Two crewmen pull a boat through a lock, as shown in Figure 5-25. One crewman pulls with a force of F1 = 155 N at an angle of θ = 32° relative to the forward direction of the raft. The second crewman, on the opposite side of the lock, pulls at an angle of 45°. With what force F2 should the second crewman pull so that the net force of the two crewmen is in the forward direction?
I understand I need to find the x component of them, but I don't understand what they mean by pulling it forward. Isn't it already heading in that direction?
In this scenario, when they mention "pulling it forward," they are referring to the net force being in the same direction as the forward motion of the boat. The boat might already have some forward motion due to external factors like the water current, but the crewmen want to exert a force that adds to this forward motion. They want to make sure that the net force points in the direction of the boat's forward motion to accelerate it in that direction or maintain its constant velocity.
To find the force F2 that the second crewman should exert, we can break down the forces into their x- and y-components. The x-component of the force is the component that adds to the forward motion. The y-component is perpendicular to the forward direction and does not contribute to the forward motion.
First, let's find the x-component of the force F1. To do this, we multiply the magnitude of F1 (155 N) by the cosine of the angle θ (32°):
F1x = F1 * cos(θ)
Next, let's find the x-component of the force F2. To do this, we multiply the unknown magnitude of F2 by the cosine of the angle of 45°:
F2x = F2 * cos(45°)
Since we want the net force to be in the forward direction, the sum of the x-components of both forces should be positive. Therefore, we can write the equation:
F1x + F2x > 0
Substituting the expressions for F1x and F2x:
F1 * cos(θ) + F2 * cos(45°) > 0
Now we can solve this equation for F2:
F2 > -(F1 * cos(θ)) / cos(45°)
Plugging in the given values (F1 = 155 N and θ = 32°) into the equation, we can calculate the minimum value for F2:
F2 > -(155 N * cos(32°)) / cos(45°)
Simplifying this expression will give you the minimum value for F2 that ensures the net force is in the forward direction.
In this scenario, "pulling it forward" refers to the direction in which the boat is moving. The boat is being pulled through the lock, which means it is moving in a specific direction, usually indicated by an arrow in the diagram.
To find the net force in the forward direction, you need to break down the forces applied by the two crewmen into their horizontal components.
Let's start with the force applied by the first crewman (F1):
F1x = F1 * cos(θ)
= 155 N * cos(32°)
Similarly, let's find the force applied by the second crewman (F2):
F2x = F2 * cos(45°)
Since the net force in the forward direction is desired, the x-components of F1 and F2 should add up to result in a positive value.
So the equation we can form is:
F1x + F2x > 0
Substitute the values of F1x and F2x:
155 N * cos(32°) + F2 * cos(45°) > 0
Now you can solve this equation to find the value of F2.