Question

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A that splits at point B and attaches to the ship at points C and . The two rope segments BC and BD angle away from the center of the ship at angles o 350 and -5 respective The boat pulls with a re o 295 0 What are the tensions Tec and TBD in the rope segments BC and BD? Express your answers numerically in pounds to three significant figures separated by a comma. View Available Hint(s) vec lb

Answer 1

As the, tugboat is moving with constant velocity. So, F = T_BC cos 15 degree + T_BD cos 35 degree equation(1) T_BD sin 35 degree = T_BC sin 15 degree equation(2) from (2): T_BD (0.57) = T_BC (0.26) T_BD = 0.451 T_BC substituting T_BD in eqn (1) we get: \r\n F = T_BC sin 35 degree + 0.451 T_BC cos 35 degree F = T_BC[sin 35 degree + 0.451 cos 35 degree] T_BC (0.943) = F T_BC = (F/0.943) T_BC = 2950 lb/0.943 T_BC = 3128.314 lb T_BD = 0.46 times T_BC = 0.46 times 3128.314 T_BD = 1439.024 lb writing neatly: T_BC = 3128.314 lb: T_BD = 1439.024 lb