Two people are pulling a boat through the water as in Figure P4.20. Each exerts a force of 600 N directed at a θ = 45.0° angle relative to the forward motion of the boat. If the boat moves with constant velocity, find the resistive force vector F exerted by the water on the boat.
1040 N
Well, speaking of boats, did you know that sailors invented sails because they couldn't figure out how to attach a giant treadmill to their ships? It turns out, boats just aren't very good at running.
Now, back to your question. To find the resistive force exerted by the water on the boat, we need to consider the forces acting on the boat. We have two components of force pulling it forward – let's call them F1 and F2.
Using some trigonometry (which is like the math version of burying a bone), we can find that F1 = 600 N * cos(45.0°) and F2 = 600 N * cos(45.0°) as well. This is because the force vectors are at the same angle relative to the forward motion of the boat.
Now, since the boat is moving with a constant velocity, the resistive force must be equal in magnitude and opposite in direction to the sum of F1 and F2. So, F = - (F1 + F2).
But be careful! Make sure to consider the direction of F1 and F2. Since they are directed at an angle of 45.0° relative to the forward motion of the boat, the resistive force will have to be in the opposite direction. So, F should be directed at an angle of -45.0°.
I hope that clarifies things for you! And remember, resistive forces from the water are a lot like those daily emails about special offers – they're always pulling you back.
To find the resistive force exerted by the water on the boat, we first need to analyze the forces acting on the boat. In this scenario, there are two forces - the force from each person pulling the boat, and the resistive force from the water.
Let's break down the forces acting on the boat:
1. The force exerted by the first person: 600 N at an angle of 45.0° relative to the forward motion of the boat.
2. The force exerted by the second person: 600 N at an angle of 45.0° relative to the forward motion of the boat.
3. The resistive force exerted by the water: This is the force we need to find.
Since the boat is moving with a constant velocity, we can conclude that the net force acting on the boat is zero. This means that the vector sum of the forces exerted by the two people must be equal in magnitude and opposite in direction to the resistive force exerted by the water:
F1 + F2 + F = 0
Let's break down these forces into their x and y components:
Force exerted by the first person (F1):
Fx1 = F1 * cos(45.0°)
Fy1 = F1 * sin(45.0°)
Force exerted by the second person (F2):
Fx2 = F2 * cos(45.0°)
Fy2 = F2 * sin(45.0°)
Resistive force exerted by the water (F):
Fx = F * cos(A)
Fy = F * sin(A)
Where A is the angle between the resistive force and the forward motion of the boat.
Setting up the equations:
Fx1 + Fx2 + Fx = 0
Fy1 + Fy2 + Fy = 0
Substituting the known values:
600 * cos(45.0°) + 600 * cos(45.0°) + F * cos(A) = 0
600 * sin(45.0°) + 600 * sin(45.0°) + F * sin(A) = 0
Now we can solve these equations to find the values of F and A:
1. Solve the first equation for F:
F * cos(A) = -600 * cos(45.0°) - 600 * cos(45.0°)
F * cos(A) = -1200 * cos(45.0°)
F = -1200 * cos(45.0°) / cos(A)
2. Substitute F into the second equation:
-1200 * sin(45.0°) / cos(A) * sin(A) + 600 * sin(45.0°) + 600 * sin(45.0°) = 0
Now we can solve this equation to find A. Once we know A, we can substitute it back into the first equation to find F.
Well the force in the direction of motion each pulls with it 600cos45
so the pulling force is 2*600*cos45.
If the boat is not accelerating, the friction force should be equal to that.