What is the minimum pushing force required

[nns91]

Homework Statement

So there are 2 boxes: 1 has a mass of 16kg and the other is 88kg. There is friction between 2 boxes, the surface which box 88kg is on is frictionless. Static friction coefficient is 0.38. Box 16kg is not on the surface, it's is push against the 88kg box. What is the minimum pushing force required in order for box 16kg not drop

Homework Equations

F=ma

The Attempt at a Solution

So I do F=ma in the y direction which is F_s - mg=0 to find F_s. Then I do in the x direction F-F_n= 0. From F_s I can get F_n by dividing F_s by 0.38.

Did I solve the problem right ?

 

[Nasa01]

Hi

It's in the middle of the night for me and I've got high fever but I think you made one single mistake. You should not divide F_s by 0.38 but rather multiply them.

Think of it this way. If you divide F_s by the friction coefficient and the friction coefficient --> 0 that would imply the friction force would --> infinity, and that dos not seem right...

Hope this helps!

Cheers

 

[nns91]

But F_s = F_n * coefficient of static friction right ? so F_n =F_{s[/SUB / 0.38 right ?}

 

[Nasa01]

Hi
I might be confused by your nomenclature. I think you might have mixed up F_s and F_n, though it might be me reading the problem or your solution wrong as well. Anyway, here's how I would solve the problem:

Known:
Mass m₁ = 16 kg
COF u = 0.38
acc. g = 9.82 m/s²

Find the friction force F_F

Equations
F = mg [1]
u = F_F/F_N [2]

My calculations:
Force normal to the plane, F_N, is found using [1] (F_N=m₁*g) [3]

[2] is rewritten as F_F=u*F_N [4]

Using [3] in [4] yields F_F=u*m₁*g

This gives a numerical value of roughly 60 N

Questions, comments?

Cheers!

 

[nns91]

probably I confused you guys.

The problem is not like that.

The 16 kg is on the left side of the 88kg and kinda pushed against the 88kg but in the air, 16kg box is not on the frictionless surface. We need to find the min F pushing the 16kg box against the other in order for the 16kg not drop. Get it ??

 

[Nasa01]

I see...
Now I understand the problem and your solution. And now I agree with you and your solution. Sorry for my confusion.

Just for clarification:
The friction force, F_F, is linearly dependent on the applied force,F_P (P for Push), on the 16kg-box:

F_F= u*F_P [1]

F_N=m₁₆*g [2]

To balance the box you need top satisfy F_N = F_F [3]

[1] and [2] in [3] gives: m₁₆*g = u*F_P [4]

And [4] can be rewritten as F_P = (m₁₆*g)/u

Cheers!

 

[nns91]

So I did it right ?

Peace ! It was on my test. I screwed up a lot so at least I did something right.