Finding Drag w/ acceleration & Boyant force

In summary, the conversation discusses solving for the drag on a speed boat that is accelerating at 5 m/s/s, given the Boyant force and thrust of the boat. The net force equation is used to determine the unknown force of friction, with the final solution being Ff = 1876 - (320)5.
  • #1
swede5670
78
0

Homework Statement


A speed boat is accelerating at 5 m/s/s. If the Boyant force (the force the prevents the boat from sinking) is 3140 N and the thrust that the motor exerts on the boat is 1876 N, what is the drag on the boat (drag in a fluid is a like friction on land).


Homework Equations


Fnet = ma


The Attempt at a Solution


We just started getting 2 part equations and I'm a little confused on what I have to solve for first. Would someone help me figure out what to solve for first?
I was thinking that since the boyant force is 3140N going up then gravity must be pulling 3,140N down right? So then the acceleration downwards is 9.81 m/s/s, so 3,140N = X * 9.81 then divide 3,140 by 9.81 and you get 320.081549kg. Does this help at all?
Fnet = ma
Fnet = (320kg * 5m/s/s)
If that is correct, what do I use for Fnet
 
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  • #2
swede5670 said:

Homework Statement


A speed boat is accelerating at 5 m/s/s. If the Boyant force (the force the prevents the boat from sinking) is 3140 N and the thrust that the motor exerts on the boat is 1876 N, what is the drag on the boat (drag in a fluid is a like friction on land).


Homework Equations


Fnet = ma


The Attempt at a Solution


We just started getting 2 part equations and I'm a little confused on what I have to solve for first. Would someone help me figure out what to solve for first?
I was thinking that since the boyant force is 3140N going up then gravity must be pulling 3,140N down right? So then the acceleration downwards is 9.81 m/s/s, so 3,140N = X * 9.81 then divide 3,140 by 9.81 and you get 320.081549kg. Does this help at all?
Fnet = ma
Fnet = (320kg * 5m/s/s)
If that is correct, what do I use for Fnet
Yes looks good so far. The boat is accelerating in the x direction. The net force is the algebraic sum of the forces acting on the boat in the x direction. What are those forces? (one is given, the other is unknown; you need to solve for it).
 
  • #3
1876 N - Ff = (320kg * 5m/s/s)

1876N/(320kg * 5m/s/s) = Ff

1.1725 is the force of friction? I'm pretty sure that's not right, what am I missing?
 
  • #4
swede5670 said:
1876 N - Ff = (320kg * 5m/s/s)

1876N/(320kg * 5m/s/s) = Ff

1.1725 is the force of friction? I'm pretty sure that's not right, what am I missing?
Your algebra!
1876-Ff = 320(5)
Ff = 1876 - (320)5
 
  • #5
Haha thanks a lot, that was a dumb mistake.
I appreciate the help!
 

FAQ: Finding Drag w/ acceleration & Boyant force

1. What is drag in relation to acceleration and buoyant force?

Drag is a force that acts opposite to the direction of motion of an object. It is caused by the interaction of the object with the surrounding medium, such as air or water. In relation to acceleration and buoyant force, drag can affect the motion of an object by slowing it down or changing its direction.

2. How does acceleration affect the drag force?

Acceleration can affect the drag force in two ways. First, an increase in acceleration will result in an increase in the drag force acting on the object. This is because the object is moving faster through the surrounding medium, causing more collisions and therefore more drag. Second, acceleration can also change the direction of the drag force, especially if the acceleration is not in the same direction as the initial motion of the object.

3. What is the relationship between buoyant force and drag force?

Buoyant force and drag force are two different forces that act on an object in a fluid. Buoyant force is the upward force that a fluid exerts on an object submerged in it, while drag force is the resistance force that acts against an object as it moves through a fluid. These forces are not directly related, but they can both impact an object's motion in a fluid.

4. How can we calculate the drag force on an object?

The drag force on an object can be calculated using the equation Fd = 1/2 * ρ * v2 * Cd * A, where ρ is the density of the fluid, v is the velocity of the object, Cd is the drag coefficient, and A is the cross-sectional area of the object. This equation can be used for both air and water, and the drag coefficient will vary depending on the shape and size of the object.

5. How can we reduce drag force on an object?

To reduce the drag force on an object, we can make changes to the object's shape or surface area. For example, a streamlined shape can reduce drag in air or water, while adding fins or other protrusions can increase drag. Additionally, decreasing the velocity of the object or decreasing the density of the surrounding fluid can also reduce the drag force acting on the object.

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