91Ó°ÊÓ

A ladder is leaning against the wall and is in static equilibrium.

We use the concept of static equilibrium, i.e., balanced torques and balanced forces to determine the magnitudes of the various forces acting on the ladder.

Formulae:

The value of the net force at equilibrium,$F_{net}=0$ (i)

The value of the torque at equilibrium, ${\mathrm{Ï„}}_{net}=0$ (ii)

The torque acting on a point, $\mathrm{τ}=r×F$ (iii)

The maximum value of the static frictional force, $F_{s,max}=\mathrm{μ}N$ (iv)

Free body diagram:

To check the equilibrium of forces in the vertical direction:

The forces on the ladder acting in the vertical direction are the weight of the ladder and the normal force of reaction (N) from the ground.

Since the ladder is in equilibrium, the vertical forces are balanced, and hence the normal force is equal to the weight of the ladder.

The weight of the ladder remains the same even when we shift the base of the ladder towards the wall.

Hence the normal force from the ground remains the same.

We first consider the balancing of the horizontal forces. The horizontal forces are the force from the wall $\left(F_w\right)$and the frictional force from the ground $\left(F_s\right)$. As the ladder is in equilibrium, the force equation considering equation (i) can be given as:

role="math" localid="1660971618430" $F_w=F_s$

$$\begin{gathered} \mathrm{The}\mathrm{moment}\mathrm{arm}\mathrm{of}{F}_w=\mathrm{height}h \\ \mathrm{The}\mathrm{moment}\mathrm{arm}\mathrm{of}\mathrm{weight}\mathrm{of}\mathrm{the}\mathrm{ladder}=a/2 \end{gathered}$$

Now, the torque about the point A at which the ladder touches the ground can be given using equation (iii) in equation (ii) as follows:

$$\begin{gathered} -(h)F_w+(aI2)mg=0 \\ {F}_w=\frac{a}{2h}mg \end{gathered}$$

But the geometry suggests that,

$\tan \mathrm{θ}=\frac{h}{a}$

Substituting the above value in the force value, we get the value as follows:

$F_w=\frac{mm}{2tan\mathrm{θ}}$

This equation indicates that the force varies inversely with the tan of the angle. So when we shift the base of the ladder towards the wall, we increase the angle from${\mathrm{θ}}_1$ to${\mathrm{θ}}_2$ .

Therefore, the force from the wall will become smaller.

As we have seen in part (b), the static frictional force from the ground is equal in magnitude to the force from the wall, as the horizontal forces are balanced.

$\begin{array}{l}{F}_s=F_w\\ =\frac{mg}{2tan\mathrm{θ}}\end{array}$

So, the static friction force also becomes smaller as the angle increases.

The coefficient of friction (μ) is a constant for equation (iv) in the given situation, and as we have seen in part (a), the normal force N also remains constant even after shifting of the ladder.

Hence, we can conclude that the maximum static frictional force remains the same.