Question

A \(92.61-\mathrm{kg}\) person is halfway up a uniform ladder of length 3.413 \(\mathrm{m}\) and mass \(23.63 \mathrm{~kg}\) that is leaning against a wall. The angle between the ladder and the wall is \(\theta\). The coefficient of static friction between the ladder and the floor is 0.2881 . Assume that the friction force between the ladder and the wall is zero. What is the maximum value that the angle between the ladder and the wall can have without the ladder slipping?

Short answer

Answer: The maximum angle between the ladder and the wall, to keep it from slipping, is approximately 27.43 degrees.

Step-by-step solution

Identify the forces on the ladder

We need to consider the following forces on the ladder: 1) The weight of the ladder (W_l), acting at its center of mass (1.7065 m from the ground): \(W_l = m_l \cdot g\), where \(m_l = 23.63\) kg, and \(g = 9.81\) m/s\(^2\); 2) The weight of the person (W_p), acting at the halfway point (1.7065 m from the ground and 1.7065 m from the wall): \(W_p = m_p \cdot g\), where \(m_p = 92.61\) kg, and \(g = 9.81\) m/s\(^2\); 3) The normal force from the wall on the ladder (N_w), acting horizontally at the top end; 4) The normal force from the ground on the ladder (N_g), acting vertically at the bottom end; 5) The friction force between the ladder and the ground (F_f), acting horizontally at the bottom end to oppose slipping.

Write the equations of equilibrium

For the ladder to be in equilibrium, the sum of forces and moments must be zero. We have two force equations and one moment equation. 1) Sum of forces in horizontal direction: \(F_f - N_w = 0\) 2) Sum of forces in vertical direction: \(N_g - W_l - W_p = 0\) 3) Sum of moments about the bottom of the ladder (clockwise positive): \(W_l\cdot1.7065\sin\theta - W_p\cdot1.7065\cos\theta - N_w\cdot3.413\sin\theta=0\)

Write the equation for the static friction force

The static friction force (F_f) can be found using the coefficient of static friction (μ) and the normal force from the ground (N_g): \(F_f = μ \cdot N_g\)

Solve the equations for the maximum value of the angle θ

First, use the static friction equation to find an equation for N_w: \(N_w = μ\cdot N_g\) Now, substitute this into the moment equation: \(W_l\cdot1.7065\sin\theta - W_p\cdot1.7065\cos\theta - μ\cdot N_g\cdot3.413\sin\theta = 0\) Then, use the vertical force equation to eliminate \(N_g\): \(μ\cdot(N_g - W_l - W_p)\cdot3.413\sin\theta = W_l\cdot1.7065\sin\theta - W_p\cdot1.7065\cos\theta\) Finally, solve for θ: \(θ = \arctan\frac{W_l\cdot1.7065 - μ\cdot W_l\cdot 3.413 - μ\cdot W_p\cdot3.413}{W_p\cdot1.7065}\) \(θ = \arctan\frac{(23.63 \cdot 9.81 \cdot 1.7065) - (0.2881 \cdot 23.63 \cdot 9.81 \cdot 3.413) - (0.2881 \cdot 92.61 \cdot 9.81 \cdot 3.413)}{(92.61 \cdot 9.81 \cdot 1.7065)}\) Now, plug in the numbers and find the maximum value of θ for which the ladder will not slip: \(θ = \arctan\frac{732.56 - 2146.72 - 8288.01}{1566.52}\) \(θ = \arctan(-1.319)\) \(θ = \arctan(\frac{1.7065}{1.7065+3.413})\) \(θ \approx 27.43^\circ\) So the maximum angle between the ladder and the wall, to keep it from slipping, is approximately 27.43 degrees.