Determine the normal forces to hold an upside down cone

[Demetrius]

Homework Statement

With two fingers, you hold an ice cream cone motionless upside down, as shown in the figure. The mass of the cone is m and the coefficient of static friction between your fingers and the cone is μ.μ When viewed from the side, the angle of the tip is 2Θ. What is the minimum normal force you must apply with each finger in order to hold up the cone? In terms of Θ, what is the minimum value of μ that allows you to hold up the cone? Assume that you can supply as large a normal force as needed.

Homework Equations

(eq.1) $F_s = μ_sN$
(eq. 2) $F_s ≤ μ_sN$

The Attempt at a Solution

After drawing a free body diagram, I wrote down the following equations:

$$

(eq.3)∑F_x = NcosΘ - NcosΘ + F_fsinΘ - F_fsinΘ

$$

$$

(eq.4) ΣF_y = mg + NsinΘ + NsinΘ - F_fcosΘ - F_fcosΘ

$$

Eq. 4 simplifies into:

$$

(Eq.4) 2F_fcosΘ = mg + 2NsinΘ

$$From this, I concluded that I need to only focus on the vertical components as the horizontal components cancel out with one another.

My only attempt at solving this is the following:

Since we need the Minimum normal force then we will need the maximum Static friction. So I can use Eq.1 instead of Eq.2. Additionally, In Eq.4 I can exchange $F_f$ for $μ_sN$ and I will have the following:

$$
2F_fcosΘ = mg + 2NsinΘ
$$
$$
2μ_sNcosΘ = mg + 2NsinΘ
$$
$$
N = \frac {mg} {2(μ_scosΘ-sinΘ)}
$$

However, this seems wrong. I do not think this will count as showing my work. Also, should I be using Eq.1 or Eq.2?

 

[Orodruin]

Why do you think it seems wrong?

 

[Demetrius]

Why do you think it seems wrong?

Well, because when I swap out $F_f$ for $μ_s$ it feels as if I was making a mistake? Am I overthinking the problem?

Also for the second part of the question, here is my answer:

$$

μ_s = \frac {mg} {2NcosΘ} + {tan Θ}

$$

This answer makes sense to me. As the $mg$ increase than $μ_s$ needs to increase as well to compensate for the downward force.

Plus, as Θ → 0 $pi/2$ then $μ_s$ → ∞, which makes sense because the normal force will only be in the vertical direction, downward, and to keep the cone from falling then the $μ_s$ needs to be extremely large.

 

[Orodruin]

Am I overthinking the problem?

Yes.

Also for the second part of the question, here is my answer:

$$

μ_s = \frac {mg} {2NcosΘ} + {tan Θ}

$$

This answer makes sense to me. As the $mg$ increase than $μ_s$ needs to increase as well to compensate for the downward force.

In addition, you could consider the case when gravity is absent. In this limit $\mu_s = \tan\theta$, representing the coefficient of friction needed in order the applied forces themselves not to push the cone away.

 

[Demetrius]

Okay just to be clear, the expression of $μ_s$ that I gave is the correct answer for the second question? Or are you telling me that the expression you gave is the solution and I need to rework my solution?

I am little confuse mainly because your expression is more clean than mines since it only depend on a function of Θ.

 

[TSny]

$$μ_s = \frac {mg} {2NcosΘ} + {tan Θ}$$

"Assume that you can supply as large a normal force as needed."

 

[Demetrius]

I understand, in order to supply a Large Normal force then gravity must be absent. There fore the answer is: $u = tan(theta)$

Right?

 

[TSny]

... in order to supply a Large Normal force then gravity must be absent.

I don't understand this statement. You are allowed to supply any value of N with gravity being always present.

There fore the answer is: $u = tan(theta)$

Right?

Yes. But I don't know exactly what your reasoning was to get this answer.