Two boxes tied together on an inclined plane. Find common acc and tension

[Twinflower]

Greetings! This is my first post on this forum :)

("Fasit" means solution in Norwegian)

Homework Statement

Inclined plane, 30 degrees from horisontal
Two boxes tied together.
BoxA = 4 kg, friction coefficient = 0,25
BoxB = 8kg, friction coefficient = 0,35

Box A is closest to the bottom of the plane. Box B is slighly above. Not touching.


A: Find each box' acceleration (the solution implies a common acceleration)
B: Find the tension between the boxes

Homework Equations

$$a = (g * sin(30)) - (\mu * g * cos(30))$$
$$\sum F = m * a$$

The Attempt at a Solution

Regarding Part A
I found both boxes' individual acceleration using the formula above:

$$A_A = 9.81 m/s^2 * sin(30) - 0.25 * 9.81 m/s^2 * cos(30) = 2.78 m/s^2$$
$$A_B = 9.81 m/s^2 * sin(30) - 0.35 * 9.81 m/s^2 * cos(30) = 1,93 m/s^2$$


However, the solution is supposed to be $$2.21m/s^2$$

What am I doing wrong?


Regarding Part b

I am getting nowhere trying to find a solution for this part.
I have tried multiplying the difference in acceleration with box A's downward force, but I am not getting anything near the solution which is supposed to be 2.27 N

 

[PeterO]

Greetings! This is my first post on this forum :)

("Fasit" means solution in Norwegian)

Homework Statement

Inclined plane, 30 degrees from horisontal
Two boxes tied together.
BoxA = 4 kg, friction coefficient = 0,25
BoxB = 8kg, friction coefficient = 0,35

Box A is closest to the bottom of the plane. Box B is slighly above. Not touching.


A: Find each box' acceleration (the solution implies a common acceleration)
B: Find the tension between the boxes

Homework Equations

$$a = (g * sin(30)) - (\mu * g * cos(30))$$
$$\sum F = m * a$$

The Attempt at a Solution

Regarding Part A
I found both boxes' individual acceleration using the formula above:

$$A_A = 9.81 m/s^2 * sin(30) - 0.25 * 9.81 m/s^2 * cos(30) = 2.78 m/s^2$$
$$A_B = 9.81 m/s^2 * sin(30) - 0.35 * 9.81 m/s^2 * cos(30) = 1,93 m/s^2$$However, the solution is supposed to be $$2.21m/s^2$$

What am I doing wrong?Regarding Part b

I am getting nowhere trying to find a solution for this part.
I have tried multiplying the difference in acceleration with box A's downward force, but I am not getting anything near the solution which is supposed to be 2.27 N

The boxes are tied together, so Box A cannot accelerate at a greater rate than Box B.

There will be some tension in the string which will reduce the acceleration of A and increase the acceleration of B so they are equal so you have to consider the forces on the masses, not just their acceleration.

 

[Doc Al]

What am I doing wrong?

If you look at each box separately, you must include all the forces on them. Including the tension in the rope.

First find the acceleration by treating both boxes together as a single system. What external forces act on the system?

 

[Twinflower]

Thanks for your reply, Peter.
I'm not really getting the hang of this Newton-chapter we're going through, but your reply seems logical (at least).

So, the difference in acceleration is
$$2.78m/s^2 - 1.93m/s^2 = 0.85 m/s^2$$
This acceleration multiplied with BoxA's weight (4kg) should result in a force (3,4N, so it's not the tension force). This force slows down A and speeds up B so that they eventually end up at 2.21m/s^2.

But there's something in between I don't know how to calculate.

 

[PeterO]

Thanks for your reply, Peter.
I'm not really getting the hang of this Newton-chapter we're going through, but your reply seems logical (at least).

So, the difference in acceleration is
$$2.78m/s^2 - 1.93m/s^2 = 0.85 m/s^2$$
This acceleration multiplied with BoxA's weight (4kg) should result in a force (3,4N, so it's not the tension force). This force slows down A and speeds up B so that they eventually end up at 2.21m/s^2.

But there's something in between I don't know how to calculate.

As Doc-AL says it is easier to treat the two masses together and add up all the forces helping and hindering the acceleration of one large system.

 

[Twinflower]

If you look at each box separately, you must include all the forces on them. Including the tension in the rope.

First find the acceleration by treating both boxes together as a single system. What external forces act on the system?

Ok, thanks. I have been experimenting somewhat on this single system part. What I did was trying to find a mean friction coffesient, but i got lost and starting googling for help.

Anyways, thanks for helping me asking the right questions and thinking in the right direction(s).


the external forces on this system would be the following:
G, being
$$F_A=(9.81m/s^2 \times 4kg \times cos(30) + F_B = (9.81m/s^2 \times 8kg \times cos(30)) = 102N$$

R, being the friction force for both boxes:
$$FF_A= (sin(30) \times 9.81m/s^2 \times 4kg \times 0.25) + FF_B=(sin(30) \times 8kg \times 9.81m/s^2 \times 0.35) = 18.63N$$

As these two forces act in opposite direction, their difference equals:
$$102N - 18.63N = 83,4N$$

Using the formula F = m*a, I obtain that a=F/m. Using my recently aqquired numbers i get:
$$\frac{83.37N}{4+8kg} = 6,94m/s^2$$...

What force did I forget.. ?

 

[PeterO]

Ok, thanks. I have been experimenting somewhat on this single system part. What I did was trying to find a mean friction coffesient, but i got lost and starting googling for help.

Anyways, thanks for helping me asking the right questions and thinking in the right direction(s).


the external forces on this system would be the following:
G, being
$$F_A=(9.81m/s^2 \times 4kg \times cos(30) + F_B = (9.81m/s^2 \times 8kg \times cos(30)) = 102N$$

R, being the friction force for both boxes:
$$FF_A= (sin(30) \times 9.81m/s^2 \times 4kg \times 0.25) + FF_B=(sin(30) \times 8kg \times 9.81m/s^2 \times 0.35) = 18.63N$$

As these two forces act in opposite direction, their difference equals:
$$102N - 18.63N = 83,4N$$

Using the formula F = m*a, I obtain that a=F/m. Using my recently aqquired numbers i get:
$$\frac{83.37N}{4+8kg} = 6,94m/s^2$$...

What force did I forget.. ?

I think you may have got your sines and cosines back to front this time? In your first attempt, the cosine was in the friction calculation, and the sine was in the "accelerating force".

 

[Twinflower]

edit: the friction was calculated by the means of the normal force, which act in the Y-direction. (i have shifted the axis, so that X-axis is parallell to the plane while the y-axis is normal to it)

 

[PeterO]

Hm, I was thinking that cosine(30) refers to the direction which is parallell to the plane, while sine(30) refers to direction outward from the plane

Is that wrong?

If the angle was zero, there would be no parallel component.

Is it sine or cosine that equals zero when the angle is zero? That is the way I remember it? - or do I really mean work it out each time?

 

[Twinflower]

see my edit above.

I like to remember it as moving in the x-direction is a lot more cozy(cosine) than moving upwards in the sine direction.

 

[Doc Al]

see my edit above.

I like to remember it as moving in the x-direction is a lot more cozy(cosine) than moving upwards in the sine direction.

What's the component of the weight parallel to the surface? Normal to the surface?

 

[Twinflower]

What's the component of the weight parallel to the surface? Normal to the surface?

I think that the normal force is directly oposite of the y-component of the weight ?

example:
$$9.81 * 4kg * \sin (30) = 19.62N$$
in the Y-direction poiting downwards
The normal force should be the same force in the oposite direction, right?
And the friction force is the normal force times friction coeffisient.

Have i misunderstood this as well?

 

[Doc Al]

I think that the normal force is directly oposite of the y-component of the weight ?

example:
$$9.81 * 4kg * \sin (30) = 19.62N$$
in the Y-direction poiting downwards
The normal force should be the same force in the oposite direction, right?
And the friction force is the normal force times friction coeffisient.

Have i misunderstood this as well?

As PeterO has already pointed out, you are mixing up your sines and cosines. Answer my questions:
What's the component of the weight parallel to the surface? Normal to the surface?

 

[Twinflower]

As PeterO has already pointed out, you are mixing up your sines and cosines. Answer my questions:
What's the component of the weight parallel to the surface? Normal to the surface?

OK, the component of weight parallel to the surface would be mass times gravity times cosine(30) i suppose.

Normal to the surface should be the same as above with sine instead of cosine.

 

[Doc Al]

OK, the component of weight parallel to the surface would be mass times gravity times cosine(30) i suppose.

Normal to the surface should be the same as above with sine instead of cosine.

That's backwards. Consider what the normal component would be if the angle of the incline were 0 degrees. Or 90 degrees.

(Interesting that you had that correct in your first post. What happened to change your mind?)

 

[Twinflower]

That's backwards. Consider what the normal component would be if the angle of the incline were 0 degrees. Or 90 degrees.

(Interesting that you had that correct in your first post. What happened to change your mind?)

Yeah.. i see. Flat surface = cosine 0 deg = 1 = no change.

I am going to do some new calculations now, and hopefully return with a sensible solution to this.

Thanks a lot for helping me out :)

 

[Twinflower]

New dawn:

First: Finding the friction force for both A and B:
$$FF_A = 0.25 \times 4kg \times 9.81m/s^2 \times \cos(30) = 8,5N$$
$$FF_B = 0.35 \times 8kg \times 9.81m/s^2 \times \cos(30) = 23,8N$$
$$FF_A + FF_B = 8,5N + 23,8N = 32,3N$$

Second: Finding the x-component of weight for both boxes:
$$F_A = 4kg \times 9.81m/s^2 \times \sin(30) = 19,6N$$
$$F_B = 8kg \times 9.81m/s^2 \times \sin(30) = 39,2N$$
$$F_A + F_B = 19,6N + 39,2N = 58,84N$$

Third: Subtracting friction force from weight:
$$58,8N - 32,3N = 26,5N$$

Fourth: Applying Newton's second law:
$$A = \frac{\sum F}{m} = \frac{26,5N}{4+8kg} = 2,21 m/s^2$$


Thank you so much, Doc Al and Peter for getting my pieces together

 

[Twinflower]

OK, I am suddenly on a roll now and figured out Problem B as well.
B: What is the tension between the boxes.



First step: Determine Box A's desired weight downwards:
$$F_A - FF_A = 19,6N - 8,5N = 11,1 N$$

Second step: Determine Box A's current weight downwards:
$$4kg \times 2.21 m/s^2 = 8,84N$$

Third step: Tension between box A and box B has to be the difference in actual and desired weight:
$$11,1N - 8,84N = 2,26N$$


woho

 

[PeterO]

OK, I am suddenly on a roll now and figured out Problem B as well.
B: What is the tension between the boxes.



First step: Determine Box A's desired weight downwards:
$$F_A - FF_A = 19,6N - 8,5N = 11,1 N$$

Second step: Determine Box A's current weight downwards:
$$4kg \times 2.21 m/s^2 = 8,84N$$

Third step: Tension between box A and box B has to be the difference in actual and desired weight:
$$11,1N - 8,84N = 2,26N$$


woho

Good Work