A ladder of mass \(37.7 \mathrm{~kg}\) and length \(3.07 \mathrm{~m}\) is leaning against a wall at an angle \(\theta\). The coefficient of static friction between ladder and floor is 0.313 ; assume that the friction force between ladder and wall is zero. What is the maximum value that \(\theta\) can have before the ladder starts slipping?

Before solving for the maximum angle θ, we must first identify all the forces acting on the ladder. These forces include: 1. Gravity (FG): This force acts downward through the center of mass of the ladder (at the midpoint of the ladder) with a magnitude of \(37.7 kg \cdot 9.8 m/s^2\). 2. Normal force from the floor (FN1): This force acts perpendicular to and away from the floor at the base of the ladder. 3. Normal force from the wall (FN2): This force acts perpendicular to and away from the wall at the top of the ladder in the horizontal direction. 4. Friction force from the floor (FF): This force acts horizontally, opposite to the direction of the force from the wall, at the base of the ladder.

As the ladder is in static equilibrium, the net force and net torque acting on the ladder must be zero. Therefore, we can set up the following equations for equilibrium conditions: 1. \(\Sigma F_x = 0 \Rightarrow FN2 - FF = 0\) 2. \(\Sigma F_y = 0 \Rightarrow FN1 - FG = 0\) 3. \(\Sigma \tau = 0 \Rightarrow FN1\cdot \frac{L\sin \theta}{2} - FF\cdot \frac{L\cos \theta}{2} - FN2 \cdot L \sin \theta = 0\) Where L is the length of the ladder and we choose the pivot point at the base of the ladder to avoid calculating the torque due to friction (which acts at the same point).

Since the ladder is at the verge of slipping, the friction force, FF, acting at the base of the ladder is maximum. This can be expressed as: \( FF = f_s \cdot FN1 \) Where \(f_s\) denotes the coefficient of static friction, and FN1 denotes the normal force from the floor.

By substituting the first equilibrium equation into the third equation, we obtain: \(FN1\cdot \frac{L\sin \theta}{2} - FN2\cdot \frac{L\cos \theta}{2} - FN2 \cdot L \sin \theta = 0\) Now, using the second equilibrium equation \(FN1 = FG\), and making the substitution for the friction force, and canceling the common terms, we get: \(FG\cdot \sin \theta = f_s\cdot FG\cdot \cos \theta + FN2 \cdot L \sin \theta\) By substituting the given values for static friction, ladder mass, and length, the equation becomes: \( 37.7 \cdot 9.8 \cdot \sin \theta = 0.313 \cdot 37.7 \cdot 9.8 \cdot \cos \theta + FN2 \cdot 3.07 \sin \theta\) We can divide both sides of the equation by \(37.7 \cdot 9.8\) to simplify: \( \sin \theta = 0.313 \cdot \cos \theta + \frac{FN2 \cdot 3.07}{37.7 \cdot 9.8} \sin \theta\) Now, we have one equation with two unknowns, θ, and FN2. To solve for θ, express FN2 in terms of θ using the first equilibrium equation: \( FN2 = FF = f_s \cdot FN1 = 0.313 \cdot FG = 0.313 \cdot 37.7 \cdot 9.8 \) Substitute the value of FN2 into the equation from Step 4: \( \sin \theta = 0.313 \cdot \cos \theta + 0.313 \cdot 3.07 \sin \theta \) We can isolate θ by dividing both sides by \(\cos \theta\): \( \tan \theta = 0.313 + 0.313 \cdot 3.07 \sin \theta\) Now, we can use an iterative method (such as the Newton-Raphson method) or a numerical solver to solve this equation for the maximum angle, θ. Alternatively, you can solve this equation graphically.

Using a numerical solver or an iterative method to solve the equation, we get: θ ≈ 75.8° Hence, the maximum value for the angle θ that the ladder can have before it starts to slip is approximately 75.8°.