Step-by-Step Solution
Step 1
We are given the force ${F_R} = {F_1} + {F_2}$,${F_1} = 250\;{\rm{lb}}$ and ${F_2} = 375\;{\rm{lb}}$.
We are asked to determine the magnitude and direction of the resultant force.
Step 2
The following is the diagram to calculate the projected component of the force.
To find the magnitude of the resultant force we will use the formula,\[{F_R} = \sqrt {{F_1}^2 + {F_1}^2 – 2{F_1}{F_2}\cos \theta } \]
On plugging the values in the above relation we get,\[\begin{array}{l} {F_R} = \sqrt {{{\left( {250\;{\rm{lb}}} \right)}^2} + {{\left( {375\;{\rm{lb}}} \right)}^2} – 2\left( {250\;{\rm{lb}}} \right)\left( {375\;{\rm{lb}}} \right)\cos 75^\circ } \\ {F_R} = \sqrt {\left( {203125\;{\rm{lb}}} \right) – \left( {48528.57\;{\rm{lb}}} \right)} \\ {F_R} = 393.2\;{\rm{lb}} \end{array}\]
Step 3
To find the direction of the resultant force we will use the law of sines,\[\frac{{{F_R}}}{{\sin 75^\circ }} = \frac{{{F_1}}}{{\sin \phi }}\]
On plugging the values in the above relation we get,\[\begin{array}{c} \frac{{393.2\;{\rm{lb}}}}{{\sin 75^\circ }} = \frac{{250\;{\rm{lb}}}}{{\sin \phi }}\\ \sin \phi = \frac{{\left( {250\;{\rm{lb}}} \right)}}{{\left( {407.07\;{\rm{lb}}} \right)}}\\ \phi = 37.9^\circ \end{array}\]
The direction of resultant force is calculated as,\[\begin{array}{l} {\theta _R} = 360^\circ – 45^\circ + \phi \\ {\theta _R} = 360^\circ – 45^\circ + 37.9^\circ \\ {\theta _R} = 352.9^\circ \left( {{\rm{CCW}}} \right) \end{array}\]