Two crewmen pull a boat through a lock, as shown in the figure. One crewman pulls with a force of F1 = 145 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?
F1*Cos32= F2*Cos45
F2=145*Cos32/Cos45
= 174N
Any value of F2 less than 174N would result in the net force in forward direction.
Well, it sounds like these crewmen are in quite a "lock-ful" situation! Let's see if we can help them out.
To find the force F2 that the second crewman should pull with, we need to break down the forces acting in the forward direction. We'll start by resolving the forces F1 and F2 into their horizontal and vertical components.
For the first crewman, the horizontal component of the force F1 is given by F1_h = F1 * cos(θ), where θ is the angle of 32°. So, F1_h = 145 N * cos(32°).
For the second crewman, the horizontal component of the force F2 is given by F2_h = F2 * cos(45°), since the angle is 45°.
Now, in order for the net force to be in the forward direction, the sum of the horizontal components should be positive. So, we have:
F1_h + F2_h > 0
Substituting the values, we get:
145 N * cos(32°) + F2 * cos(45°) > 0
Simplifying this equation will give us the answer for F2. You can plug the numbers in and calculate it!
Remember, it's always important to lend a "hand" when it comes to these boat-related conundrums. Good luck to the crewmen, and happy pulling!
To find the force (F2) that the second crewman should pull with so that the net force of the two crewmen is in the forward direction, we need to break down the given forces into their components.
Let's start with the force F1:
- The force F1 is 145 N.
- The angle θ is 32 degrees relative to the forward direction.
To break this force down, we can use trigonometric functions. The horizontal component (F1x) can be found using cosine, and the vertical component (F1y) can be found using sine.
F1x = F1 * cos(θ) = 145 N * cos(32°)
F1y = F1 * sin(θ) = 145 N * sin(32°)
Now let's move on to the force F2:
- The angle for the second crewman is 45 degrees.
Again, we can break down this force into its horizontal (F2x) and vertical (F2y) components using cosine and sine.
F2x = F2 * cos(45°)
F2y = F2 * sin(45°)
To find the net force in the forward direction, we need to add up the horizontal components of the forces:
Net force in the x-direction: Fnet_x = F1x + F2x
Since we want the net force to be in the forward direction, the vertical components of the forces should cancel out:
Net force in the y-direction: Fnet_y = F1y + F2y = 0
Now, let's solve for F2. Since Fnet_y must be zero, we can write the equation:
F1y + F2y = 0
Substituting the previously found expressions for F1y and F2y:
145 N * sin(32°) + F2 * sin(45°) = 0
Solving for F2:
F2 * sin(45°) = -145 N * sin(32°)
F2 = (-145 N * sin(32°)) / sin(45°)
Calculating this expression, the value of F2 is approximately -128.2 N.
Since we are looking for a positive force (magnitude), we can take the absolute value of F2:
F2 ≈ | -128.2 N | ≈ 128.2 N
Therefore, the second crewman should pull with a force of approximately 128.2 N so that the net force of the two crewmen is in the forward direction.
To find the force F2 that the second crewman should pull with, we need to consider the vector components of the forces applied by both crewmen.
Let's break down each force into its horizontal and vertical components. The force applied by the first crewman can be written as:
F1x = F1 * cos(θ)
F1y = F1 * sin(θ)
Similarly, the force applied by the second crewman can be written as:
F2x = F2 * cos(45°)
F2y = F2 * sin(45°)
Since we want the net force to be in the forward direction, the sum of the horizontal forces (F1x + F2x) should be greater than the sum of the vertical forces (F1y + F2y).
F1x + F2x > F1y + F2y
Substituting the equations for F1x, F1y, F2x, and F2y, we get:
F1 * cos(θ) + F2 * cos(45°) > F1 * sin(θ) + F2 * sin(45°)
Now we substitute the given values: F1 = 145 N and θ = 32°, and solve for F2:
145 * cos(32°) + F2 * cos(45°) > 145 * sin(32°) + F2 * sin(45°)
To solve this equation, we need the values of cos(32°), cos(45°), sin(32°), and sin(45°). We can look up these values in a trigonometric table or use a scientific calculator to find them.
Once we have these values, we can set up the equation and solve for F2:
145 * cos(32°) + F2 * cos(45°) > 145 * sin(32°) + F2 * sin(45°)
Calculate the right-hand side and then rearrange the equation to isolate F2:
F2 * cos(45°) - F2 * sin(45°) > 145 * sin(32°) - 145 * cos(32°)
Factor out F2:
F2 * (cos(45°) - sin(45°)) > 145 * (sin(32°) - cos(32°))
Finally, divide both sides of the equation by (cos(45°) - sin(45°)) to solve for F2:
F2 > 145 * (sin(32°) - cos(32°)) / (cos(45°) - sin(45°))
Plug in the values for sin(32°), cos(32°), cos(45°), and sin(45°), and evaluate the right-hand side expression to find the value of F2.