Explanation:
It is given that,
Force acting due north direction, \(F_N=390\ N\)
Force acting due east direction, \(F_E=180\ N\)
Mass of the boat, m = 270 kg
Let F is the magnitude of force acting on the sailboat. The resultant force acting on the sailboat is given by :
\(F=\sqrt{F_N^2+F_E^2}\)
\(F=\sqrt{390^2+180^2}\)
F = 429.53 N
Since, F = ma
\(a=\dfrac{F}{m}\)
\(a=\dfrac{429.53\ N}{270\ kg}\)
\(a=1.59\ m/s^2\)
Let \(\theta\) is the direction of the boat's acceleration. It is given by :
\(tan\theta=\dfrac{F_N}{F_E}\)
\(tan\theta=\dfrac{390}{180}\)
\(\theta=65.22^{\circ}\)
Hence, this is the required solution.
