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* N_(ground) \\ =0.25*240 \\ =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)(240)/(60) \\ =75.9^(\circ) \end{gathered}`