Question

Two tugboats pull a barge across the harbor. One boat exerts a force of 7.5 x 10^4N north, while the second boat exerts a force of 9.5 x 10^4N at 15 degrees north of west. Precisely, in what direction does the barge move?

Answer 1

The barge moves at an angle 47.3°North of West.

The tug boats exert forces in different directions on the barge and the barge moves in the direction of the resultant force.

Draw a diagram representing the forces, as shown in the figure attached. The tug boat 1 exerts a force F₁ along North while the tug boat 2 exerts a force F₂ at an angle of 15° North of west.

Resolve the forces along the North and the West.

The force acting along the North is given by,

$F_N=F_1+F_2sin15\\ =(7.5*10^4N)+(9.5*10^4 N*sin15)\\ =9.96*10^4N$

The force along the West is given by,

$F_W=F_2cos15=(9.5*10^4N)cos15\\ =9.18*10^4N$

The resultant force F makes an angle θ North of West.

Therefore, from the diagram,

$tan\theta =\frac{F_N}{F_W} \\ =\frac{9.96*10^4N}{9.18*10^4N} \\ =1.0849$

Hence,

$\theta =tan^-^1(1.00849)\\ =47.3^o$

Thus, the barge moves in a direction 47.3° North of West.