Question

Two blocks A and B have a weight of 10 lb and 6 lb , respectively. They are resting on the incline for which the coefficients of static friction are ?A = 0.14 and ?B= 0.24. Determine the incline angle ? for which both blocks begin to slide. Also find the required stretch or compression in the connecting spring for this to occur. The spring has a stiffness of k = 1.9 lb/ft
Part A: Determine the incline angle ? for which both blocks begin to slide.

Part B: Determine the change of the length of the spring. Assume this change to be positive if the spring is stretched and negative if the spring is compressed.

Answer 1

Answer:

Part A- $10.06^\circ$

Part B- 0.1946ft

Explanation:

Block A

Apply force equilibrium equation along x direction.

$\begin{array}{l}\\\sum {{F_x}} = 0\\\\{F_A} + {F_S} - 10\sin \theta = 0\\\end{array}$

Here, ${F_A$} is friction force on block A and ${F_S$}is spring force, and \theta is inclination angle.

By substitution

$0.14\,{N_A} + \left( {1.9x} \right)\,{\rm{lb}} \cdot {\rm{ft}} - 10\sin \theta = 0$

Similarly, apply force equilibrium equation along y direction.

$\begin{array}{l}\\\sum {{F_y}} = 0\\\\{N_A} - 10\cos \theta = 0\\\\{N_A} = 10\cos \theta \\\end{array}$

By substitution

$\begin{array}{l}\\0.14\,\left( {10\cos \theta } \right) + \left( {1.9x} \right) - 10\sin \theta = 0\\\\\left( {1.9x} \right) = 10\sin \theta - 1.4\cos \theta \\\end{array}$

Block B

Apply force equilibrium equation along x direction.

$\begin{array}{l}\\\sum {{F_x}} = 0\\\\{F_B} - {F_S} - 6\sin \theta = 0\\\end{array}$

By substitution

$0.24\,{N_A} - \left( {1.9x} \right) - 6\sin \theta = 0$

Apply force equilibrium equation along y direction.

$\begin{array}{l}\\\sum {{F_y}} = 0\\\\{N_B} - 6\cos \theta = 0\\\\{N_B} = 6\cos \theta \\\end{array}$

$\begin{array}{l}\\0.24\,\left( {6\cos \theta } \right) - \left( {1.9x} \right) - 6\sin \theta = 0\\\\\left( {1.9x} \right) = 1.44\cos \theta - 6\sin \theta \\\end{array}$

Calculate the inclination angle when both blocks begin to slide.

$\begin{array}{l}\\10\sin \theta - 1.4\cos \theta = 1.44\cos \theta - 6\sin \theta \\\\16\sin \theta = 2.84\cos \theta \\\\\tan \theta = \frac{{2.84}}{{16}}\\\\\theta = {\tan ^{ - 1}}\left( {\frac{{2.84}}{{16}}} \right)\\\end{array}$

Calculate the change in length of the spring.

$\begin{array}{l}\\\left( {1.9x} \right) = 1.44\cos \left( {10.06^\circ } \right) - 6\sin \left( {10.06^\circ } \right)\\\\1.9x = 0.3697\\\\x = 0.1946\,{\rm{ft}}\\\end{array}$