Object exerts an opposing force of equal magnitude

[Ukitake Jyuushirou]

now i have to solve for the coefficient of kinetic friction. kinetic friction force is supposed to be proportional to the Fn. but from wat is given, how do i work out the friction force?

 

[arunbg]

What is the constant of proportionality in the equation relating kinetic frictional force and normal force ?
What are the conditions for horizontal equilibrium ?
As Doc Al suggested, write down the math step by step .

 

[Ukitake Jyuushirou]

frictional force is related to the Fn.

Fn = 156, Fx = 69.2. coefficient of friction is 69.2 divided by 156 = 0.44

am i correct?

 

[Doc Al]

w=mg
196=20x9.8

Fn = 196 (no vertical force)

196-40 = 156 (force of 40N in +y-component)
Fn = 156

Good. But just for the record, here's how I would do it:
(Sum of the forces in the vertical direction) = +Fn -mg +80sin(30) = 0

That's the equation for vertical equilibrium. Solving it (trivial, yes) gives:
Fn = mg - 80sin(30)

Now do the same for horizontal forces.

 

[Ukitake Jyuushirou]

Good. But just for the record, here's how I would do it:
(Sum of the forces in the vertical direction) = +Fn -mg +80sin(30) = 0

That's the equation for vertical equilibrium. Solving it (trivial, yes) gives:
Fn = mg - 80sin(30)

Now do the same for horizontal forces.

using ur formula


Fx = 80cos30 = 69.2

 

[Doc Al]

OK. Again, the way I would do it is:
(Sum of the forces in the horizontal direction) = -Ff +80cos(30) = 0

Solving gives: Ff = 80cos(30)

(The reason for sticking to a systematic approach is: (1) It drives home the key physics principles & (2) It allows you to solve much harder problems.)

Now you can apply the other relationship you know about kinetic friction to solve for $\mu_k$ systematically.

 

[Ukitake Jyuushirou]

OK. Again, the way I would do it is:
(Sum of the forces in the horizontal direction) = -Ff +80cos(30) = 0

Solving gives: Ff = 80cos(30)

(The reason for sticking to a systematic approach is: (1) It drives home the key physics principles & (2) It allows you to solve much harder problems.)

Now you can apply the other relationship you know about kinetic friction to solve for $\mu_k$ systematically.

yes ur right. i'd try to use this approach for my other qns

frictional force is related to the Fn.

Fn = 156, Fx = 69.2. coefficient of friction is 69.2 divided by 156 = 0.44

am i correct?

 

[Doc Al]

Looks good. I'd write it this way. The key relationship for kinetic friction is:
$F_f = \mu_k F_n$

thus:
$\mu_k = F_f /F_n$

then just plug in the numbers

 

[Ukitake Jyuushirou]

whew...one question down...many many more to go... thanks for everyone's help and advice

 

[Doc Al]

I think we beat that one to death!

 

[arunbg]

We sure did