# How to find the the force necessary to pull from a cylinder so it keeps in place?

#### Chemist116

The problem is as follows:

A cylinder of $\textrm{500 grams}$ in mass has a very thin flexible non elastic tin wire of negligible weight winded around it as shown in the figure from below. By how much a force must be applied to pull the wire so that the cylinder spins and keeps in place?. Assume that the coefficient of friction is $0.3$ and the acceleration due gravity is $9.8\,\frac{m}{s^2}$

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&\textrm{3 N}\\ 2.&\textrm{2.5N}\\ 3.&\textrm{1.5N}\\ 4.&\textrm{0.15N}\\ 5.&\textrm{4.5N }\\ \end{array}$

I'm not sure exactly if I'm understanding this problem correctly. What I've attempted to do here was to assume that the condition which must be met is given by:

$\sum ^n_{i=1}\tau_{i}=0$

Therefore:

$-F\cos 37^{\circ}\cdot R-F\sin 37^{\circ}\cdot R + f_R\cdot R = 0$

Hence:

$f_R=F\cos 37^{\circ}+F\sin 37^{\circ}=F\frac{4}{5}+F\frac{3}{5}$

$f_R=\frac{7}{5}F$

But:

$f_R=\mu N$

$N=mg-F\sin 37^{\circ}$

$f_R=\frac{3}{10}(mg-F\sin 37^{\circ})=\frac{3}{10}\left(0.5\times 9.8-\frac{3F}{5}\right)$

Solving this thing yield:

$\frac{3}{10}\left(0.5\times 9.8-\frac{3F}{5}\right)=\frac{7}{5}F$

Hence:

$F=0.93\,N$

But it doesn't check with any of the alternatives given. What did I missunderstood?. Can someone help me here please?.

#### skeeter

Math Team
Translational equilibrium ...

$F\cos(37) = \mu \cdot N$

$F\cos(37) = \mu[mg - F\sin(37)]$

$0.8 F = 0.3 (0.5g - 0.6F) \implies F = 1.5 \text{ N}$

rotational dynamics ...

$f = 0.3(.5g - 1.5 \cdot 0.6) = 1.2 \text{ N}$

$\tau_{net} = r(F - f) > 0$

#### Chemist116

Translational equilibrium ...

$F\cos(37) = \mu \cdot N$

$F\cos(37) = \mu[mg - F\sin(37)]$

$0.8 F = 0.3 (0.5g - 0.6F) \implies F = 1.5 \text{ N}$

rotational dynamics ...

$f = 0.3(.5g - 1.5 \cdot 0.6) = 1.2 \text{ N}$

$\tau_{net} = r(F - f) > 0$
I think it would help to have a free body diagram for this thing. But If I'm understanding correctly the value of the force is only given by horizontal component not the vertical one and by such reason the force necesary to pull is $1.5\,N$. But the second part is where I'm a bit confused. Why do I need to have the value for the frictional force?. The radius isn't known in this problem.

This condition $\tau_{net}>0$ what is it for?.

$r(1.5-1.2)>0$

$r>0$

This part is obvious the radius has to be greater than zero but I don't know what other conclusion can be taken from it?. Can you please explain this later part please?.

#### skeeter

Math Team
I think it would help to have a free body diagram for this thing.
When are you going to learn how to sketch free-body diagrams on your own? This is a shortcoming you need to address with your instructor.

But If I'm understanding correctly the value of the force is only given by horizontal component not the vertical one and by such reason the force necesary to pull is $1.5\,N$. But the second part is where I'm a bit confused. Why do I need to have the value for the frictional force?. The radius isn't known in this problem.

This condition $\tau_{net}>0$ what is it for?.

$r(1.5-1.2)>0$

$r>0$

This part is obvious the radius has to be greater than zero but I don't know what other conclusion can be taken from it?. Can you please explain this later part please?.
reference the original problem statement ...

A cylinder of 500 grams in mass has a very thin flexible non elastic tin wire of negligible weight winded around it as shown in the figure from below. By how much a force must be applied to pull the wire so that the cylinder spins and keeps in place?.
Having the cylinder stay in place requires a state of translational equilibrium.

Spinning the cylinder requires a net torque. It isn't necessary to calculate it, but it is necessary to know $\tau_{net} > 0$, causing the cylinder to spin.

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Chemist116

#### Chemist116

When are you going to learn how to sketch free-body diagrams on your own? This is a shortcoming you need to address with your instructor.

View attachment 11097
Thanks for helping me with this. I must say that is not that I'm not capable to do the FBD on my own, but it seems that I'm struggling in diagrams when one body is supported below another (except the floor). Which isn't in this case. I wished my instructor had explained this better but not. Thus I had to learn most of this on the go and by my own. Sorry if this may had caused you troubles. Btw, I arrived to the exact FBD you had drawn on the right. But I was confused on why there was a necesity of having to compute the net torque. Thus your first diagram on the left made it all clear. Now that I think on it was just obvious. Net torque has to be greater than zero when this object is spinning.

reference the original problem statement ...

Having the cylinder stay in place requires a state of translational equilibrium.

Spinning the cylinder requires a net torque. It isn't necessary to calculate it, but it is necessary to know $\tau_{net} > 0$, causing the cylinder to spin.
As I mentioned in the above paragraph. Yes I noticed this. My initial confusion was why did you put this at the end of the solution?. But it turns out that it was only to specify that when there isn't a rotational equilibrium the net torque isn't zero. But now is much clearer.