Question

The main gate of building is $2.6\mathrm{m}$ broad. It can be opened by applying a force of $80\mathrm{N}$ at the middle of the gate. Where should the least force be applied?

Answer 1

Step 1- Given data

Length of the gate of main building is $\mathrm{L}=2.6\mathrm{m}$

Force used to open the door is $\mathrm{F}=80\mathrm{N}$

Step 2- Formulae used

1. Moment of force is defined as the product of force and the perpendicular distance between the line of action of force and the point about which the body rotates, i.e., moment arm. $\mathrm{Moment}\mathrm{of}\mathrm{force}=\mathrm{Force}\mathrm{x}\mathrm{Moment}\mathrm{arm}$
2. Since the force is acting at the middle of the door, so the distance between the point of application of force and the pivot point is $\mathrm{l}=\frac{2.6}{2}=1.3\mathrm{m}$

Step 3- Calculations

1. For the given situation, the moment of force is calculated when the force is acting at the mid point of the door. $$\begin{gathered} \mathrm{Moment}\mathrm{of}\mathrm{force}=\mathrm{Force}\mathrm{x}\mathrm{Moment}\mathrm{arm} \\ \mathrm{M}=\mathrm{F}\mathrm{x}\mathrm{l} \\ \mathrm{M}=80\mathrm{N}\mathrm{x}1.3\mathrm{m} \\ \mathrm{M}=104\mathrm{Nm} \end{gathered}$$
2. According to the given question for the force to be minimum, it must be applied at the end point and at this location the moment arm will be of maximum length. $\mathrm{Moment}\mathrm{of}\mathrm{force}=\left(\mathrm{Minimum}\right)\mathrm{force}\mathrm{x}\left(\mathrm{maximum}\right)\mathrm{moment}\mathrm{arm}$
3. The maximum moment arm can only be the full length of the door, i.e., $\mathrm{L}=2.6\mathrm{m}$
4. The moment of force required to open the door is calculated as $\mathrm{M}=104\mathrm{Nm}$
5. $$\begin{gathered} \mathrm{Moment}\mathrm{of}\mathrm{force}=\left(least\right)\mathrm{force}\mathrm{x}\left(maximum\right)\mathrm{moment}\mathrm{arm} \\ M=F^{\text{'}}xL \\ Least\mathrm{Force}=\frac{\mathrm{Moment}\mathrm{of}\mathrm{force}}{\mathrm{Maximum}\mathrm{Moment}\mathrm{arm}} \\ {\mathrm{F}}^{\text{'}}=\frac{\mathrm{M}}{\mathrm{L}} \\ {\mathrm{F}}^{\text{'}}=\frac{104\mathrm{Nm}}{2.6\hspace{0.17em}\mathrm{N}} \\ {\mathrm{F}}^{\text{'}}=40\mathrm{N} \end{gathered}$$
6. The location of the least force will be the maximum moment arm, i.e., $2.6\mathrm{m}$

Hence, the least force $40N$ is applied at $2.6\mathrm{m}$ i.e. at the free end.