# How Do You Calculate Net Force on a Block on a Ramp?

• MysticDude
In summary, to calculate the net force on a block of mass m that is accelerating down a ramp, you will need to use the equation: (mg)sin(θ) - μkmg*cos(θ) = ma, where θ is the angle of the ramp. This equation takes into account the component of the weight force down the ramp and the friction force acting against the block's motion. It is important to draw a final free body diagram with all the forces included in order to accurately set up and solve for the net force.
MysticDude
Gold Member

## Homework Statement

Write the equation used to calculate net force on a block of mass m that is accelerating down a ramp.

## The Attempt at a Solution

Okay I think that we are going to have to use the equation F - f = ma. Since the question says to find the net force, does this mean the I'm going to have to add all of the vector forces and get the magnitude? Here is my free body diagram:
[PLAIN]http://img7.imageshack.us/img7/4929/nov24physicsnum4.jpg

Last edited by a moderator:
MysticDude said:

## Homework Statement

Write the equation used to calculate net force on a block of mass m that is accelerating down a ramp.

## The Attempt at a Solution

Okay I think that we are going to have to use the equation F - f = ma. Since the question says to find the net force, does this mean the I'm going to have to add all of the vector forces and get the magnitude? Here is my free body diagram:
[PLAIN]http://img7.imageshack.us/img7/4929/nov24physicsnum4.jpg[/QUOTE]

That looks pretty close. The sum of the forces will equal ma for the block. The forces are gravity down, ramp pushing back at an angle, and friction retarding the acceleration. I prefer to keep the FBD in the original orientation, though, but that's up to your preference. (Well, if a human is grading a test paper and trying to understand your work, they would probably prefer the traditional orientation too...)

Last edited by a moderator:
berkeman said:
That looks pretty close. The sum of the forces will equal ma for the block. The forces are gravity down, ramp pushing back at an angle, and friction retarding the acceleration. I prefer to keep the FBD in the original orientation, though, but that's up to your preference. (Well, if a human is grading a test paper and trying to understand your work, they would probably prefer the traditional orientation too...)

Well, do you have any hints?

I remember learning that Fnet is all of the forces acting on a body combined. That is why I thought that I was going to have to add up all of the vector forces. Isn't that right?

Is something pulling the block down the ramp (other than gravity)?
If not, you should omit the applied force vector.
I would use the unrotated diagram, and I would show the mg vector split into a component along the ramp plus a component into the ramp. The component into the ramp is exactly canceled by the normal force (since there is no acceleration in that direction).

The net force is the sum of the other forces and the F in F = ma.

Delphi51 said:
Is something pulling the block down the ramp (other than gravity)?
If not, you should omit the applied force vector.
I would use the unrotated diagram, and I would show the mg vector split into a component along the ramp plus a component into the ramp. The component into the ramp is exactly canceled by the normal force (since there is no acceleration in that direction).

The net force is the sum of the other forces and the F in F = ma.

I only have the information which the question stated, so I guess you are right and I should take out the applied forces vector. You kind of confused me with the mg vector component part.

In your last part, should I use F - f = ma? If I do, then my equation would be ma - μkmgcos(θ) = ma. Which means μkmgcos(θ)= 0. I'm not sure what to do for this question as I'm confused big time.

Yeah, get rid of the applied force, and your mg X component will take care of that role for you.

Your equation shouldn't include F as one of its forces, so that's what's messing it up. :)
In the X direction, is Fnet the same as the F in F=ma?

Last edited:
should I use F - f = ma
Not as you have interpreted it, anyway. You need
component of mg down the ramp - friction force = ma

Delphi51 said:
Not as you have interpreted it, anyway. You need
component of mg down the ramp - friction force = ma

So what you are saying is (mg)cos(θ) - μkmg = ma which is mg(cos(θ) - μk) = ma. OR is it mgcos(θ)(1 - μk) = ma?

So, is this correct?

Maybe (mg)cos(θ) - μkmg*sin(θ) = ma
Has to be sin(θ) in one place, cos in the other. No diagram, can't tell what you are calling θ.

Delphi51 said:
Maybe (mg)cos(θ) - μkmg*sin(θ) = ma
Has to be sin(θ) in one place, cos in the other. No diagram, can't tell what you are calling θ.
Okay sorry about that. If we are using Normal diagram, in other words, the way things are without me changing the axes, would it then be (mg)sin(θ) - μkmg*cos(θ) = ma?

i would recommend drawing a final FBD breaking mg into components making the mg cos theta and mg sin theta

I too would draw a final FBD so that you can take inti account your additional x and y forces representing sin theta and cos theta and from there you can set your forces equal to one another

## 1. What is a force equation?

A force equation is a mathematical representation of the relationship between the force applied to an object and the resulting motion or acceleration of that object. It is typically written as F=ma, where F is force, m is mass, and a is acceleration.

## 2. How do I write a force equation?

To write a force equation, you first need to identify the force being applied to the object and the mass of the object. Then, you can use the formula F=ma to calculate the resulting acceleration. Make sure to use consistent units throughout the equation.

## 3. What are some common forces that can be included in a force equation?

Some common forces that can be included in a force equation include gravity, friction, tension, and normal force. Other forces such as air resistance, buoyancy, and magnetic force may also be included depending on the specific situation.

## 4. Can a force equation be used for both stationary and moving objects?

Yes, a force equation can be used for both stationary and moving objects. For stationary objects, the equation can be used to determine the forces needed to keep the object in place. For moving objects, the equation can be used to calculate the acceleration resulting from the applied forces.

## 5. Are there any limitations to using a force equation?

There are some limitations to using a force equation. It assumes that the object being studied is a point mass, meaning that it has no size or shape. It also assumes that the forces acting on the object are constant and that there are no external factors, such as air resistance, affecting the motion. In real-world situations, these assumptions may not always hold true, so the force equation may not always accurately predict the motion of an object.

• Introductory Physics Homework Help
Replies
56
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
11
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
10
Views
1K
• Introductory Physics Homework Help
Replies
15
Views
2K
• Introductory Physics Homework Help
Replies
41
Views
767
• Introductory Physics Homework Help
Replies
39
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
1K