# Horizontal force

1. Sep 29, 2006

### vu10758

Two blocks (M1 = 5kg and M2= 10kg) are pushed so that M1 is not touching the ground (in the air). The horizontal surface is frictionless but there is a friction between the two blocks (mu_k = .4 and mu_s = .5). Find the minimum force F needed so that M1 will remain "in the air" as the blocks move to the right.

I drew free body diagrams for these.

M1 has F pushing to the right, F_s along its vertical surface that touches M2, and M1*g pulling down. It also has a Normal force (N12) pushing left against it.

M2 has a Normal force (N2) pushing up, M2*g pushing down, N12 pushing right against it. It also has F_s along its vertical surface that touches M1.

I then use these relationships

M1

x) F-N12=(M1+M2)a
y) M1*g - F_s = 0

M2

x) F-N12 = (M1+M2)a
y) N2-M2*g = 0

Are these correct so far? If yes, how do I find F from here not knowing "a"?

2. Sep 29, 2006

### Andrew Mason

I am not sure what "remain in the air" means. Are you trying to find the minimum force on M2 that will cause M1 and M2 to separate?

What is the maximum force that M2 can apply to M1? At what point does the force applied to M2 cause this maximum force to be exceeded?

Your expressions for force are incorrect. The horizontal force on M2 must supply the acceleration to both blocks when they are together, so:

$$F_{applied} = (M_1 + M_2)a$$

If both blocks have horizontal acceleration a, and the only horizontal force on M1 is supplied only by M2, what is the force that M2 applies to M1?

AM

3. Sep 29, 2006

### vu10758

Sorry for not explaining it well. There is a picture that comes with this problem. I scanned the original problem and posted have it at this link

http://viewmorepics.myspace.com/ind...&MyToken=0161aa82-9321-4c55-bfa7-c39e8e9c3d77

4. Sep 29, 2006

### Andrew Mason

Ok. Now it makes sense. I thought that M1 was on top of M2.

The forces on M1 are the applied force (to the right, horizontally), the force of gravity (down), the force of friction (up) and the force of M2 on M1 (to the left, horizontally).

Since the two are accelerating at the same rate, the force of M1 on M2 is $F_{12} = M_2a$ This means that the force of M2 on M1 is equal and opposite: $F_{21} = -M_2a$. This enables you to determine the (maximum) vertical force of M2 on M1 (supplied by static friction).

What is the condition for M1 to stay up?

Apply that condition and express a in terms of the applied force (see previous post) and you will have your answer.

AM

5. Nov 7, 2009

### waldvocm

I am trying to solve a similar problem. I understand the equations that I have to use but how do I solve them without knowing the acceleration?

What is the normal force? is it still N=mg?

Last edited: Nov 7, 2009
6. Nov 7, 2009

### Andrew Mason

As I reconstruct this problem, I think the second block was in front of the first and the first was being accelerated in the direction of the second. The force was such that the force of friction of the first on the second kept the second from sliding down. So the normal force for the purpose of determining friction was actually the sideways force of M1 on M2.

AM

7. Nov 12, 2009

### waldvocm

I am still confused.

Here is my problem: In the figure shown, the coefficients of friction between the two blocks are s=0.553 and k=0.336 There is no friction on the horizontal surfaces.

a) What is the minimum force F(N) must be exerted on the block M in order for the block m to not fall?

b) What is the resulting acceleration of the blocks?

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Last edited: Nov 12, 2009
8. Nov 12, 2009

### waldvocm

To find the acceleration I take

Fapplied=(M+m)a

To find Fapplied I take
F-s*N=M+m

9. Nov 12, 2009

### Andrew Mason

The force that must be supplied by static friction to M2 is M2's weight: M2g. What is the normal force that is required in order to have that much static friction between M2 and M1?

$$F_s = \mu_sN$$

What is N in terms of the applied force?

AM

10. Nov 13, 2009

### waldvocm

When you say M2, I am assuming you mean m (the little block that is suspended).

So, to find the minimum force F that must be exerted on M (the big block) for m not to fall is;

The weight of m (the small block) times the coefficient of static friction?

Fs=.553*98=54.194 So that is the force of static friction that can not be exceeded or else the block m will move. To find the applied force

F=Fs+Mg

The to find acceleration I would use F=(M+m)a

11. Nov 13, 2009

### Andrew Mason

The Normal force is the force of the large block on the smaller block. What is that caused by and what is its magnitude?

AM

12. Nov 16, 2009

### waldvocm

The Normal Force is mg (of the large block)+coefficient of static friction?

So is the Normal force the same as the applied force in this situation?

13. Nov 16, 2009

### Andrew Mason

The force of the large block on the smaller is the force that causes m to accelerate: F = ma . This is the normal force that determines the maximum static friction force between M and m: [itex]F_s = \mu_sN = \mu_sma[/tex]. How large does the static friction force have to be to prevent m from sliding down?

AM

14. Nov 16, 2009

### waldvocm

F=100a
Fs=.553ma

Fs>F

The force that needs to be supplied by static friction is m's weight mg=98

98=.553N
N=177.2

So the minimum force must be 177.2N and the acceleration is

F=ma 177.2=(M+m)a 177.2=(110)a a=1.61

Is this correct??? :)

15. Nov 16, 2009

### Andrew Mason

Not quite.

You are correct that the 10 kg block has to have a frictional force of 98 N which means that the normal force must be 177.2 N. That is the force that accelerates m, so what must m's acceleration be according to Newton's second law? That must, obviously, be the same acceleration of the first block. Use the total mass (M+m) to determine the total force that must be applied to M.

AM