A boat is pulled along a straight canal by two ropes, so that the resultant pull is 1500 N along the canal (the x-axis). Rope 1 makes 30° above the x-axis and rope 2 makes 20° below the x-axis, both pulling forward. Find the tension in each rope.

Answer: \(T_1=669.7\) N, \(T_2=979.1\) N.

The resultant has no y-component, so

\(\sum F_y=0:\ T_1\sin30^\circ-T_2\sin20^\circ=0\) ...(1)

\(\sum F_x=1500:\ T_1\cos30^\circ+T_2\cos20^\circ=1500\) ...(2)

Using the triangle of forces (sine rule), the angle between the two ropes is \(30^\circ+20^\circ=50^\circ\):

\[\frac{T_1}{\sin20^\circ}=\frac{T_2}{\sin30^\circ}=\frac{R}{\sin50^\circ}=\frac{1500}{0.7660}=1958.1\]

\(T_1=1958.1\times0.3420=669.7\) N, \(T_2=1958.1\times0.5=979.1\) N.

Check (1): \(669.7(0.5)-979.1(0.342)=334.9-334.9=0\) ✓; (2): \(579.9+920.1=1500\) N ✓.