Answer:
Part A- [tex]10.06^\circ[/tex]
Part B- 0.1946ft
Explanation:
Block A
Apply force equilibrium equation along x direction.
[tex]\begin{array}{l}\\\sum {{F_x}} = 0\\\\{F_A} + {F_S} - 10\sin \theta = 0\\\end{array}[/tex]
Here, [tex]{F_A[/tex]} is friction force on block A and [tex]{F_S[/tex]}is spring force, and \theta is inclination angle.
By substitution
[tex]0.14\,{N_A} + \left( {1.9x} \right)\,{\rm{lb}} \cdot {\rm{ft}} - 10\sin \theta = 0[/tex]
Similarly, apply force equilibrium equation along y direction.
[tex]\begin{array}{l}\\\sum {{F_y}} = 0\\\\{N_A} - 10\cos \theta = 0\\\\{N_A} = 10\cos \theta \\\end{array}[/tex]
By substitution
[tex]\begin{array}{l}\\0.14\,\left( {10\cos \theta } \right) + \left( {1.9x} \right) - 10\sin \theta = 0\\\\\left( {1.9x} \right) = 10\sin \theta - 1.4\cos \theta \\\end{array}[/tex]
Block B
Apply force equilibrium equation along x direction.
[tex]\begin{array}{l}\\\sum {{F_x}} = 0\\\\{F_B} - {F_S} - 6\sin \theta = 0\\\end{array}[/tex]
By substitution
[tex]0.24\,{N_A} - \left( {1.9x} \right) - 6\sin \theta = 0[/tex]
Apply force equilibrium equation along y direction.
[tex]\begin{array}{l}\\\sum {{F_y}} = 0\\\\{N_B} - 6\cos \theta = 0\\\\{N_B} = 6\cos \theta \\\end{array}[/tex]
[tex]\begin{array}{l}\\0.24\,\left( {6\cos \theta } \right) - \left( {1.9x} \right) - 6\sin \theta = 0\\\\\left( {1.9x} \right) = 1.44\cos \theta - 6\sin \theta \\\end{array}[/tex]
Calculate the inclination angle when both blocks begin to slide.
[tex]\begin{array}{l}\\10\sin \theta - 1.4\cos \theta = 1.44\cos \theta - 6\sin \theta \\\\16\sin \theta = 2.84\cos \theta \\\\\tan \theta = \frac{{2.84}}{{16}}\\\\\theta = {\tan ^{ - 1}}\left( {\frac{{2.84}}{{16}}} \right)\\\end{array}[/tex]
Calculate the change in length of the spring.
[tex]\begin{array}{l}\\\left( {1.9x} \right) = 1.44\cos \left( {10.06^\circ } \right) - 6\sin \left( {10.06^\circ } \right)\\\\1.9x = 0.3697\\\\x = 0.1946\,{\rm{ft}}\\\end{array}[/tex]
