Finding acceleration using FORCE equations

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Homework Help Overview

The discussion revolves around a physics problem involving a 12 kg box being pushed along a horizontal surface by a force of 24 N at an angle of 40 degrees, with a given kinetic frictional force of 6 N. Participants are exploring how to calculate the acceleration of the box using force equations.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to determine the net forces acting on the box, particularly focusing on the horizontal forces. Questions arise regarding the relevance of gravitational force (Fg) and normal force (Fn) in the context of the problem.

Discussion Status

Some participants are clarifying the necessity of considering gravitational force in their calculations, while others suggest that the provided frictional force simplifies the problem. There is an ongoing exploration of the components of the applied force and their impact on the overall force analysis.

Contextual Notes

There is some confusion regarding the angle of 40 degrees, specifically whether it is measured from the vertical or horizontal, which may affect the interpretation of the forces involved.

helpme2011
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Homework Statement


A 12 kg box is pushed along a horizontal surface by a 24N force at 40 degrees. The frictional force (kinetic) acting on the object is 6 N. What is the acceleration of the object?

Homework Equations


Ff= uFn
Fn= Fg + Ft ??

The Attempt at a Solution



My Fn = 136 N
Ff = 16.3 N
ma= 24cos40 - uFn

---

I had this question on a test and I don't know how to solve it!
 
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helpme2011 said:

Homework Statement


A 12 kg box is pushed along a horizontal surface by a 24N force at 40 degrees. The frictional force (kinetic) acting on the object is 6 N. What is the acceleration of the object?


Homework Equations


Ff= uFn
Fn= Fg + Ft ??


The Attempt at a Solution



My Fn = 136 N
Ff = 16.3 N
ma= 24cos40 - uFn

---

I had this question on a test and I don't know how to solve it!

The acceleration is going to be horizontal, right? That means you only need to know the forces acting in the horizontal direction. What are they?
 
So you don't consider Fg at all ? :S
Hmm...what if Fg and Fn are equal...and Fn is 24cos40?
I'm really confused :(
 
helpme2011 said:
So you don't consider Fg at all ? :S
Hmm...what if Fg and Fn are equal...and Fn is 24cos40?
I'm really confused :(

No, you don't need to. Look, they give you that the frictional force is 6N. You don't need Fg or Fn to find that, they gave it to you. The only thing you need to do is find the other tangential (horizontal) force.
 
helpme2011 said:
So you don't consider Fg at all ? :S
Hmm...what if Fg and Fn are equal...and Fn is 24cos40?
I'm really confused :(

If Fg means the weight force, then yes.

You were given the friction force, you didn't have to work it out. That is why you didn't need Fg.

Had you been told just the co-efficient of friction, then Fg and the vertical component of the applied force would be used in the calculation of the size of the friction Force. The question setter just made it easy fro you by saying that answer is 6N

By the Way, was the 40 degrees in the original question 40degrees to the vertical or 40 degrees to the horizontal?

Peter
 
PeterO said:
If Fg means the weight force, then yes.

You were given the friction force, you didn't have to work it out. That is why you didn't need Fg.

Had you been told just the co-efficient of friction, then Fg and the vertical component of the applied force would be used in the calculation of the size of the friction Force. The question setter just made it easy fro you by saying that answer is 6N

By the Way, was the 40 degrees in the original question 40degrees to the vertical or 40 degrees to the horizontal?

Peter

oh ok. 40 to the horizontal. i think i get i now though :)
 

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