Step-by-Step Solution
Step 1
We are given the mass of the pipe as $M = 800\,{\rm{kg}}$, the acceleration of the truck as ${a_t} = 0.5\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}$, the angle of the cable $AB$ with the horizontal as $\phi = 45^\circ $, the radius of the pipe s $r = 0.4\,{\rm{m}}$, and the coefficient of kinetic friction between the pipe and the ground as ${\mu _k} = 0.1$.
We are asked to determine the angle $\theta $ and the tension in the cable.
Step 2
The free body diagram of the given system can be drawn as,
Step 3
Find the force produced by the pipe due to acceleration using the following relation.\[{F_p} = M{a_t}\]
On substituting the known values in the above equation we get,\[\begin{array}{c} {F_p} = \left( {800\,{\rm{kg}}} \right)\left( {0.5\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\\ = \left( {400\,{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {\frac{{1\,{\rm{N}}}}{{1\,{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \right)\\ = 400\,{\rm{N}} \end{array}\]
Step 4
Find the angle of cord $AB$ with the extended line connecting $A$ and $G$ using the following relation.\[\alpha = \phi – \theta \]
On substituting the known values in the above equation we get,\[\alpha = 45^\circ – \theta \,……\left( 1 \right)\]
Step 5
Find the relation of tension by balancing force along $x$-axis using the following relation.\[T\cos \phi – {f_C} = {F_p}\]
On substituting the known values in the above equation we get,\[\begin{array}{c} T\cos 45^\circ – {f_C} = 400\,{\rm{N}}\\ T = \frac{{400\,{\rm{N}} + {f_C}}}{{0.7071}}\,……\left( 2 \right) \end{array}\]
Step 6
Find the relation of tension by balancing force along $y$-axis using the following relation.\[T\sin \phi + {N_C} – Mg = 0\]
On substituting the known value of equation (2) in the above equation we get,\[\left( {\frac{{400\,{\rm{N}} + {f_C}}}{{0.7071}}} \right)\sin \phi + {N_C} – Mg = 0\]
On substituting the known values in the above equation we get,\[\begin{array}{c} \left( {\frac{{400\,{\rm{N}} + {f_C}}}{{0.7071}}} \right)\sin 45^\circ + {N_C} – \left( {800\,{\rm{kg}}} \right)\left( {9.81\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) = 0\\ 400\,{\rm{N}} + {f_C} + {N_C} – \left( {7848\,{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {\frac{{1\,{\rm{N}}}}{{1\,{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \right) = 0\\ {f_C} = 7448\,{\rm{N}} – {N_C}\,……\left( 3 \right) \end{array}\]
Step 7
Find the normal force at point $C$ using the following relation.\[{f_C} = {\mu _k}{N_C}\]
On substituting the known value of equation (3) in the above equation we get,\[7448\,{\rm{N}} – {N_C} = {\mu _k}{N_C}\]
On substituting the known values in the above equation we get,\[\begin{array}{c} 7448\,{\rm{N}} – {N_C} = \left( {0.1} \right){N_C}\\ {N_C} = 6771\,{\rm{N}} \end{array}\]
Step 8
Find the tension in the cord $AB$ using the following relation.\[T = \frac{{400\,{\rm{N}} + {f_C}}}{{0.7071}}\]
On substituting the known value of equation (3) in the above equation we get,\[T = \frac{{400\,{\rm{N}} + \left( {7448\,{\rm{N}} – {N_C}} \right)}}{{0.7071}}\]
On substituting the known values in the above equation we get,\[\begin{array}{c} T = \frac{{400\,{\rm{N}} + \left( {7448\,{\rm{N}} – 6771\,{\rm{N}}} \right)}}{{0.7071}}\\ = \left( {1523.12\,{\rm{N}}} \right) \times \left( {\frac{{{{10}^{ – 3}}\,{\rm{kN}}}}{{1\,{\rm{N}}}}} \right)\\ = 1.523\,{\rm{kN}} \end{array}\]
Step 9
Find the angle $\theta $ by taking moment about the centre of the pipe using the following relation.\[\left( {T\sin \alpha } \right) \times \left( r \right) – \left( {{f_C}} \right) \times \left( r \right) = 0\]
On substituting the known values of equation (1) and equation (3) in the above equation we get,\[\begin{array}{c} \left( {T\sin \left( {45^\circ – \theta } \right)} \right) \times \left( r \right) – \left( {7448\,{\rm{N}} – {N_C}} \right) \times \left( r \right) = 0\\ \left( {T\sin \alpha – 7448\,{\rm{N}} + {N_C}} \right) \times \left( r \right) = 0 \end{array}\]
On substituting the known values in the above equation we get,\[\begin{array}{c} \left( {\left( {1.523\,{\rm{kN}}} \right)\sin \left( {45^\circ – \theta } \right) – 7448\,{\rm{N}} + 6771\,{\rm{N}}} \right) \times \left( {0.4\,{\rm{m}}} \right) = 0\\ \sin \left( {45^\circ – \theta } \right) = \frac{{\left( {677\,{\rm{N}}} \right) \times \left( {\frac{{{{10}^{ – 3}}\,{\rm{kN}}}}{{1\,{\rm{N}}}}} \right)}}{{1.523\,{\rm{kN}}}}\\ \left( {45^\circ – \theta } \right) = 26.39^\circ \\ \theta = 18.61^\circ \end{array}\]