Object and Friction on the object

[dexter90]

Good Morning

I am new user on Forum and I'm sorry for all mistake. I am living in Poland and My English is on the low level.

My question concers friction on the objects. We have formula:

$$T=μ \cdot N$$, where

$$μ$$ - coefficient on fraction
$$N=m\cdot g$$

and formula:

$$F=m\cdot a$$

I think that:

$$F=T \Rightarrow m\cdot a=μ \cdot m\cdot g$$

This is true?? Coefficient $$m$$ is mass object?? If on the object interacts more physical force ( example Friction's ) = $$m \cdot a=m\cdot μ\cdot (N_{1}+N_{2}+...+N_{n})$$??

I greet

 

[tiny-tim]

welcome to pf!

good morning, dexter90! welcome to pf!

I think that: $F=T \Rightarrow m\cdot a=μ \cdot m\cdot g$

This is true?? Coefficient $m$ is mass object?? If on the object interacts more physical force ( example Friction's ) = $m \cdot a=m\cdot μ\cdot (N_{1}+N_{2}+...+N_{n})$??

(btw, we usually write "T" only for "tension" )

one object only has one normal force (N), and one friction force (µN) with a particular surface

for example, if a rope with tension T is pulling an object with mass m along a horizontal surface, then the F = ma equation is:

T - µmg = ma ​

(your equation ma = µmg is impossible … if the only force is friction, then the object won't move!)

 

[dexter90]

Thanks you for quickly answer.

I understand, one object -> one force ( friction on the surface ). Friction is force which retard object. If on the object interacts more force's then coefficient retard is bigger, example: air resistance, then We have from formula:

$$m\cdot a=m\cdot g\cdot μ$$, next

$$a=g\cdot μ + \lambda$$, where:

$$\lambda$$ - coefficient air resistance in [N].

Yes?

I greet.

 

[tiny-tim]

hi dexter90!

… If on the object interacts more force's then coefficient retard is bigger, example: air resistance, then We have from formula:

$$m\cdot a=m\cdot g\cdot μ$$, next

$$a=g\cdot μ + \lambda$$, where:

$$\lambda$$ - coefficient air resistance in [N].

Yes?

not exactly …

air resistance is not proportional to mass (it's proportional to cross-section area, and it also depends on speed, but not mass, see http://en.wikipedia.org/wiki/Drag_equation),

so the equation will be something like -ma = µmg + kAv²,

or a = -(µg + kAv²/m)

 

[dexter90]

:)

I have a question to this formula:

$$-ma...$$

Not exist $$-$$ in $$ma=μmg$$. Now, it is present because force air resistance is returned in the opposite direction??

I apologize for the grammar :)

 

[tiny-tim]

both the friction and the air resistance are retarding forces (they both reduce the acceleration of a moving object)

so it should have been -ma = µmg

 

[dexter90]

Ok.

I can add forces in the same way??

 

[tiny-tim]

what do you mean?

 

[dexter90]

Ok.

I understand all, thanks you.

I greet! :-)