Question

Two canal workers pull a barge along the narrow waterway at a constant speed. One worker pulls with a force of \(105 \mathrm{N}\) at an angle of \(28^{\circ}\) with respect to the forward motion of the barge and the other worker, on the opposite tow path, pulls at an angle of \(38^{\circ}\) relative to the barge motion. Both ropes are parallel to the ground. (a) With what magnitude force should the second worker pull to make the sum of the two forces be in the forward direction? (b) What is the magnitude of the force on the barge from the two tow ropes?

Short answer

Answer: (a) The second worker should pull with a force of approximately \(72.63 \mathrm{N}\), and (b) the total magnitude of the force on the barge from the two tow ropes is approximately \(148.39 \mathrm{N}\).

Step-by-step solution

Understand the problem and draw a diagram

To begin, it's always good to visualize the problem by drawing a diagram. Here, we can represent the two workers, the barge, and the forces exerted by them on the barge, along with the respective angles.

Analyze the forces acting on the barge

Next, we need to analyze the forces acting on the barge and separate them into their horizontal and vertical components. Let F be the force exerted by the second worker. In order for the total force to remain in the forward direction, the vertical components of the forces by the first and second workers must be equal, canceling out each other's effects.

Separate the forces into their components

Now, we will separate the forces into their horizontal and vertical components. For the first worker, who pulls with a force of \(105 \mathrm{N}\) at an angle of \(28^{\circ}\), the horizontal and vertical components are: Horizontal component: \(105 \cos(28^{\circ})\) Vertical component: \(105 \sin(28^{\circ})\) For the second worker, with a force of \(F\) and an angle of \(38^{\circ}\): Horizontal component: \(F \cos(38^{\circ})\) Vertical component: \(F \sin(38^{\circ})\)

Equate the vertical components to find the magnitude of F

In order for the forces to sum in the forward direction, the vertical components must be equal, which means: \(105 \sin(28^{\circ}) = F \sin(38^{\circ})\) To find F, we can divide both sides by \(\sin(38^{\circ})\): \(F = \frac{105 \sin(28^{\circ})}{\sin(38^{\circ})} \approx 72.63 \mathrm{N}\)

Find the total horizontal force

Now we need to find the total horizontal force on the barge from the two tow ropes. The horizontal components of both forces acting on the barge sum together: Total horizontal force = \(105 \cos(28^{\circ}) + F \cos(38^{\circ}) \approx 105 \cos(28^{\circ}) + 72.63 \cos(38^{\circ}) \approx 148.39 \mathrm{N}\) So, (a) the second worker should pull with a force of approximately \(72.63 \mathrm{N}\) and (b) the total magnitude of the force on the barge from the two tow ropes is approximately \(148.39 \mathrm{N}\).