Finding acceleration using FORCE equations

[helpme2011]

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

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

 

[Dick]

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?

 

[helpme2011]

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 :(

 

[Dick]

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.

 

[PeterO]

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

 

[helpme2011]

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 :)