• Support PF! Buy your school textbooks, materials and every day products Here!

Irodov mechanics problem

  • Thread starter tejas
  • Start date
  • #1
7
0
two masses, m1 and m2 are connected with an undeformed light spring and lie on a surface. the coefficents of friction between the masses and the surface are k1 and k2, respectively. What is the minimum constant force F, that needs to be applied in the horizontal direction to the m1, to shift the other mass?

http://irodovsolutionsmechanics.blogspot.com/2008_02_16_archive.html


here is the official solution to the problem, it's same as in Singh's book and the answer is also correct. But I simply cannot understand it. How can the force F be smaller than the sum of the sliding friction forces?

Also, if we take m1 to be negligable, then the force necessary would be the Frictional force on m2/2. How does that work out? Also, that would me that to move the m1 would be harder, if we applied the same force to m2. Help please, I'm puzzled :(
 

Answers and Replies

  • #2
Forget the given solution.

Lets break the problem down. There are two blocks, each with the frictional force acting on it. On one block, we want to apply a force F, so we factor that into the FBD. At the same time, there is the spring force opposing the force F.

Now, the frictional force will be opposite to the direction of F, because it needs to oppose the force F. The spring force on the block m1 also opposes F.

The only forces on the block m2 are the frictional force and the spring force.

Now, lets say the block m1 moves a distance x. Now, lets, for the moment, assume that the block m2 is heavy enough that the spring must expand a little before the mass m2 moves.

Can you work through the equations up till this point?
 
  • #3
7
0
well, the total force on m1 is F - kx- m1gk. the force on m2 is simply kx, until kx = m2gk.

anyways, now about the work. A = integral (Fdx) (sry I don't know how to post equations)

then A = Fx - kx^2/2 - m1gkx. is that zero? dunno. This is the work done on m1 by the sum of the forces on it. it must go somewhere. Where it goes? into -kx^2/2? as we take the kinetic energy zero.
 
  • #4
7
0
anyone?
 
  • #5
31
0
well, the total force on m1 is F - kx- m1gk. the force on m2 is simply kx, until kx = m2gk.

anyways, now about the work. A = integral (Fdx) (sry I don't know how to post equations)

then A = Fx - kx^2/2 - m1gkx. is that zero? dunno. This is the work done on m1 by the sum of the forces on it. it must go somewhere. Where it goes? into -kx^2/2? as we take the kinetic energy zero.
the work done by external force goes partly in increasing the elastic potential energy in spring and partly used up in overcoming friction. 'A' becomes 0 when x is the extreme position of the spring of max stretch
 
  • #6
7
0
Ok I understand that, but how it solves the problem that only half of the frictional force is needed to move the m2 when m1 tends to 0?
 

Related Threads on Irodov mechanics problem

  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
1
Views
545
  • Last Post
Replies
6
Views
8K
  • Last Post
Replies
10
Views
2K
Replies
50
Views
2K
Replies
2
Views
2K
Replies
7
Views
2K
Top