Help with Equation Concerning Maximum Force

[johnhuntsman]

A block of mass mt=4.0 kg is put on top of a block of mass mb=5.0 kg. To cause the top
block to slip on the bottom one while the bottom one is held fixed, a horizontal force of at
least 12N must be applied to the top block. The assembly of blocks is now placed on a
horizontal, frictionless table (Fig. 6-51). Find the magnitudes of (a) the maximum horizontal
force F that can be applied to the lower block so that the blocks will move together and (b)
the resulting acceleration of the blocks.

Link to page with question:
http://www.ifm.liu.se/edu/coursescms/TFYA16/lessons/Le-3-extra.pdf
It's on the 5th page.

I don't understand their equation for maximum horizontal force (M a = M μ_s g) and would appreciate if someone would explain it to me.

 

[Simon Bridge]

I don't understand their equation for maximum horizontal force (M a = M μ_s g) and would appreciate if someone would explain it to me.

Hmmm... homework...

Have you tried working the problem yourself?
Do you get a different equation?

If you show your reasoning it will pinpoint your confusion for us and we can hep you out a lot faster.
Otherwise what you want basically involves doing the problem for you and that is against the rules.

 

[johnhuntsman]

Hmmm... homework...

Have you tried working the problem yourself?
Do you get a different equation?

If you show your reasoning it will pinpoint your confusion for us and we can hep you out a lot faster.
Otherwise what you want basically involves doing the problem for you and that is against the rules.

I solved the problem a different way, after trying to understand the way it was done here for a long time. I did:

m_t * a = F_net
4.0 * a = 12 <-------from horizontal force acting on m_t
a = 3.0 m/s^2

m_sys * a = F_net
9.0 * 3.0 = F_net
27N = F_net

I just want to know how they solved it so that if ever I'm testing I can have more than one way of solving a similar problem.

 

[SammyS]

A block of mass mt=4.0 kg is put on top of a block of mass m_b=5.0 kg. To cause the top block to slip on the bottom one while the bottom one is held fixed, a horizontal force of at least 12N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table (Fig. 6-51). Find the magnitudes of (a) the maximum horizontal force F that can be applied to the lower block so that the blocks will move together and

(b) the resulting acceleration of the blocks.

Link to page with question:
http://www.ifm.liu.se/edu/coursescms/TFYA16/lessons/Le-3-extra.pdf
It's on the 5th page.

I don't understand their equation for maximum horizontal force (M a = M μ_s g) and would appreciate if someone would explain it to me.

Below is the solution you refer to. I'll look at it further, when I get a chance. In the meantime, perhaps someone else will take a look. (My initial impression is that it is not a very clearly written solution.)

Solution:
(a) Using F = μ_smg , the coefficient of static friction for the surface between the two blocks is μ_s = (12 N)/(39.2 N) = 0.31, where m_t g = (4.0 kg)(9.8 m/s²)=39.2 N is the weight of the top block.

Let M = m_t + m_b = 9.0 kg, be the total system mass, then the maximum horizontal force has a magnitude Ma = Mμ_s g = 27 N.

(b) The acceleration (in the maximal case) is a = μ_sg =3.0 m/s².

 

[Simon Bridge]

I solved the problem a different way, after trying to understand the way it was done here for a long time. I did:

m_t * a = F_net
4.0 * a = 12 <-------from horizontal force acting on m_t
a = 3.0 m/s^2

m_sys * a = F_net
9.0 * 3.0 = F_net
27N = F_net

I just want to know how they solved it so that if ever I'm testing I can have more than one way of solving a similar problem.

You haven't shown you reasoning above though. In fact, as written, your working, above, is internally contradictory: you seem to be saying that 27N=12N because F_net appears twice in (hopefully) different contexts. On top of that, you don't seem to have used μ_s at all.

You seem to be asserting that the maximum acceleration of the two blocks stuck together should be the same as the acceleration of the top block, by itself, on a frictionless surface, under the maximum force. I think you need to clarify your reasoning for this method to get accepted. (see below)

Below is the solution you refer to.

(a) Using F = μ_smg , the coefficient of static friction for the surface between the two blocks is μ_s = (12 N)/(39.2 N) = 0.31, where m_t g = (4.0 kg)(9.8 m/s²)=39.2 N is the weight of the top block.

Let M = m_t + m_b = 9.0 kg, be the total system mass, then the maximum horizontal force has a magnitude Ma = Mμ_s g = 27 N.

(b) The acceleration (in the maximal case) is a = μ_sg =3.0 m/s².

The first thing they did was work out the coefficient of static friction. Presumably you can follow that bit.

Then they appear to have been implicitly using two free-body diagrams when it should have been explicit:
The diagram for the top block has one force, friction, pointing to the right: $f=\mu_s m_t g = m_t a$ ...(1)
The diagram for the bottom block has two forces: $F-f=m_b a$ ...(2)
This is two equations with two unknowns ... you have to find F and a.

Solve for a in the first equation: $a=\mu g$ and substitute it into the bottom one.

Actually, (1) is two equations.
(1a) $f=\mu_s m_t g$
(1b) $f=m_t a$

For f to be maximal, $a=\mu_s g$ all right - but you already know the value of the maximal force!

So what you have done is noticed that you don't need (1a).

(caveat: I have not checked the numbers so it is possible I have slipped up someplace.)