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

735 Nm, √(122.5)^2 + (588)^2 N

Answer 2

Let L be the length of the ladder, L = 6.5 m.
Let h be the height reached on the wall, h = 6.0 m.
Let x be the distance of the base of the ladder from the wall. By Pythagorean theorem, $L^2 = h^2 + x^2$.
$x^2 = (6.5)^2 - (6.0)^2 = 42.25 - 36.00 = 6.25$.
$x = \sqrt{6.25} = 2.5$ m.

Let A be the upper end of the ladder (on the wall) and B be the lower end (on the ground).
Let's set up a coordinate system with the origin at the base of the wall on the ground. The ground is the x-axis and the wall is the y-axis.
The coordinates of A are (0, h) = (0, 6.0).
The coordinates of B are (x, 0) = (2.5, 0).

The man of mass $m_m = 60$ kg stands halfway up the ladder. Let M be the position of the man. M is the midpoint of AB.
The coordinates of M are $(\frac{0+2.5}{2}, \frac{6.0+0}{2}) = (1.25, 3.0)$.
The force exerted by the man on the ladder is his weight, $W_m = m_m g$. Let's use $g = 9.8$ m/s$^2$.
$W_m = 60 \times 9.8 = 588$ N.
The force vector is $\vec{F}_m = (0, -W_m) = (0, -588)$ N.

(a) Find the torque of the force exerted by the man on the ladder about the upper end of the ladder.
The point about which the torque is calculated is A (0, 6.0).
The position vector from A to M is $\vec{r}_{AM} = M - A = (1.25 - 0, 3.0 - 6.0) = (1.25, -3.0)$ m.
The torque $\vec{\tau}_A$ is given by $\vec{\tau}_A = \vec{r}_{AM} \times \vec{F}_m$.
$\vec{\tau}_A = (1.25 \hat{i} - 3.0 \hat{j}) \times (-588 \hat{j})$
$\vec{\tau}_A = (1.25 \hat{i}) \times (-588 \hat{j}) + (-3.0 \hat{j}) \times (-588 \hat{j})$
$\vec{\tau}_A = 1.25 \times (-588) (\hat{i} \times \hat{j}) + 0$
$\vec{\tau}_A = -735 \hat{k}$ Nm.
The magnitude of the torque is $|\vec{\tau}_A| = 735$ Nm. The direction is clockwise (into the page).

(b) Find the force exerted by the ground on the ladder.
Let's consider the forces acting on the ladder:

1. Weight of the man $W_m = 588$ N acting vertically downwards at M.
2. Weight of the ladder $W_l$, which is negligible ($W_l \approx 0$).
3. Force exerted by the wall on the ladder at A. Since the wall is smooth, this force is horizontal. Let it be $\vec{N}_A = (N_A, 0)$, acting away from the wall (positive x direction).
4. Force exerted by the ground on the ladder at B. This force has a vertical component (normal force $\vec{N}_B$) and a horizontal component (friction force $\vec{f}_B$). Let the total force be $\vec{F}_B = \vec{f}_B + \vec{N}_B$. $\vec{N}_B = (0, N_B)$ acting upwards (positive y direction), and $\vec{f}_B = (f_B, 0)$ acting towards the wall (positive x direction) to prevent slipping. So $\vec{F}_B = (f_B, N_B)$.

The ladder is in equilibrium, so the net force is zero and the net torque about any point is zero.
Force equilibrium:
Horizontal: $N_A - f_B = 0 \implies N_A = f_B$.
Vertical: $N_B - W_m = 0 \implies N_B = W_m = 588$ N.

Torque equilibrium: Let's take torque about point B to eliminate the force $\vec{F}_B$.
The forces producing torque about B are $W_m$ and $N_A$.
The position vector from B to M is $\vec{r}_{BM} = M - B = (1.25 - 2.5, 3.0 - 0) = (-1.25, 3.0)$ m.
The force $\vec{F}_m = (0, -588)$ N.
The torque due to $W_m$ about B is $\vec{\tau}_m = \vec{r}_{BM} \times \vec{F}_m = (-1.25 \hat{i} + 3.0 \hat{j}) \times (-588 \hat{j})$.
$\vec{\tau}_m = (-1.25 \hat{i}) \times (-588 \hat{j}) + (3.0 \hat{j}) \times (-588 \hat{j}) = (-1.25)(-588) (\hat{i} \times \hat{j}) + 0 = 735 \hat{k}$ Nm. (Counter-clockwise)

The position vector from B to A is $\vec{r}_{BA} = A - B = (0 - 2.5, 6.0 - 0) = (-2.5, 6.0)$ m.
The force $\vec{N}_A = (N_A, 0) = N_A \hat{i}$.
The torque due to $N_A$ about B is $\vec{\tau}_A = \vec{r}_{BA} \times \vec{N}_A = (-2.5 \hat{i} + 6.0 \hat{j}) \times (N_A \hat{i})$.
$\vec{\tau}_A = (-2.5 \hat{i}) \times (N_A \hat{i}) + (6.0 \hat{j}) \times (N_A \hat{i}) = 0 + (6.0)(N_A) (\hat{j} \times \hat{i}) = 6.0 N_A (-\hat{k}) = -6.0 N_A \hat{k}$ Nm. (Clockwise)

For equilibrium, the net torque about B is zero: $\vec{\tau}_m + \vec{\tau}_A = 0$.
$735 \hat{k} - 6.0 N_A \hat{k} = 0$.
$735 - 6.0 N_A = 0$.
$N_A = \frac{735}{6.0} = 122.5$ N.

From horizontal force equilibrium, $f_B = N_A = 122.5$ N.
From vertical force equilibrium, $N_B = 588$ N.
The force exerted by the ground on the ladder is $\vec{F}_B = (f_B, N_B) = (122.5, 588)$ N.
The magnitude of this force is $|\vec{F}_B| = \sqrt{(122.5)^2 + (588)^2}$.
$|\vec{F}_B| = \sqrt{15006.25 + 345744} = \sqrt{360750.25}$.
$|\vec{F}_B| \approx 600.625$ N.

The force exerted by the ground is a vector $(122.5 \hat{i} + 588 \hat{j})$ N, or its magnitude is $\sqrt{(122.5)^2 + (588)^2}$ N. The question asks for "the force", which usually implies the vector or its magnitude depending on context. Given the context of forces in mechanics, providing the components or the magnitude and direction is appropriate. Let's provide the components and the magnitude.

Force exerted by the ground = $(122.5 \hat{i} + 588 \hat{j})$ N.
Magnitude of the force = $\sqrt{(122.5)^2 + (588)^2}$ N.

Let's re-calculate the magnitude.
$122.5 = \frac{245}{2}$. $588 = \frac{1176}{2}$.
$|\vec{F}_B| = \sqrt{(\frac{245}{2})^2 + (\frac{1176}{2})^2} = \frac{1}{2} \sqrt{245^2 + 1176^2}$.
$245 = 5 \times 49 = 5 \times 7^2$.
$1176 = 2 \times 588 = 2^2 \times 294 = 2^3 \times 147 = 2^3 \times 3 \times 49 = 2^3 \times 3 \times 7^2$.
$245^2 = (5 \times 7^2)^2 = 5^2 \times 7^4 = 25 \times 2401 = 60025$.
$1176^2 = (2^3 \times 3 \times 7^2)^2 = 2^6 \times 3^2 \times 7^4 = 64 \times 9 \times 2401 = 576 \times 2401 = 1382976$.
$245^2 + 1176^2 = 60025 + 1382976 = 1443001$.
$|\vec{F}_B| = \frac{1}{2} \sqrt{1443001}$.
$\sqrt{1443001} \approx 1201.25$.
$|\vec{F}_B| \approx \frac{1201.25}{2} \approx 600.625$ N.

Let's check the angle of the ladder with the ground. Let it be $\theta$.
$\tan \theta = \frac{h}{x} = \frac{6.0}{2.5} = \frac{12}{5}$.
$\sin \theta = \frac{h}{L} = \frac{6.0}{6.5} = \frac{12}{13}$.
$\cos \theta = \frac{x}{L} = \frac{2.5}{6.5} = \frac{5}{13}$.

Let's recalculate torque about B using lever arms.
Horizontal distance of M from the wall is $x_M = 1.25$ m. This is the horizontal distance from B to the line of action of $W_m$.
Torque due to $W_m$ about B = $W_m \times x_M = 588 \times 1.25 = 735$ Nm (counter-clockwise).
Vertical height of A from the ground is $y_A = 6.0$ m. This is the vertical distance from B to the line of action of $N_A$.
Torque due to $N_A$ about B = $N_A \times y_A = N_A \times 6.0$ Nm (clockwise).
Equating magnitudes: $N_A \times 6.0 = 735 \implies N_A = \frac{735}{6} = 122.5$ N.

This confirms $f_B = 122.5$ N and $N_B = 588$ N.
The force exerted by the ground is the vector sum of the friction and the normal force.
$\vec{F}_B = \vec{f}_B + \vec{N}_B$.

Final Answer Check:
(a) Torque about the upper end A.
$\vec{r}_{AM} = \frac{1}{2} \vec{r}_{AB}$. $\vec{r}_{AB} = B - A = (2.5, -6.0)$.
$\vec{r}_{AM} = (1.25, -3.0)$. This matches.
Force $\vec{F}_m = (0, -588)$.
$\vec{\tau}_A = (1.25, -3.0) \times (0, -588) = (1.25)(-588) - (-3.0)(0) = -735$ (in z direction).
Magnitude is 735 Nm. This matches.

(b) Force by ground.
$N_B = W_m = 588$ N (upwards).
$f_B = N_A$.
Torque about B: $W_m \times (L/2 \cos \theta) - N_A \times (L \sin \theta) = 0$.
Wait, the angle $\theta$ is with the horizontal.
Horizontal distance from B to M is $(L/2) \cos \theta = (6.5/2) \times (5/13) = 3.25 \times 5/13 = (13/4) \times 5/13 = 5/4 = 1.25$ m. This matches $x_M$.
Vertical height of A from B is $L \sin \theta = 6.5 \times (12/13) = 0.5 \times 12 = 6.0$ m. This matches $y_A$.
So, $W_m \times 1.25 - N_A \times 6.0 = 0$.
$588 \times 1.25 = N_A \times 6.0$.
$735 = N_A \times 6.0$.
$N_A = \frac{735}{6} = 122.5$ N.
$f_B = N_A = 122.5$ N (towards the wall).
$N_B = W_m = 588$ N (upwards).
The force exerted by the ground is the vector sum of friction and normal force.
$\vec{F}_B = (122.5 \hat{i} + 588 \hat{j})$ N.
Magnitude $|\vec{F}_B| = \sqrt{(122.5)^2 + (588)^2} = \sqrt{15006.25 + 345744} = \sqrt{360750.25} \approx 600.625$ N.

The question asks for "the force exerted by the ground". This could mean the vector force or its magnitude. Given the context of typical problems, the components or the magnitude are usually expected.

Final Answer for (a): Magnitude of torque is 735 Nm.
Final Answer for (b): The force exerted by the ground has a horizontal component of 122.5 N (towards the wall) and a vertical component of 588 N (upwards). The magnitude of this force is $\sqrt{(122.5)^2 + (588)^2} \approx 600.6$ N.

Let's provide the magnitude as the final answer for (b) unless specified otherwise.