Question

problem 5: two sailors are pulling a barge through a canal from both sides. one sailor is pulling with a force of 300 n at an angle of 37°. the second sailor exerts a force at an angle of -48°. find the magnitude of the force exerted by the second sailor if the barge is moving straight at 0 angle.

Explanation:

Step1: Resolve forces in the y - direction

Since the barge is moving straight (net force in y - direction is 0), for the first force $F_1 = 300N$ at an angle $\theta_1=37^{\circ}$ and the second force $F_2$ at an angle $\theta_2 = - 48^{\circ}$, the sum of the y - components of the forces is zero. The y - component of a force $F$ is given by $F_y=F\sin\theta$. So, $F_1\sin\theta_1+F_2\sin\theta_2 = 0$.
$300\sin37^{\circ}+F_2\sin(-48^{\circ})=0$

Step2: Solve for $F_2$

First, note that $\sin(-48^{\circ})=-\sin48^{\circ}$. Then we can rewrite the equation as $300\sin37^{\circ}-F_2\sin48^{\circ}=0$. Rearranging to solve for $F_2$ gives $F_2=\frac{300\sin37^{\circ}}{\sin48^{\circ}}$.
We know that $\sin37^{\circ}\approx0.6$ and $\sin48^{\circ}\approx0.743$.
$F_2=\frac{300\times0.6}{0.743}=\frac{180}{0.743}\approx242.26N$

Answer:

$242.26N$