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

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?.

• This question would be more appropriate for the Physics Stack Exchange. Mar 24 '20 at 9:33

The module of the torque is $$F·R·\sin\alpha$$ being $$\alpha$$ the angle between the force and the arm and this is not what you wrote that it rather seems the decomposition of some force. So, the condition of rotation equilibrium is:

$$f_R·R-F·R=0$$

But this doesn't matter because one of your implicit assumptions is wrong. You assumed that it is a problem of statics, but it isn't. If the (center of mass of the) cylinder has to be in place, it has to have an angular acceleration around its center of mass: the above equations doesn't hold.

By Hypothesis, the center of mass is not moving, thus, the total force along an horizontal axis is zero:

$$F\cos 37º-f_R=0$$

Then

$$f_R=\frac{4F}{5}$$

that, with the equation you got,

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

Gives $$F=1.5$$, approx.

We can get the angular acceleration from here:

$$I\alpha=F·R-f_R·R$$

Being $$\alpha$$ the angular acceleration and $$I$$ the moment of inertia of the cylinder around its symmetry axis.

• How did you assume that there is no net horizontal force acting on it and yet that it has non-zero torque? The only forces acting on the cylinder with non-zero arms are those on the horizontal axis and they add up to zero. Mar 23 '20 at 10:38
• @Km356 Of course it's possible zero net force and non zero torque! How if not can you spin a wheel in place? Mar 23 '20 at 11:39
• Maybe I am confused. To spin a wheel in place, it must be attached to something that prevents the movement of its center of mass so that some force(which doesn't produce torque) opposes the horizontal force applied to the wheel. So this force stops the horizontal movement but doesn't affect the torque. This doesn't happen in this case since the opposing force itself (which is friction) produces a torque in the opposite direction. Mar 23 '20 at 11:52
• @Km356 You can have two forces producing torque and preventing to move the center of mass. Here we have a nice example of this. Mar 23 '20 at 12:18
• OK I understand, but this is still not the case here, the applied force AND the friction force produce torques in the opposite directions. They do stop the center of mass but they act at the same point on the cylinder in opposite directions and this forces them to stop it from spinning too. Mar 23 '20 at 12:31