example: 3.the uniform 10 - kg ladder in fig rests against the smooth wall at b ,and the end a rests on the…

Explanation:

Step1: Analyze forces in x - direction

Let the normal reaction at B be $N_B$ and the normal reaction at A be $N_A$. The frictional force at A is $F_f=\mu_sN_A$. In the x - direction, $\sum F_x = 0$, so $F_f - N_B=0$, which gives $N_B=\mu_sN_A$.

Step2: Analyze forces in y - direction

In the y - direction, $\sum F_y = 0$. The weight of the ladder $W =mg$, where $m = 10$ kg and $g=9.81$ m/s². So $N_A - mg=0$, then $N_A=mg = 10\times9.81=98.1$ N.

Step3: Calculate $N_B$

Since $N_B=\mu_sN_A$ and $\mu_s = 0.3$, $N_B=0.3\times98.1 = 29.43$ N.

Step4: Take moment about A

The length of the ladder $L = 4$ m. The weight acts at the mid - point of the ladder. $\sum M_A=0$. $N_B\times L\sin\theta-mg\times\frac{L}{2}\cos\theta = 0$. Substitute $N_B=\mu_sN_A$ and $N_A = mg$ into it. $\mu_smg\times L\sin\theta-mg\times\frac{L}{2}\cos\theta = 0$. Cancel out $mg$ and $L$, we get $2\mu_s\tan\theta= 1$.

Step5: Solve for $\theta$

$\tan\theta=\frac{1}{2\mu_s}$. Substitute $\mu_s = 0.3$ into it, $\tan\theta=\frac{1}{2\times0.3}=\frac{1}{0.6}\approx1.667$. So $\theta=\arctan(1.667)\approx59.04^{\circ}$.

Answer:

The angle of inclination $\theta\approx59.04^{\circ}$ and the normal reaction at B is $N_B = 29.43$ N.