an $80\,\text{kg}$ non-uniform ladder that is $3.2\,\text{m}$ long is placed against a frictionless wall. the center of gravity of the ladder is at a point $x = 1.30\,\text{m}$ from the base of the ladder. The coefficient of static friction at the base of the ladder is $0.450$ . What is the minimum angle the ladder makes with the floor for the ladder to not slip and fall.

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Q. an

80\,\text{kg}

non-uniform ladder that is

3.2\,\text{m}

long is placed against a frictionless wall. the center of gravity of the ladder is at a point

x = 1.30\,\text{m}

from the base of the ladder. The coefficient of static friction at the base of the ladder is

0.450

. What is the minimum angle the ladder makes with the floor for the ladder to not slip and fall.

Find Torque Weight: We need to find the minimum angle at which the ladder will not slip. To do this, we will use the principles of static equilibrium, where the sum of the forces and the sum of the moments (torques) about any point must be zero. The forces acting on the ladder are the force of gravity (acting at the center of gravity), the normal force from the floor, and the frictional force at the base of the ladder. The ladder will start to slip when the torque due to the frictional force is just able to balance the torque due to the weight of the ladder.
Calculate Frictional Force: First, let's calculate the torque due to the weight of the ladder. The torque ( $\tau$ ) is given by $\tau = r \cdot F \cdot \sin(\theta)$ , where $r$ is the distance from the pivot point to the point where the force is applied, $F$ is the force, and $\theta$ is the angle between the force and the lever arm. In this case, the pivot point is at the base of the ladder, the force is the weight of the ladder ( $W = m\cdot g$ , where $m$ is the mass and $g$ is the acceleration due to gravity), and $r$ is the distance from the base to the center of gravity ( $x = 1.30\,\text{m}$ ). The angle $\theta$ is the angle between the ladder and the floor, which we are trying to find.
Calculate Torque Friction: The torque due to the weight of the ladder is $\tau_{\text{weight}} = x \cdot W \cdot \cos(\theta)$ . We use $\cos(\theta)$ because the weight acts vertically and we want the perpendicular component of the distance $x$ to the direction of the weight. The weight of the ladder is $W = mg = 80 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 784 \, \text{N}$ .
Set Equilibrium Condition: Now let's calculate the maximum static frictional force $f_s$ that can act at the base of the ladder before it starts to slip. This force is given by $f_s = \mu_s \times N$ , where $\mu_s$ is the coefficient of static friction and $N$ is the normal force. Since there are no vertical accelerations, $N = W = 784\,\text{N}$ . Therefore, $f_s = \mu_s \times N = 0.450 \times 784\,\text{N} = 352.8\,\text{N}$ .
Simplify Equation: The torque due to the frictional force is $\tau_{\text{friction}} = f_s \cdot L \cdot \sin(\theta)$ , where $L$ is the length of the ladder. However, since the frictional force acts at the base of the ladder and is parallel to the floor, the lever arm for the frictional force is actually the length of the ladder ( $L = 3.2\,\text{m}$ ). Therefore, $\tau_{\text{friction}} = 352.8\,\text{N} \cdot 3.2\,\text{m} \cdot \sin(\theta)$ .
Use Trigonometric Identity: For the ladder to be in equilibrium and not slip, the torque due to the weight must be equal to the torque due to the frictional force. Therefore, we set $\tau_{\text{weight}} = \tau_{\text{friction}}$ and solve for $\theta$ : $\newline$ $x \cdot W \cdot \cos(\theta) = f_s \cdot L \cdot \sin(\theta)$ $\newline$ $1.30 \, \text{m} \cdot 784 \, \text{N} \cdot \cos(\theta) = 352.8 \, \text{N} \cdot 3.2 \, \text{m} \cdot \sin(\theta)$
Calculate Angle: We can simplify the equation by dividing both sides by the weight of the ladder ( $784\,\text{N}$ ) and the length of the ladder ( $3.2\,\text{m}$ ): $\newline$ $\$\frac{1.30\,\text{m}}{3.2\,\text{m}}$ * $\cos(\theta)$ = \frac{ $352$ . $8$ \,\text{N}}{ $784$ \,\text{N}}\) * $\sin(\theta)$ $\newline$ $0.40625 * \cos(\theta) = 0.450 * \sin(\theta)$
Calculate Angle: We can simplify the equation by dividing both sides by the weight of the ladder $784 \, \text{N}$ and the length of the ladder $3.2 \, \text{m}$ : $\left(\frac{1.30 \, \text{m}}{3.2 \, \text{m}}\right) \cdot \cos(\theta) = \left(\frac{352.8 \, \text{N}}{784 \, \text{N}}\right) \cdot \sin(\theta)$ $0.40625 \cdot \cos(\theta) = 0.450 \cdot \sin(\theta)$ To solve for $\theta$ , we can use the trigonometric identity $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ , which gives us: $\tan(\theta) = \frac{0.450}{0.40625}$ $\theta = \arctan\left(\frac{0.450}{0.40625}\right)$
Calculate Angle: We can simplify the equation by dividing both sides by the weight of the ladder $784 \, \text{N}$ and the length of the ladder $3.2 \, \text{m}$ : $(1.30 \, \text{m} / 3.2 \, \text{m}) \cdot \cos(\theta) = (352.8 \, \text{N} / 784 \, \text{N}) \cdot \sin(\theta)$ $0.40625 \cdot \cos(\theta) = 0.450 \cdot \sin(\theta)$ To solve for $\theta$ , we can use the trigonometric identity $\tan(\theta) = \sin(\theta) / \cos(\theta)$ , which gives us: $\tan(\theta) = 0.450 / 0.40625$ $\theta = \arctan(0.450 / 0.40625)$ Now we calculate the angle $\theta$ using a calculator: $\theta = \arctan(0.450 / 0.40625) \approx \arctan(1.1077) \approx 47.8 \, \text{degrees}$