Feynman's Lectures exercise about Newton's laws

[sefatod]

TL;DR Summary: Solutions to an exercise coming from the famous "Exercises for the Feynman lectures on physics".

I'm stuck on an exercise.
"A mass $m$ (kg) hangs on a cord suspended from an elevator which is descending with an acceleration of $0.1g$. What is the tension T in the cord in newtons ?"
If would have say that $T=0.98m$ because we can consider that the only force applied to the cord is the force of tension applied by the mass.
By Newton we have that $F=ma$. Hence, $F=m\times 0.1g= 0.98m N$.
But the solution in the book is $8.8m N$, why ?

 

[kuruman]

In Newton's second law, F is the net force acting on the mass, i.e. the sum of all the forces, which is what you calculated. Here you are looking for the tension in the cord not the vector sum of the tension and the force of gravity.

 

[sefatod]

Ok so it is $T=mg+ma=9.8m-m0.98=m(9.8-0.98)=8.82m N$ which is the correction ! Thank you !

 

[kuruman]

You got the answer that you know is correct, but I am not sure you understand what you are doing and why. You are given that $a=0.1g~$ and you write $$T=mg+ma=9.8m-m0.98=m(9.8-0.98).$$Why do you replace $g$ with $-9.8~\text{m/s}^2$ in one of terms in the expression and not both? Aren't the acceleration of the mass and the acceleration of gravity in the same direction?

 

[sefatod]

Oh yes you're right...
I rethought about it and found that $\vec{T}=-\vec{B}$ with $\vec{B}$ the force applied to the mass $m$ because the point at the end of the cord is motionless (in the elevator referential) so by the Newton's first law, we have that $\sum \vec{F_{ext}}=0$. The only force applied to this point is the tension and $\vec{B}$. Again by Newton, we have that $\vec{B}=m\vec{a}$ which implies that $||\vec{B}||=0.98m N$
So $||\vec{T}||=-0.98m N$ which is not the answer

 

[sefatod]

Oh dang I forgot the gravitational force applied to the mass. It is $F_{grav}=9.8m$ so $||\vec{B}||=0.98m + 9.8m$ so $||\vec{T}||=-||\vec{B}||=-(0.98m+9.8m)=11m N$ !

 

[kuruman]

You still seem to be confused about Newton's second law. It is an equation describing how Nature works not a formula, like the area of a circle, that allow you to calculate things. The way to use it is this
1. Decide which object in the Universe you wish to study. That is your system.
2. Decide which other objects in the Universe exert forces on it and write expressions for them. Then add them as vectors to find the net force. It will be a number that has a magnitude and a direction. It is called the net force.
3. Now put your system on a scale and get a number for its mass. Also, get a ruler and a stopwatch and measure its acceleration. Multiply the mass and the acceleration (also a vector) and you get a second number that has magnitude and direction.

Newton's second law asserts that the two numbers that you got using completely different methods are the same. Let's see how this applies to your question.

1. The system is mass $m$, assumed known.
2. The objects of the Universe that exert forces on it are (a) the Earth that exerts the force of weight and (b) the cord that exerts the force of tension. To add them as vectors we have to find their directions. We define up as positive and down as negative. Let $T$ be the ma
gnitude of the tension (a positive number) and $mg$ the magnitude of the weight (also a positive number). We write the net force as (I use boldface to represent vectors)$$\mathbf{F}_{\text{net}}=T~(\mathbf{up})+mg~(\mathbf{down}).$$Since "down" is the opposite of "up", we can write $(\mathbf{down})=-(\mathbf{up})$ which gives,$$\mathbf{F}_{\text{net}}=T~(\mathbf{up})-mg~(\mathbf{up})=(T-mg)~(\mathbf{up}).$$3. Now for the acceleration. We are told that its magnitude is $0.1g$ and its direction is down. So the mass times acceleration vector is $$m\mathbf{a}=0.1mg~(\mathbf{down})=-0.1mg~(\mathbf{up}).$$ We have calculated the two sides of Newton's second law equation separately. We now assert that they are equal as demanded by the law and write $$(T-mg)~\cancel{(\mathbf{up})}=-0.1mg~\cancel{(\mathbf{up})}\implies T-mg=-0.1mg\implies T=mg(1-0.1).$$See how it works?

 

[PeroK]

@sefatod many students seem to struggle with this problem. As explained above, applying Newton's second law is the way to go. That said, here is my insight:

If an object of mass $m$ is hanging by a rope or string, then the tension in the rope must be the full weight the object: $mg$. If the whole system is accelerating upwards, then it should be clear that there is more tension in the rope, because now the rope not only supports the object's weight, but also must provide the upwards acceleration. If the upward acceleration is $a$, then the tension must be $ma + mg$.

The confusing case is where the whole system is accelerating downwards. It's tempting to think that the rope still supports the full weight of the object. In fact, it does not. If the system is in free fall, then the tension would be zero. And, in general, for a downwards acceleration of $a$, the tension will be $mg -ma$. This may be a case where your intuition is insufficient and you need to fall back on Newton's laws to make this calculation.

 

[Mister T]

$F_{net}=ma$
$mg-T=ma$

 

[sefatod]

Thank you all for your answers !
You're right, I was confused.
You really helped me with this one, thank you

 

[Steve4Physics]

I’m probably going to regret this but it’s annoying me, so …

Sign-convention: ‘up is positive’.

Conventionally, $g$ is taken to be positive: $g = 9.8m/s^2$ approx. For example, the vertical acceleration of a free-falling object is $a = -g = -9.8m/s^2$.

A mass $m$ (kg) hangs on a cord suspended from an elevator which is descending with an acceleration of $0.1g$.

Consider these 2 possibilities:

a) The elevator could be going down with speed decreasing. In this case the direction of acceleration is up; acceleration is positive.

b) The elevator could be going down with speed increasing. In this case, the direction of acceleration is down; acceleration is negative.

The question says the acceleration is '$0.1g$'. This should be interpreted as a positive value, so a) applies.

But the official answer (8.8m N) is based on b) and is therefore wrong.

 

[kuruman]

But the official answer (8.8m N) is based on b) and is therefore wrong.

The statement of the problem is

"A mass m (kg) hangs on a cord suspended from an elevator which is descending with an acceleration of 0.1g."

I agree that 0.1g is the the magnitude of the acceleration, but what about the direction of the acceleration? We are told that the elevator is "descending" which means that the velocity is down. We are not told whether that velocity and the acceleration are in the same (down) or opposite (up) direction.

Most readers (myself included) would assume that "descending with an acceleration of 0.1g" implies that the acceleration is also down. Clearly, this was the intent of the problem's author. I wouldn't say that the answer is wrong, but I would agree that the question is ambiguous because it admits two answers as you correctly pointed out.

 

[Mister T]

The question says the acceleration is '0.1g'. This should be interpreted as a positive value,

Only if your convention of 'up is positive' is adopted.

What you've revealed is that question is poorly worded as we don't know if the elevator is speeding up or slowing down.

 

[haruspex]

Only if your convention of 'up is positive' is adopted.

What you've revealed is that question is poorly worded as we don't know if the elevator is speeding up or slowing down.

It does say "descending with acceleration ….". Since it is irrelevant whether it is ascending or descending, it is reasonably clear that the intention is that the acceleration is positive downwards.