Engineering Mechanics (Dynamics)
Chapter 12
Exercise 220
Given: vB/w​=16 km/h, vW​=(−4j^​) km/h. Boat's velocity is:
vB​vB​cos70°i^+vB​sin70°j^​vB​cos70°vB​sin70°cos70°16sinθ​⋅sin70°43.96sinθ43.96⋅1−cos2θ​1932.5(1−cos2θ)2188.5cos2θ−128cosθcos2θ−0.0585cosθcosθ1,2​cosθ1​cosθ2​​=vW​+vB/W​=−4j^​+16sinθi^+16cosθj^​=16sinθ⇒vB​=cos70°16sinθ​=−4+16cosθ=−4+16cosθ=−4+16cosθ=−4+16cosθ/2=16−128cosθ+256cos2θ−1916.5=0/:2188.5−0.876=0=20.0585±0.00342+3.5​​=20.0585±1.881​=0.97θ1​=14.13°=−0.911θ2​=155.68°​
The bearing angle θ needed for the boat to stay on course is 14.1°
Answer
θ=14.1°.