Question

3.0-m-long ladder, as shown in Figure 12.35, leans against a frictionless wall. The coefficient of static friction between the ladder and the floor is 0.40. What is the minimum angle the
ladder can make with the floor without slipping?

Short answer

The minimum angle the ladder can make with the floor without slipping is 51.34^o

Step-by-step solution

Given information

3.0-m-long ladder, leans against a frictionless wall.

The coefficient of static friction between the ladder and the floor is 0.40.

Explanation

Lets first draw the diagram to understand the and solve the problem

First find the force due to friction is

f_s=f_smax=μ_sN_{2...................................................}(1)

For the ladder to be in equilibrium the net resultant horizontal and vertical forces must zero and Torque about origin should be zero

$$\begin{gathered} \sum F_x=0 \\ {N}_1-f_{s_{\max }}=0 \\ {f}_{s_{\max }}=N_1\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(2\right) \end{gathered}$$

And

$$\begin{gathered} \sum F_y=0 \\ {N}_2-W=0 \\ {N}_2=W\dots .....................................\left(3\right) \end{gathered}$$

Equate Torque

$$\begin{gathered} \sum {\tau }_O=0 \\ {N}_1(3mSin\theta )+W(1.5mCos\theta )-N_2(3mCos\theta )=0 \end{gathered}$$

Solve for angle we get

$\theta ={\tan }^{-1}\left(\frac{5}{4}\right)=51.34°$