Solve question 2 in English

Okay Nike🙃, let's solve question 2.

Here's the information given for the non-uniform ladder:

• Length of ladder $L = 8.0 \text{ m}$
• Mass of ladder $m = 10.0 \text{ kg}$
• Center of gravity (CG) from the bottom B: $d_{CG} = 2.0 \text{ m}$
• Angle with the vertical wall: $\alpha = 30^\circ$. This means the angle with the horizontal ground is $\theta = 90^\circ - 30^\circ = 60^\circ$.
• Upper end (A) against a smooth wall (no friction).
• Lower end (B) on rough ground (friction present).
• Acceleration due to gravity $g = 9.81 m/s^2$.

a) Draw a free body diagram of the ladder.

Imagine the ladder AB leaning against a vertical wall with its base on horizontal ground.

• Draw the ladder inclined at $60^\circ$ to the horizontal. Label the top end A and the bottom end B.
• Weight (W): A vertical arrow pointing downwards from the center of gravity G. G is located 2.0 m from B along the ladder. The magnitude is $W = mg$.
• Normal force at the wall ($N_A$): A horizontal arrow pointing away from the wall (to the right), acting at point A. Since the wall is smooth, there is no friction here.
• Normal force at the ground ($N_B$): A vertical arrow pointing upwards, acting at point B.
• Frictional force at the ground ($F_f$): A horizontal arrow pointing towards the wall (to the left), acting at point B. This force opposes the tendency of the ladder to slip away from the wall.

b) Calculate:

First, let's calculate the weight of the ladder: $W = mg = (10.0 kg)(9.81 m/s^2) = 98.1 N$

For the ladder to be in equilibrium, the net force and net torque must be zero.

• Sum of forces in the horizontal (x) direction = 0: $N_A - F_f = 0 \implies N_A = F_f \quad (Equation 1)$
• Sum of forces in the vertical (y) direction = 0: $N_B - W = 0 \implies N_B = W = 98.1 N \quad (Equation 2)$
• Sum of moments about point B (bottom of the ladder) = 0: Taking moments about B eliminates $N_B$ and $F_f$ from the moment equation. The weight $W$ creates a clockwise moment. The perpendicular distance from B to the line of action of W is $d_{CG} \cos(\theta) = (2.0 m) \cos(60^\circ)$. The normal force $N_A$ creates an anti-clockwise moment. The perpendicular distance from B to the line of action of $N_A$ is $L \sin(\theta) = (8.0 m) \sin(60^\circ)$. $W (d_{CG} \cos(60^\circ)) = N_A (L \sin(60^\circ))$ $(98.1 N) (2.0 m \times 0.5) = N_A (8.0 m \times \frac{\sqrt{3}}{2})$ $98.1 = N_A (4.0 \sqrt{3})$ $N_A = \frac{98.1}{4.0 \sqrt{3}} = \frac{98.1}{6.9282} \approx 14.159 N$

i) The frictional force between the ladder and the ground ($F_f$). From Equation 1, $F_f = N_A$. $F_f = 14.159 N$ Rounding to three significant figures: $F_f = 14.2 \text{ N}$

ii) The resultant force at the bottom of the ladder. The resultant force at the bottom (point B) is the vector sum of the normal force $N_B$ and the frictional force $F_f$. Since these forces are perpendicular, we use the Pythagorean theorem. $R_B = \sqrt{N_B^2 + F_f^2}$ $R_B = \sqrt{(98.1 N)^2 + (14.159 N)^2}$ $R_B = \sqrt{9623.61 + 200.477} = \sqrt{9824.087} \approx 99.116 N$ Rounding to three significant figures: $R_B = 99.1 \text{ N}$

iii) The coefficient of friction $\mu$. Assuming the ladder is in limiting equilibrium (or we are finding the minimum coefficient of static friction required to prevent slipping), the frictional force is given by $F_f = \mu N_B$. $\mu = \frac{F_f}{N_B}$ $\mu = \frac{14.159 N}{98.1 N} \approx 0.14433$ Rounding to three significant figures: $\mu = 0.144$

iv) The angle of friction $\theta$. The angle of friction $\theta$ is the angle whose tangent is the coefficient of friction. $\tan \theta = \mu$ $\theta = \arctan(\mu) = \arctan(0.14433)$ $\theta \approx 8.209^\circ$ Rounding to one decimal place: $\theta = 8.2^\circ$