Question

FIGURE 1 shows a uniform ladder PQ of weight 240 N leans on a smooth wall and resting on a rough floor with a minimum inclination angle. The coefficient of friction between the ladder and floor is 0.25. With the aid of a force diagram, calculate the forces acting on the ladder at point P and point Q. Find resultant forces at point Q. Find magnitude and direction at point Q.

Answer 1

The weight of the ladder is,

$W=240\text{ N}$

The coefficient of friction between the ladder and the floor is,

$\mu=0.25$

The diuagram of the forces is given below:

The force on point Q is equal to the weight of the ladder. we can write,

$\begin{gathered} N_{ground}=W \\ =240\text{ N} \end{gathered}$

The force at point P will be equal to the frictional force. we can write,

$\begin{gathered} F_{friction}=N_{wall} \\ =\mu\times N_{ground} \\ =0.25\times240 \\ =60\text{ N} \end{gathered}$

The resultant of these two forces is,

$\begin{gathered} F=\sqrt[]{240^2+60^2} \\ =247\text{ N} \end{gathered}$

The angle with the horizontal is,

$\begin{gathered} \emptyset=\tan ^{-1}\frac{240}{60} \\ =75.9^{\circ} \end{gathered}$