A uniform rod \( KJ \) of weight \( w \) shown in the figure rests against a frictionless vertical wall at the point \( K \) and a rough horizontal surface at point \( J \). It is given that \( w = 10 \, \text{kN} \), \( a = 4 \, \text{m} \), and \( b = 3 \, \text{m} \).
The minimum coefficient of static friction that is required at the point \( J \) to hold the rod in equilibrium is _________ (round off to three decimal places).
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The static friction required to hold a rod in equilibrium is determined by balancing moments about a pivot point, using the rod's geometry and weight.
For the rod to be in equilibrium, the sum of forces and the sum of moments must be zero. Considering the forces acting on the rod:
1. The weight \( w = 10 \, \text{kN} \) acts vertically downward through the center of mass of the rod.
2. The friction force \( f \) at point \( J \) acts horizontally, and the normal force \( N \) at point \( J \) acts vertically upward.
To find the minimum static friction coefficient \( \mu \), we first use the moment equation about point \( K \) (taking counterclockwise as positive):
\[
\text{Moment due to } w = w \times \frac{a}{2}
\]
The moment due to friction \( f \) is:
\[
\text{Moment due to } f = f \times b
\]
For equilibrium:
\[
w \times \frac{a}{2} = f \times b
\]
Substitute \( f = \mu N \), and \( N = w \), as the vertical forces balance:
\[
w \times \frac{a}{2} = \mu w \times b
\]
Solve for \( \mu \):
\[
\mu = \frac{a}{2b}
\]
Substitute the given values \( a = 4 \, \text{m} \) and \( b = 3 \, \text{m} \):
\[
\mu = \frac{4}{2 \times 3} = \frac{4}{6} = 0.6667
\]
Thus, the minimum coefficient of static friction is:
\[
\boxed{0.667}
\]
