Question

A 6.5 m long ladder rests against a vertical wall reaching a height of 6.0 m. A 60 kg man stands half way up the ladder. (a) Find the torque of the force exerted by the man on the ladder about the upper end of the ladder. (b) Assuming the weight of the ladder to be negligible as compared to the man and assuming the wall to be smooth, find the force exerted by the ground on the ladder.

Answer 1

(a) The torque of the force exerted by the man on the ladder about the upper end of the ladder is 735 Nm (clockwise). (b) The force exerted by the ground on the ladder has a vertical component (normal force) of 588 N upwards and a horizontal component (friction force) of 122.5 N towards the wall. The magnitude of the total force exerted by the ground is approximately 600.6 N.

Answer 2

The problem consists of two parts.

Part (a): Find the torque of the force exerted by the man on the ladder about the upper end of the ladder.

Let the ladder have length $L = 6.5$ m. It rests against a vertical wall, reaching a height $h = 6.0$ m. Let A be the upper end of the ladder (touching the wall) and B be the base of the ladder (touching the ground). The horizontal distance from the wall to the base of the ladder is $b = \sqrt{L^2 - h^2} = \sqrt{(6.5)^2 - (6.0)^2} = \sqrt{42.25 - 36.00} = \sqrt{6.25} = 2.5$ m.

The man has a mass $m_m = 60$ kg. His weight is $W_m = m_m g = 60 \times 9.8 = 588$ N. This force acts vertically downwards. The man stands halfway up the ladder, so his position M is at a distance $L/2 = 6.5/2 = 3.25$ m from the base B and also $L/2 = 3.25$ m from the upper end A.

We need to find the torque of the man's weight about the upper end A. The torque $\vec{\tau}$ is given by $\vec{\tau} = \vec{r} \times \vec{F}$, where $\vec{r}$ is the position vector from the pivot point (A) to the point where the force is applied (M), and $\vec{F}$ is the force (man's weight). The vector $\vec{r}$ is the vector $\vec{AM}$, which has magnitude $AM = L/2 = 3.25$ m and points along the ladder from A to M (downwards along the ladder). The force $\vec{F}$ is the man's weight $\vec{W}_m$, acting vertically downwards.

The magnitude of the torque is $|\vec{\tau}| = |\vec{r}| |\vec{F}| \sin \theta$, where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$. The angle between the ladder and the vertical is $\beta$. The vector $\vec{r}$ is along the ladder, and $\vec{F}$ is vertical. So, the angle between $\vec{r}$ and $\vec{F}$ is $\beta$. We can find $\cos \beta$ or $\sin \beta$ from the geometry of the ladder. The angle $\beta$ is the angle between the ladder (hypotenuse $L$) and the vertical wall (side $h$). So, $\cos \beta = h/L = 6.0/6.5 = 60/65 = 12/13$. The angle between the ladder and the horizontal ground is $\alpha$. $\sin \alpha = h/L = 12/13$, $\cos \alpha = b/L = 2.5/6.5 = 5/13$. Note that $\beta = 90^\circ - \alpha$, so $\sin \beta = \cos \alpha = 5/13$.

The magnitude of the torque is $\tau_m = (L/2) W_m \sin \beta$. $\tau_m = (3.25 \text{ m}) \times (588 \text{ N}) \times (5/13)$. $\tau_m = (6.5/2) \times 588 \times (5/13) = (13/4) \times 588 \times (5/13) = (1/4) \times 588 \times 5 = 147 \times 5 = 735$ Nm.

Alternatively, the torque magnitude is the force times the perpendicular distance from the pivot (A) to the line of action of the force. The line of action of the man's weight is a vertical line passing through M. The horizontal distance of M from the wall (vertical line through A) is half the horizontal distance of the base from the wall, i.e., $b/2 = 2.5/2 = 1.25$ m. This horizontal distance is the perpendicular distance from A to the vertical line of action of the man's weight. $\tau_m = W_m \times (b/2) = 588 \text{ N} \times 1.25 \text{ m} = 588 \times (5/4) = 147 \times 5 = 735$ Nm. The torque tends to rotate the ladder clockwise about A.

Part (b): Find the force exerted by the ground on the ladder.

The system (ladder + man) is in equilibrium. We assume the weight of the ladder is negligible ($W_L \approx 0$). The wall is smooth, so it exerts only a normal force horizontally. The ground exerts both a normal force and a frictional force.

Let the forces acting on the ladder be:

1. Man's weight $W_m = 588$ N acting vertically downwards at M.
2. Normal force from the wall $\vec{N}_w$ acting horizontally at A. Let the wall be along the y-axis, and the ground along the x-axis. A is at (0, h), B is at (b, 0), M is at (b/2, h/2). The wall pushes away from itself, so $\vec{N}_w = N_w \hat{i}$.
3. Force from the ground at B. This force has a vertical component (normal force from ground, $N_g$) and a horizontal component (friction force from ground, $f_s$). The base tends to slip away from the wall, so friction acts towards the wall. $\vec{F}_g = -f_s \hat{i} + N_g \hat{j}$.

For equilibrium, the net force is zero: Horizontal components: $\sum F_x = N_w - f_s = 0 \implies N_w = f_s$. Vertical components: $\sum F_y = N_g - W_m = 0 \implies N_g = W_m = 588$ N.

The force exerted by the ground on the ladder is $\vec{F}_g = -f_s \hat{i} + N_g \hat{j} = -N_w \hat{i} + 588 \hat{j}$. We need to find $f_s$ (or $N_w$).

For equilibrium, the net torque about any point is zero. Let's take torque about the base B (b, 0) to eliminate the ground force $\vec{F}_g$. Forces producing torque about B:

1. Man's weight $W_m$ at M (b/2, h/2). The position vector from B to M is $\vec{r}_M = \vec{M} - \vec{B} = (b/2 - b, h/2 - 0) = (-b/2, h/2)$. The force is $\vec{W}_m = (0, -W_m)$. The torque is $\vec{\tau}_m = \vec{r}_M \times \vec{W}_m = (-b/2 \hat{i} + h/2 \hat{j}) \times (-W_m \hat{j}) = (-b/2)(-W_m) (\hat{i} \times \hat{j}) + (h/2)(-W_m) (\hat{j} \times \hat{j}) = (b/2) W_m \hat{k}$. Magnitude $\tau_m = (b/2) W_m = (2.5/2) \times 588 = 1.25 \times 588 = 735$ Nm (counter-clockwise). Alternatively, the perpendicular distance from B to the line of action of $W_m$ (vertical line through M) is the horizontal distance of M from B, which is $b/2$. $\tau_m = W_m \times (b/2) = 588 \times 1.25 = 735$ Nm.
2. Wall normal force $\vec{N}_w$ at A (0, h). The position vector from B to A is $\vec{r}_A = \vec{A} - \vec{B} = (0 - b, h - 0) = (-b, h)$. The force is $\vec{N}_w = (N_w, 0)$. The torque is $\vec{\tau}_w = \vec{r}_A \times \vec{N}_w = (-b \hat{i} + h \hat{j}) \times (N_w \hat{i}) = (-b)(N_w) (\hat{i} \times \hat{i}) + (h)(N_w) (\hat{j} \times \hat{i}) = 0 + h N_w (-\hat{k}) = -h N_w \hat{k}$. Magnitude $\tau_w = h N_w = 6.0 N_w$ (clockwise). Alternatively, the perpendicular distance from B to the line of action of $N_w$ (horizontal line through A) is the vertical distance of A from B, which is $h$. $\tau_w = N_w \times h = N_w \times 6.0$.

For rotational equilibrium, the net torque about B is zero: $\sum \vec{\tau}_B = \vec{\tau}_m + \vec{\tau}_w = 0$ $735 \hat{k} - 6.0 N_w \hat{k} = 0$ $735 - 6.0 N_w = 0$ $N_w = \frac{735}{6.0} = \frac{735}{6} = 122.5$ N.

From force equilibrium, $f_s = N_w = 122.5$ N. The force exerted by the ground on the ladder is $\vec{F}_g = -f_s \hat{i} + N_g \hat{j} = -122.5 \hat{i} + 588 \hat{j}$. The magnitude of this force is $|\vec{F}_g| = \sqrt{(-122.5)^2 + (588)^2} = \sqrt{15006.25 + 345744} = \sqrt{360750.25} \approx 600.625$ N.

The question asks for "the force exerted by the ground on the ladder", which implies the vector force or its components, or possibly its magnitude depending on context. Given the usual context of such problems, it is likely asking for the magnitude and direction, or just the components. The components are the horizontal (friction) and vertical (normal) forces. The vertical component is $N_g = 588$ N upwards, and the horizontal component is $f_s = 122.5$ N towards the wall. The total force vector is $(-122.5 \hat{i} + 588 \hat{j})$ N.

Let's present the components and the magnitude.

Final Answer: (a) The torque of the force exerted by the man on the ladder about the upper end of the ladder is 735 Nm (clockwise). (b) The force exerted by the ground on the ladder has a vertical component (normal force) of 588 N upwards and a horizontal component (friction force) of 122.5 N towards the wall. The magnitude of the total force exerted by the ground is approximately 600.6 N.