How Is Acceleration Calculated in a Friction-Affected System on an Incline?

[Masrat_A]

Homework Statement

In Figure 4, the coefficient of kinetic friction between block 1 and the incline is 0.2. Determine the acceleration of the system, assuming a frictionless policy.

Figure: http://i.imgur.com/jbTo1Nnh.jpg

Homework Equations

Please see below.

The Attempt at a Solution

Could anyone please review my work and point out any mistakes?

$F_1 = 0.2(2)(9.8)(cos37^o) = 3.13$
$F_2 = 0.2(2)(9.8)(sin37^o) = 2.36$

$F_t = \sqrt{F_1^2+F_2^2}$
$F_t = \sqrt{9.8+5.6}$
$F_t = 4N$

$F_t = Ma$
$4 = 2a$
$a = 2 m/s^2$

 

[kuruman]

Major mistakes
1. You did not write down Newton's second law equations for each mass.
2. You did not take friction into account correctly.

 

[Masrat_A]

Will you please give me an idea of how either of the Newton's second law equations would look? How would I be able to incorporate friction?

 

[kuruman]

1. Start by identifying all the separate forces acting on each mass.
2. Read about kinetic friction here http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html
3. Draw them in a diagram for each mass.
4. Add all the forces as vectors to get the net force for each mass.
5. Set the net force equal to mass times acceleration (this is Newton's Second Law). You get two equations and two unknowns, the tension in the string and the acceleration.
6. Solve for the acceleration.

 

[Masrat_A]

I've been trying to follow the steps from my understanding, but I'm still a little stuck here.

$T - M_1(g)(Fk) = M_1(a)$
$M_2(g)(sin\theta) - T = M_2(a)$

$M_1(a) + M_1(g)(Fk) = M_2(a) + M_2(g)(sin\theta)$
$2(a) + 2(10)(0.2) = 2(a) + 2(10)(0.6)$
$2a + 4 = 2a + 12$

 

[kuruman]

It looks like you are confused about the forces acting on each mass.

Please identify the origin of the forces acting on each, using names, not equations. Identifying the origin of each force will help you understand them better. Imagine what is in contact with each mass and don't forget to include the "at a distance force" exerted by the Earth.

1. For mass M₁ on the incline, what pieces of the Universe act on it?
2. For hanging mass M₂, what pieces of the Universe act on it?

 

[Masrat_A]

Please let me know if there's anything I'm missing.

1. For $M_1$, I believe the forces are applied, frictional, normal, tension, and gravity.
2. For $M_2$, they are only tension and gravity.

 

[kuruman]

In the case of M₂, the forces are directed either up or down and so is the acceleration. The Earth exerts the force of gravity and the string exerts the force of tension.
In the case of M₁, the Earth exerts the force of gravity, the string exerts the force of tension and the incline exerts the normal and frictional forces. Can you explain what piece of the Universe, something we can see and touch, exerts the force that you have called "applied"?

 

[Masrat_A]

Applied force would occur when an object (or person) causes another object to move. In this case, I was assuming $M_2$ to be applying force upon $M_1$ via the string, but I may be confusing this with tension.

 

[YoshiMoshi]

You can solve the system without "worrying" about tension. The system has a mass of 2m so draw a square with mass 2m. In one direction you have force of gravity on block m2. In the "opposite direction" you have the force of friction on block m1 AND the force of gravity on block m1.

This is just an alternative way of doing it.

 

[Masrat_A]

You can solve the system without "worrying" about tension. The system has a mass of 2m so draw a square with mass 2m. In one direction you have force of gravity on block m2. In the "opposite direction" you have the force of friction on block m1 AND the force of gravity on block m1.

This is just an alternative way of doing it.

Is this how it would look?

$M_1(g)(Fk) = M_1(a)$
$M_2(g) = M_2(a)$

$M_1(a) + M_1(g)(Fk) = M_2(a) + M_2(g)$

 

[YoshiMoshi]

Is this how it would look?

$M_1(g)(Fk) = M_1(a)$
$M_2(g) = M_2(a)$

$M_1(a) + M_1(g)(Fk) = M_2(a) + M_2(g)$

No.

Do you agree with below statement?
In one direction you have force of gravity on block m2. In the "opposite direction" you have the force of friction on block m1 AND the force of gravity on block m1.
Put this into mathematical relation.

Net force on system = mass of system * acceleration of system = force of gravity on block m2 - force of friction on block m1 - force of gravity on block m1

solve for acceleration of system, than plug in everything.
Do you know the force of gravity on block m2?
Do you know the force of gravity of block m1 in the "down ramp" direction?
Do you know the force of friction on block m1 in the "down ramp" direction?

We sure do.

Sorry I shouldn't say "don't worry about tension" to get acceleration of the system, but you can you come up with relation for acceleration of the system without calculating individual tension, well explicitly

 

[Masrat_A]

I do agree, but at this moment, it's a little difficult for me to put them into an equation. Here is another shot; please do comment.

$F_{net} = 2m * Ma = M_2(g) - M1(g)(Fk) - M1(g)(sinθ)$
$2 * Ma = 20 - 4 - 12$
$2 * Ma = 4$
$Ma = 2$

However, the result seems to be the same as my first try.

 

[YoshiMoshi]

1. 2m*Ma -> Why are you multiplying by "M"? Net force is mass of system times acceleration of system. Mass of system is "2m" if you are using "m" for mass of one block correct. But why do you say acceleration of system is "Ma"?
2. The value of "12" does not seem to coincide with the value you calculated for "M1(g)(sin theta)" in your first post
3. I'm not sure you defined values for the the block mass
4. Forget values and come up with generic formula first than plug in values last. Well you don't have to do this but I can't tell where you got these numbers from
5. If you can't come up with the equation, than it doesn't matter what values you use because it'll be wrong

Net force on system = mass of system * acceleration of system = force of gravity on block m2 - force of friction on block m1 - force of gravity on block m1

how do you solve for "acceleration of system" term in equation above? Use variables only for now. Then substitute knowledge that mass of block of m1 = mass of block m2. Substitute Mass of system = mass of block 1 AND mass of block 2

Oh your using "Ma" for acceleration of system?
1. Put "mass of system" in terms of variables
2. Put "force of gravity on block m2" in "up ramp direction" in terms of variables
3. Put "force of gravity on block m1" in "down ramp direction" in terms of variables
4. Put "force of friction on block m1" in "down ramp direction" in terms of variables
5. Solve for "acceleration of system"
6. Simply using knowledge of "mass of system" and "mass of block 1" = "mass of block 2"

 

[Masrat_A]

Thank you; I'm understanding this better. Does the following result look appropriate?

$a_{sys} = (Fg - m_1gsin\theta - Fn)/m_{tot}$

$Fg = m_2g = 2(10) = 20$
$m_1gsin\theta = 2(10)(sin37^o) = 12.04$
$Fn = umg_1cos\theta = 0.2(2)(cos37^o) = 0.32$
$m_{tot} = m_1 + m_2 = 2 + 2 = 4$

$(20-12.04-0.32)/4 = 1.91$
$a_{sys} = 1.91$

 

[YoshiMoshi]

Looks ok to me. Just don't forget to put units and specify direction. I mean it's implied from your equations which is the positive direction, but id just spell it out your saying positive acceleration, which direction is that?

Also instead of finding each term numerically and plugging into the equation, you can simplyify the equation further. The acceleration of the system is directly proportional to gravity, so your final formula should reflect that.

Also consider kuruman method.