# Static Friction Required to Keep the System from Moving (Two Boxes)

• BurpHa
BurpHa
Homework Statement
For the system of Fig. 4–32 (see below), how large
a mass would box A have to have to prevent any motion
from occurring? Assume the coefficient of the static friction = 0.30.
Relevant Equations
Newton's Third Law, Newton's Second Law.
Ok, logically, it must be that the static friction force of block A equal to the force of gravity on block B, so mass of block A is:

m_A * 9.8 * 0.30 = m_B * 9.8

m_A * 2.94 = 2 * 9.8

m_A * 2.94 = 19.6

m_A \approx 6.7 kg.

However, when I look at block A individually, there is one thing confuses me.

Say block A's mass is x, then in the frictionless scenario, tension force on block A is:

\frac 19.6 {x + 2} * x
I wonder how about if the static friction of block A is equal to the tension force on block A, because then block A should be stopped. Then:

9.8 * x * 0.3 = \frac 19.6 {x + 2} * x

9.8 * 0.3 = \frac 19.6 {x + 2}

2.94 = \frac 19.6 {x + 2}

x \approx 4.7 kg.

According to Newton's second law, then block A cannot move, because the force of tension and the force of static friction cancel out, thus, block B cannot move. So what is wrong about this thinking?

This is why it is confusing. I can understand how all of this work at the bird's eye view, but when I get to the individual level, it starts to confuse. I know my question is kind of silly, but please clarify for me.

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Homework Helper
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You know that block B is at rest. For that to happen the net force on it must be zero, which means that the tension ##T## is equal to the weight of B, ##T=m_B~g##. That is a given no matter what the mass ##x## of block A is.

Now block A is also at rest. For that to happen, the net force on it must also be zero. You already know from above that tension ##T=m_B~g## to the right acts on it. How much static friction and in what direction is acting on block A? Start from there. Then think about it this way.

With hanging mass of 2.0 kg, if block A had the mass of a battleship, would it slide? Answer: No.
With hanging mass of 2.0 kg, if block A had the mass of a feather, would it slide? Answer: Yes.

So somewhere between a battleship and a feather is a threshold mass for A such that it is not sliding but if it is decreased it by a teeny tiny little bit, it will slide. You are asked to find that threshold value for the mass of A.

• • MatinSAR, BurpHa and Lnewqban
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Homework Statement:: For the system of Fig. 4–32 (see below), how large
a mass would box A have to have to prevent any motion
from occurring? Assume the coefficient of the static friction = 0.30.
Relevant Equations:: Newton's Third Law, Newton's Second Law.
I find that posting the full size image is often helpful. • kuruman
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Say block A's mass is x, then in the frictionless scenario, tension force on block A is:

\frac 19.6 {x + 2} * x
Agreed. This is the tension when A and B are both accelerating. It is less than the tension when A and B are not accelerating.

I wonder how about if the static friction of block A is equal to the tension force on block A, because then block A should be stopped. Then:

9.8 * x * 0.3 = \frac 19.6 {x + 2} * x
The above equation is wrong.

The left side of the equation, “9.8 * x * 0. 3”, is the limiting static frictional force. It is the (maximum) magnitude of the frictional force on block A when A is just on the point of slipping but still has zero acceleration.

The right side of the equation is the tension in the absence of any friction, when the system has maximum acceleration.

There is no justification for equating the two quantities, they apply to very different situations. In fact the left side is always bigger than the right side.
________

PS. If you want to display equations using LaTeX, enclose the LaTeX code between a pair of double hash signs, ##\#\#LaTeX code\#\###, or between a pair of double dollar signs, ##\$\$LaTeX code\$\$##.

For example, \frac {19.6} {x + 2} x, with hashes, renders as ##\frac {19.6} {x + 2} x##. And with dollar signs it renders as $$\frac {19.6} {x + 2} x$$The preview toggle button on the right side of the edit toolbar (top of edit window) lets you check how it looks before you post.

For more detailed LaTeX help, try the link at the bottom left of the edit window.

• • MatinSAR and BurpHa
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Say block A's mass is x, then in the frictionless scenario, tension force on block A is:

##\frac {19.6} {x + 2} * x##
Yes, but that is because, without friction, the acceleration is ##\frac {m_B }{m_A+m_B}g##.
More generally, ##a=\frac {m_B-\mu_km_A }{m_A+m_B}g## and ##T=\frac{m_Am_B}{m_A+m_B}g(1+\mu_k)##.
So as ##\mu_k## increases the acceleration decreases and the tension increases. The acceleration does not reduce to zero until ##\mu_k=\frac{m_B}{m_A}##, making the tension ##m_Bg##.

Of course, the question concerns static friction and kinetic is often a bit less.

Btw, you can make the arithmetic easier by cancelling the 'g's. There are many advantages in working entirely algebraically, deferring plugging in numbers until the end. Spotting cancellations is one of them.

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• • MatinSAR and BurpHa
BurpHa
You know that block B is at rest. For that to happen the net force on it must be zero, which means that the tension ##T## is equal to the weight of B, ##T=m_B~g##. That is a given no matter what the mass ##x## of block A is.

Now block A is also at rest. For that to happen, the net force on it must also be zero. You already know from above that tension ##T=m_B~g## to the right acts on it. How much static friction and in what direction is acting on block A? Start from there. Then think about it this way.

With hanging mass of 2.0 kg, if block A had the mass of a battleship, would it slide? Answer: No.
With hanging mass of 2.0 kg, if block A had the mass of a feather, would it slide? Answer: Yes.

So somewhere between a battleship and a feather is a threshold mass for A such that it is not sliding but if it is decreased it by a teeny tiny little bit, it will slide. You are asked to find that threshold value for the mass of A.
Alright, I have understood.

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• Lnewqban
BurpHa
In fact the left side is always bigger than the right side.
Did you want to say the right side is always larger then the left side?

BurpHa
I find that posting the full size image is often helpful.

View attachment 319420
You mean my image is not clear? Because other things are not related. ;)

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Did you want to say the right side is always larger then the left side?
Apologies for expressing myself badly. I should have said something like this:

“The actual force represented by the left side of the equation (limiting friction) is always bigger than the actual force represented right side of the equation (tension when system accelerates). Equating these two unequal forces leads to a nonsensical equation.”

Edt. I should add that in the offending equation
“9.8 * x * 0.3 = \frac 19.6 {x + 2} * x”
it is worth noting that the right side appears to be the tension when the system accelerates with zero friction acting on A.

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• • MatinSAR and BurpHa
BurpHa
Apologies for expressing myself badly. I should have said something like this:

“The actual force represented by the left side of the equation (limiting friction) is always bigger than the actual force represented right side of the equation (tension when system accelerates). Equating these two unequal forces leads to a nonsensical equation.”

Edt. I should add that in the offending equation
“9.8 * x * 0.3 = \frac 19.6 {x + 2} * x”
it is worth noting that the right side appears to be the tension when the system accelerates with zero friction acting on A.
Wait, I thought that the force on the right side (tension) is larger than the force on the left side (static friction.) That is why block A still moves forward with an acceleration.

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Wait, I thought that the force on the right side (tension) is larger than the force on the left side (static friction.) That is why block A still moves forward with an acceleration.
You are confusing static and kinetic friction. You have static friction when the two surfaces are at rest relative to each other. When they are rubbing together so that one moves relative to the other, you have kinetic friction.

• MatinSAR and Lnewqban
BurpHa
You are confusing static and kinetic friction. You have static friction when the two surfaces are at rest relative to each other. When they are rubbing together so that one moves relative to the other, you have kinetic friction.
I mean that when the static friction force on block A equal to ##\frac {19.6} {x + 2} * x##, the tension force on block A is still larger and therefore be able to win over the maximum static friction force.

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Wait, I thought that the force on the right side (tension) is larger than the force on the left side (static friction.) That is why block A still moves forward with an acceleration.
I know use of symbols is generally preferred, but here’s a numerical example which might help to explain what I mean. Sorry it’s a bit long.

Let’s suppose:
##g = 10m/s^2## and ##\mu_s = 0.4## (I’ve chosen convenient values to make the arithmetic simpler).
A consists of a stack of 10 metal discs, each with a mass of 1kg (weight of a disc = 10N).
B’s mass is 2kg (weight = 20N).

When the system is in equilibrium (not accelerating), the string’s tension equals the weight of B, i.e. T = 20N.

The limiting static frictional force is ##\mu_s m_Ag = 0.4 \times10 \times 10 = 40##N. That means a tension of more than 40N is required to make A slip. So A does not slip; the frictional force acting on A simply equals the tension, 20N.

Now we reduce the weight of A by removing a disc. The limiting static frictional force is now ##\mu_s m_A g = 0.4 \times 9 \times10 = 36##N. That means a tension of more than 36N is required to make A slip. So A still does not slip; the frictional force acting on A is still 20N.

We keep reducing A’s weight by removing discs. Eventually we reach the point where A’s mass is such that the limiting frictional force (##\mu_s m_Ag##) equals the tension (20N). That means ##0.4 m_A g = 20##, so ##m_A= 5##kg. When there are only 5 discs left, A is on the verge of slipping - but the tension is still 20N.

Now we remove another disc. The limiting static frictional force is ##\mu m_Ag = 0.4 \times 4 \times 10 = 16##N.

For an instant (but only an instant) the tension (20N) is greater than the limiting frictional force (16N) and the masses accelerate. This is a ‘discontinuity’, where the system changes from being in equilibrium to accelerating. Consider what happens next.

A’s mass is now 4kg. The system accelerates. The acceleration depends on (amongst other things) the kinetic frictional force. But for simplicity and illustration purposes, let’s assume kinetic friction is negligible.

If we imagine the string ‘straightened out’, we see the system is equivalent to a total mass of (4kg + 2kg) accelerated by B’s weight (20N). The acceleration is ##a(= \frac Fm)= \frac {20}6 ≈ 3.33 m/s^2##

Considering A, the new tension (T’) is the accelerating force acting on A, so T’ (= ma) = ##4 \times 3.33## ≈ 13.3N.

The tension when accelerating (13.3N) is less than the tension when stationary (20N).

A more accurate calculation, taking kinetic friction into count, would give the tension when accelerating closer to (but still less than) 20N.

• • MatinSAR and BurpHa
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I mean that when the static friction force on block A equal to ##\frac {19.6} {x + 2} * x##, the tension force on block A is still larger and therefore be able to win over the maximum static friction force.
Larger that what? You got it backwards. As long as the block doesn't slide, the static friction does not have a fixed value; it is equal to the tension. It is zero if you hang no mass, 9.8 N if you hang a 1-kg mass, 19.6 N if you hand a 2-kg mass, 196 N if you hang a 20-kg mass, etc. If you increase the hanging mass, static friction will also increase to match the tension. However, it cannot increase indefinitely. There will be a point at which sliding occurs. You need a separate equation that gives the upper limit of static friction beyond which it can no longer increase. What is that equation?

• MatinSAR, BurpHa and SammyS
BurpHa
Larger that what? You got it backwards. As long as the block doesn't slide, the static friction does not have a fixed value; it is equal to the tension. It is zero if you hang no mass, 9.8 N if you hang a 1-kg mass, 19.6 N if you hand a 2-kg mass, 196 N if you hang a 20-kg mass, etc. If you increase the hanging mass, static friction will also increase to match the tension. However, it cannot increase indefinitely. There will be a point at which sliding occurs. You need a separate equation that gives the upper limit of static friction beyond which it can no longer increase. What is that equation?
What is that equation?
The equation is ##9.8 * 0.3 * m_A##.

BurpHa
The tension when accelerating (13.3N) is less than the tension when stationary (20N)
I finally got this point.
Homework Statement:: For the system of Fig. 4–32 (see below), how large
a mass would box A have to have to prevent any motion
from occurring? Assume the coefficient of the static friction = 0.30.
Relevant Equations:: Newton's Third Law, Newton's Second Law.

Ok, logically, it must be that the static friction force of block A equal to the force of gravity on block B, so mass of block A is:

m_A * 9.8 * 0.30 = m_B * 9.8

m_A * 2.94 = 2 * 9.8

m_A * 2.94 = 19.6

m_A \approx 6.7 kg.

However, when I look at block A individually, there is one thing confuses me.

Say block A's mass is x, then in the frictionless scenario, tension force on block A is:

\frac 19.6 {x + 2} * x
I wonder how about if the static friction of block A is equal to the tension force on block A, because then block A should be stopped. Then:

9.8 * x * 0.3 = \frac 19.6 {x + 2} * x

9.8 * 0.3 = \frac 19.6 {x + 2}

2.94 = \frac 19.6 {x + 2}

x \approx 4.7 kg.

According to Newton's second law, then block A cannot move, because the force of tension and the force of static friction cancel out, thus, block B cannot move. So what is wrong about this thinking?

This is why it is confusing. I can understand how all of this work at the bird's eye view, but when I get to the individual level, it starts to confuse. I know my question is kind of silly, but please clarify for me.

After reviewing the thread and my study materials, I could nearly understand the matter. Thank you anyone who has lent me a hand in rewiring my thinking 🤞✌💪💪

• MatinSAR, Steve4Physics and kuruman
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