Direction of friction for three bodies stacked one on top of the other

• lorenz0
In summary, when applying Newton's Second Law to a system of Boxes with no friction, the middle mass (box 3) moves downwards while the other two boxes stay still.
lorenz0
Homework Statement
Three boxes of mass ##m_1=10\ kg##, ##m_2=5\ kg## and ##m_3= 30\ kg## are arranged as shown in the figure and move by sliding down the inclined plane (##m_3##) and one on the other (##m_1## and ##m_2##).
The dynamic friction coefficient between the case of mass ##m_3## and the inclined plane is ##\mu_{d3} =0.15## while the dynamic friction coefficient between the ##m_2## and the other two boxes is ##\mu_d = 0.10##.
Find the acceleration of the boxes and the tension of the rope assuming that the rope is ideal and the pulley of negligible mass. Assume that the plane is inclined by an angle ##\alpha= \frac{\pi}{6}## and stationary with respect to the ground and that the boxes, do not fall over but remain one on top of the other during the observation period.
Relevant Equations
##\sum_{i}\vec{F}_i=m\vec{a}##, ##F_{fr}=\mu_d N##
I have drawn three free body diagrams, one for each box and then I applied Newton's Second Law after choosing a reference frame rotated clockwise by ##\alpha##, with ##x## pointing south-east and ##y## pointing north-east and I got:

##\begin{cases}m_{1x}: -T+m_1g\sin(\alpha)+F_{fr_{12}}=m_1 (-a)\\ m_{2x}: -F_{fr_{12}}+F_{fr_{23}}+m_2g\sin(\alpha)=m_2 a_2\\ m_{3x}: -T-F_{fr_{23}}-F_{fr_3}+m_3g\sin(\alpha)=m_3 a\\ m_{1y}: N_1-m_1g\cos(\alpha)=0,\ |F_{fr_{12}}|=\mu_d N_1\\ m_{2y}: N_2-N_1-m_2g\cos(\alpha)=0,\ |F_{fr_{23}}|=\mu_d N_2\\ m_{3y}: N_3-N_2-m_3g\cos(\alpha)=0,\ |F_{fr_{3}}|=\mu_{d3}N_3\end{cases} ##

where ##F_{fr_{ij}}## denotes the friction force between objects ##i## and ##j##, ##a_1=a_2=a, T_1=T_3=T## since the rope is ideal and the pulley of negligible mass.

Now, I checked my solution against that of the text I am learning from and the text says that ##F_{fr_{23}}## on ##m_2## should point to the left and on ##m_3## to the right!

It says (all other equations coinciding with mine) that for ##m_2## and ##m_3## we have:
\begin{cases}
m_{2x}: -F_{fr_{12}}-F_{fr_{23}}+m_2g\sin(\alpha)=m_2 a_2\\ m_{3x}: -T+F_{fr_{23}}-F_{fr_{12}}+m_3g\sin(\alpha)=m_3 a
\end{cases}

How is that possible? As far as I can see as soon as the system is set in motion ##m_3## should go down the inclined plane and so the friction force with ##m_2## should prevent it from sliding, isn't it? If this is not the case then it makes me question the method I have always used up to now to find the direction of the friction: I try to see where the point of contact "is trying to move to" and I know that the friction force points in the opposite direction... is there another (better) way to do this? Thanks.

Attachments

• boxes.png
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Last edited:
Delta2
in your first three equations you have not been consistent wrt signs.
You started off taking each new force and acceleration (except for T) as positive down the slope. In the third, you have taken friction from the ground on m3 as positive up the slope (ok, but confusing) but more seriously you have both m1 and m3 with acceleration a down the slope.

Another problem is your Ff=μN equations. Since you don't yet know which way friction acts, these should be |Ff|=μN.

lorenz0
haruspex said:
in your first three equations you have not been consistent wrt signs.
You started off taking each new force and acceleration (except for T) as positive down the slope. In the third, you have taken friction from the ground on m3 as positive up the slope (ok, but confusing) but more seriously you have both m1 and m3 with acceleration a down the slope.
Thanks, I have corrected that and a couple of other typos.

lorenz0 said:
Thanks, I have corrected that and a couple of other typos.
See my edit.

haruspex said:
See my edit.
The problem is that I thought I had gotten the correct signs for the friction forces between the boxes.

lorenz0 said:
The problem is that I thought I had gotten the correct signs for the friction forces between the boxes.
Even my edit did not quite capture it. Originally you had assumed a2 was strictly between a and -a. Now you are assuming only that it is not equal to either of those. I should have written |Ff|≤μN.

haruspex said:
Even my edit did not quite capture it. Originally you had assumed a2 was strictly between a and -a. Now you are assuming only that it is not equal to either of those. I should have written |Ff|≤μN.
Isn't that the condition for static friction? I assumed (and the book agrees with me) that the boxes are sliding relative to one another so I used the formula for dynamic friction which, if I am not mistaken, is (in magnitude) constant and equal to ##F_{fr_d}=\mu_d N##.

lorenz0 said:
Isn't that the condition for static friction? I assumed (and the book agrees with me) that the boxes are sliding relative to one another so I used the formula for dynamic friction which, if I am not mistaken, is (in magnitude) constant and equal to ##F_{fr_d}=\mu_d N##.
In practice, it is highly likely that the middle mass moves together with one of the other two, but since you are not given a static coefficient we must either assume it is the same or that it is all kinetic.

Consider the case where there is no friction. How would the three move?

haruspex said:
In practice, it is highly likely that the middle mass moves together with one of the other two, but since you are not given a static coefficient we must either assume it is the same or that it is all kinetic.

Consider the case where there is no friction. How would the three move?

Considering only friction between ##m_3## and the inclined plane intuition tells me that m_2 should start sliding downwards with respect to ##m_3## so that's why the friction force on ##m_2## relative to ##m_3## should point to the left and, by Newton's third law, why the friction force on ##m_3## relative to ##m_2## should point downwards.

If I try to set up Newton's third law this way I get:
##\begin{cases}
(1): -T+m_1 g\sin(\alpha)=m_1 (-a)\\
(2): m_2g\sin(\alpha)=m_2a_2\\
(3): -T+m_3g\sin(\alpha)-F_{fr_3}=m_3 a\\
(4): N_1-m_1g\cos(\alpha)=0\\
(5): N_2-N_1-m_2g\cos(\alpha)=0\\
(6): N_3-N_2-m_3g\cos(\alpha)=0\\
(7): F_{fr_3}=\mu_dN_3
\end{cases}##

##(8)=(3)-(1): g\sin(\alpha)(m_3-m_1)-F_{fr_3}=(m_3+m_1)a##;
##(9): (4)\to (5)\to (6): N_3=(m_1+m_2+m_3)g\cos(\alpha)##;
##(10): (9)\to (7)\to (8): a=\frac{1}{m_3+m_1}\left( (m_3-m_1)g\sin(\alpha)-\mu_d (m_1+m_2+m_3)g\cos(\alpha)\right)=\frac{9.8}{40}\left( 10-\frac{45\sqrt{3}}{2} \right)\approx 1.50\ \frac{m}{s^2}##

while, from ##(2)##, we get ##a_2=g\sin(\alpha)=4.9\ \frac{m}{s^2}##.
Now, this result also appears to suggest that ##m_2## will slide downwards to the right and so the friction force between ##m_2## and ##m_3## is as I said in the above paragraph.

I think I understand what the boxes are doing now. Is my reasoning correct? Thanks.

Last edited:
Rookie question here. I was under the impression that you don't have to choose the direction of the frictional forces correctly so long as you are consistent in applying Newtons 3rd law between bodies for which ever direction you do choose?

i.e. if you choose ## \rightarrow ## for ## f_{r_{21}}## you must choose ## \leftarrow ## for ##f_{r_{12}}##. In the end you will have the directions revealed to you w.r.t the directions you chose?

erobz said:
Rookie question here. I was under the impression that you don't have to choose the direction of the frictional forces correctly so long as you are consistent in applying Newtons 3rd law between bodies for which ever direction you do choose?
Quite so, but the relationship between normal force and kinetic frictional force is ##|F_f|=|N|\mu_k##. Assuming N is chosen positive in its actual direction, that simplifies to ##|F_f|=N\mu_k##.

erobz
lorenz0 said:
Considering only friction between ##m_3## and the inclined plane intuition tells me that m_2 should start sliding downwards with respect to ##m_3## so that's why the friction force on ##m_2## relative to ##m_3## should point to the left and, by Newton's third law, why the friction force on ##m_3## relative to ##m_2## should point downwards.

If I try to set up Newton's third law this way I get:
##\begin{cases}
(1): -T+m_1 g\sin(\alpha)=m_1 (-a)\\
(2): m_2g\sin(\alpha)=m_2a_2\\
(3): -T+m_3g\sin(\alpha)-F_{fr_3}=m_3 a\\
(4): N_1-m_1g\cos(\alpha)=0\\
(5): N_2-N_1-m_2g\cos(\alpha)=0\\
(6): N_3-N_2-m_3g\cos(\alpha)=0\\
(7): F_{fr_3}=\mu_dN_3
\end{cases}##

##(8)=(3)-(1): g\sin(\alpha)(m_3-m_1)-F_{fr_3}=(m_3+m_1)a##;
##(9): (4)\to (5)\to (6): N_3=(m_1+m_2+m_3)g\cos(\alpha)##;
##(10): (9)\to (7)\to (8): a=\frac{1}{m_3+m_1}\left( (m_3-m_1)g\sin(\alpha)-\mu_d (m_1+m_2+m_3)g\cos(\alpha)\right)=\frac{9.8}{40}\left( 10-\frac{45\sqrt{3}}{2} \right)\approx -7.1\ \frac{m}{s^2}##

while, from ##(2)##, we get ##a_2=g\sin(\alpha)=4.9\ \frac{m}{s^2}##.
Now, this result, while it seems absurd relative to the pair ##m_1##, ##m_3## also appears to suggest that ##m_2## will slide downwards to the right and so the friction force between ##m_2## and ##m_3## is as I said in the above paragraph.

I think I understand what the boxes are doing now. Is my reasoning correct? Thanks.
you seem to have dropped ##\mu_d## in your evaluation of (10).
But if it were to turn out that a is negative, it would mean your equation 7 is wrong. Either the system is static or 7 should have the sign flipped (see post #11).

lorenz0
I would approach this by allowing for static friction and assuming it equals the kinetic value (it cannot be less). Then I would consider the ways m2 might move in relation to m1 and m3. A priori, there are five:
1. Moves up faster than m1
2. Moves with m1
3. Moves up less than m1 but down less than m3
4. Moves down with m3
5. Moves down faster than m3.
One can be discounted immediately. Another is very unlikely, requiring a precise combination of parameters (and produces an indeterminate result).
Start with the middle of the remaining possibilities. If it violates one of the friction limits, that should point the way to the correct case.

Edit:
Thinking that through again, it is easy to show that m2 moves down.

Last edited:
lorenz0, jbriggs444 and Lnewqban
haruspex said:
you seem to have dropped ##\mu_d## in your evaluation of (10).
But if it were to turn out that a is negative, it would mean your equation 7 is wrong. Either the system is static or 7 should have the sign flipped (see post #11).
I computed (10) again including ##\mu_d## and I got ##a=1.50\ \frac{m}{s^2}## so it appears that indeed ##m_2## is sliding relative to ##m_3##. Thanks again for your help.

Lnewqban

What is the direction of friction for three bodies stacked one on top of the other?

The direction of friction for three bodies stacked one on top of the other depends on the orientation and weight distribution of the bodies. Friction always acts in the opposite direction of the applied force, so if the bodies are stacked horizontally, the direction of friction will be vertical. However, if the bodies are stacked at an angle, the direction of friction will be parallel to the surface they are stacked on.

How does the weight distribution affect the direction of friction for three bodies stacked one on top of the other?

The weight distribution of the bodies can greatly impact the direction of friction. If the weight is evenly distributed among the three bodies, the direction of friction will be more balanced. However, if one body is significantly heavier than the others, the direction of friction will be more towards that body.

Can the direction of friction change while the bodies are stacked on top of each other?

Yes, the direction of friction can change while the bodies are stacked on top of each other. This can happen if the orientation of the bodies changes or if the weight distribution changes. For example, if one body is moved to a different position on top of the other two, the direction of friction may shift as well.

How can the direction of friction be determined for three bodies stacked one on top of the other?

The direction of friction can be determined by analyzing the forces acting on the bodies and applying the laws of friction. It may also be helpful to draw a free body diagram to visualize the forces and their directions. Additionally, conducting experiments with different orientations and weight distributions can help determine the direction of friction.

Is the direction of friction the same for all three bodies when they are stacked on top of each other?

No, the direction of friction may not be the same for all three bodies when they are stacked on top of each other. This is because the weight distribution and orientation of each body can vary, which can affect the direction of friction. It is important to consider each body individually when determining the direction of friction.

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