Question

Three forces in the x-y pkane act on a 5.70Kg mass. 16.90N directed at 12deg, 13.40N directed at 116deg, and 5.90N directed at 243deg. Calculate de magnitud of the acceleration. Help!!!!!!

Answer 1

please I rewrite all of your three forces using their components:
\[\begin{gathered}
{{\mathbf{F}}_{\mathbf{1}}} = \left( {16.9\cos \left( {12} \right),16.9\sin \left( {12} \right)} \right) \hfill \\
{{\mathbf{F}}_{\mathbf{2}}} = \left( {13.4\cos \left( {116} \right),13.4\sin \left( {116} \right)} \right) \hfill \\
{{\mathbf{F}}_{\mathbf{3}}} = \left( {5.9\cos \left( {243} \right),5.9\sin \left( {243} \right)} \right) \hfill \\
\end{gathered} \]

Answer 2

so, their resultant is:
\[{{\mathbf{F}}_{\mathbf{1}}} + {{\mathbf{F}}_{\mathbf{2}}} + {{\mathbf{F}}_{\mathbf{3}}}\]
and the x-component of that resultant forcre is:
\[16.9\cos \left( {12} \right) + 13.4\cos \left( {116} \right) + 5.9\cos \left( {243} \right)\]
whereas the y-component of that resultant is:

\[16.9\sin \left( {12} \right) + 13.4\sin \left( {116} \right) + 5.9\sin \left( {243} \right)\]