{}^\circ } \\ & \mu \ge 0\textrm{.}289\text{ (to 3sf)} \\ \end{align}