Question

1) 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 θmin. The coefficient of 

friction between the ladder and floor is 0.25. 

(i)

With the aid of a force diagram, calculate the forces acting on the ladder at 

point P and point Q.

(ii)

Determine the value of θmin. 

Answer 1

From the equilibrium of forces:

$$\Sigma F_x = 0,\\ \Sigma F_x = F_\text{friction} – N_\text{wall},\\ N_\text{wall} = F_\text{friction}.\\\space\\ \Sigma F_y = N_\text{ground} – W = 0,\\ N_\text{ground}=W=240\text{ N}.$$

From the equilibrium of moments:

$$\frac12WL\cos\theta-N_\text{wall}L\sin\theta=0,\\\space\\ \tan\theta=\frac{W}{2N_\text{wall}}=\frac{W}{2F_\text{friction}}.\\\space\\ F_\text{friction}=\mu N_\text{ground},\\\space\\ \tan\theta=\frac{W}{2\mu N_\text{ground}}=\frac{W}{2\mu W}=\frac1{2\mu}.\\\space\\ \theta=\arctan\frac1{2\mu}=63°.$$