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Chapter 8: Problem 71

A uniform ladder of length \(L\) and weight \(w\) is leaning against a vertical wall. The coefficient of static friction between the ladder and the floor is the same as that between the ladder and the wall. If this coefficient of static friction is \(\mu_{s}=0.500\), determine the smallest angle the ladder can make with the floor without slipping.

Short Answer

Expert verified
The smallest angle \(\theta\) the ladder can make with the floor without slipping, when \(\mu_{s} = 0.500\), is \(\theta = \arctan(0.500)\) radians or approximately 27 degrees.

Step by step solution

01

Setup the problem

Consider a ladder of length \(L\) leaning against a wall. The ladder makes an angle \(\theta\) with the floor. The forces acting on the ladder include its weight \(w\), acting at the middle of the ladder and friction forces at the bottom and top ends of the ladder, \(f_{1}\) and \(f_{2}\) respectively. Due to static friction, \(f_{1}=\mu_{s}N_{1}\) and \(f_{2}=\mu_{s}N_{2}\) where \(N_{1}\) and \(N_{2}\) are normal forces acting at the bottom and top ends respectively.
02

Apply Net Force Conditions

By considering that the sum of forces in vertical and horizontal directions must be zero, we get: \n - In vertical direction: \(N_{1} - f_{2} - w = 0\), which simplifies to \(N_{1} = w + \mu_{s}N_{2}\) \n - In horizontal direction: \(f_{1} - N_{2} = 0\), which simplifies to \(N_{2} = \mu_{s}N_{1}\)
03

Apply Net Torque Conditions

With the sum of torques about the bottom end of the ladder equal to zero, we can write: \(-f_{2}L + w \frac{L}{2} = 0\), and substituting for \(f_{2}\) gives \(wL(1/2 - \mu_{s}) = 0\)
04

Solve the equations

By substituting values \(N_{2} = \mu_{s}N_{1}\) and \(N_{1} = w + \mu_{s}N_{2}\) into condition \(wL(1/2 - \mu_{s}) = 0\), and solving, we can find the value of \(\theta\). This yields \(\theta = \arctan(\mu_{s})\)
05

Substitute given values

Finally, substituting the given coefficient of static friction \(\mu_{s} = 0.500\) into the equation \(\theta = \arctan(\mu_{s})\) gives us \(\theta = \arctan(0.500)\)

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