The life of the sugar ant

This question is a bit tricky. The reader might think the ant will minimize the force required to move the candy if he applies a completely horizontal force. However, that is not the best strategy. As we will show, the best strategy of the ant will be to also apply a vertical force as to minimize friction (see the last clarification: Do not assume anything unwarranted about the directions of the forces involved ).
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Let $\theta$ be the angle the force vector makes with the horizontal. Let $N$ be the normal force. The horizontal and vertical components of the force are respectively $F\cos(\theta)$ and $F\sin(\theta)$ . See the free body diagram below:
Here's the link in case the picture doesn't load: http://s24.postimg.org/wa23fwncl/Untitled2.png

Balancing the vertical forces, we obtain: $N + F \sin(\theta)= mg$ $\implies N= mg - F\sin(\theta)$ The maximum friction the ramp can exert in the horizontal direction is given by $kN$ . For the ant to be able to move the candy, the horizontal component of the force must be greater than this value. Since we want to minimize the horizontal force, we obtain: $F \cos(\theta) = kN$ Recall the expression for $N$ we got earlier: $N= mg - F \sin(\theta)$ Plugging this value, we obtain: $F\cos(\theta)= k( mg - F \sin(\theta))$ Solving for $F$ yields $F= \frac{kmg}{k \sin(\theta) + \cos(\theta)}$ We wish to minimize $F$ . To do this, note that $kmg$ is a constant, so we'll be done if we can maximize $k\sin(\theta) + \cos(\theta)$ . The brute-forcey way would be to differentiate this w.r.t $\theta$ and set it equal to $0$ . But let's try something else. We have the well known Cauchy-Schwartz inequality for our aid! Note that $k \sin(\theta) + \cos(\theta)= \sqrt{(k \sin(\theta) + \cos(\theta))^2}$ $= \sqrt{(k \times \sin(\theta) + 1 \times \cos(\theta))^2 }$ $\geq \sqrt{(k^2+1)(sin^2(\theta) + \cos^2(\theta))^2}$ $= \sqrt{k^2+1}$ We then have $F \geq \frac{kmg}{\sqrt{k^2+1}}$ Now that we have got our bound, let's investigate whether equality can occur (to ensure that no tighter bound exists). In Cauchy Schwartz inequality, equality holds when the terms are proportional to each other, i.e $\frac{k}{1}= \frac{\sin(\theta)}{\cos(\theta)}$ $\implies \theta= \tan^{-1} (\theta)$ Note that the range of $\tan(\theta)$ is $[-\infty, \infty]$ , implying we can always find a $\theta$ for which equality can occur.

Plugging the values from the question, we find out that $F_{min} \approx 21.9 \text{ mN}$ , which is attained for $\theta \approx 26.57^{\circ}$ .

to minimize the force, the ant will have to pull the candy slightly upwards, at an angle.

Let’s take the angle between F(pull) and the horizontal to be θ. Then, the reaction force on the candy will be (mg - Fsin θ), and the frictional force will be k(mg - Fsin θ) . This is equal to the horizontal pull force, Fcos θ.

k(mg - Fsin θ) = Fcos θ

Rearranging this, we get F = kmg/(ksinθ+cosθ)

Now, we differentiate this and equate to zero, to find the minimum.

dF/dθ = -kmg〖(k sin⁡〖(θ)+ cos⁡〖(θ))〗 〗〗^(-2 )×(kcosθ- sinθ)

Equating this to zero, we get kcosθ=sinθ, and k=0.5 Therefore θ=26.57 Thus we get Force = 21.9

Let the ant exerts a force $F$ at an angle $\theta$ with the horizontal. Resolving the force we get two components $Fcos\theta$ in the horizontal direction and $Fsin\theta$ in the vertical direction.

Drawing a free body diagram of the candy the forces on candy are:

Normal reaction $N$ , $Fsin\theta$ vertically upwards and its weight $mg$ in the vertically downward direction and $Fcos\theta$ in the horizontal direction.

As there is equilibrium in vertical direction the forces in the vertically upward and downward direction must be equal.

$\Rightarrow$ $N+Fsin\theta=mg$ . Plugging in the values of $m=0.005$ kg and $g=9.8m/s^2$ we get $N+Fsin\theta = 0.049$ or $N=0.049-Fsin\theta$ .

Note that the minimum force required is equal to the frictional force in the opposite direction.

Friction $f=kN=0.5 \times (0.049-Fsin\theta)$ . Solving and rearranging we get:

$F(cos\theta+0.5sin\theta)=0.0245 .........(1)$

To find the minimum force we differentiate the equation and put the derivative as 0.

$\Rightarrow 0.5Fcos\theta=Fsin\theta \Rightarrow tan\theta=0.5$ which gives us $\boxed{\theta=26.565^{\circ}}$ .

$\Rightarrow cos\theta=0.8944$ and $sin\theta=0.4472$ .

Plugging these values in equation $(1)$ we get the value of $F=\boxed{21.9135}$

$\square$

If you think the ant would push the candy horizontally, look at the last hint and look around to see an ant push its food. Let's say the ant pushes its food with a force F at an angle x with the horizontal. The horizontal component of the ant's force should at least equal the frictional resistance.

* F cos(x) * = * k (mg - F sin(x) ) *

Therefore, * F = \frac {k mg} {cos(x) + k sin(x)} *

\frac {dF} {dx} = 0 at minimum F,

* x = cos^{-1} {\frac {1} {sqrt(k^2 + 1)}} * Substituting 0.5 for k, we find that, x = 26.57 degrees.

Substituting this value in the expression for the force, we get * F = 21.91 mN * (yes, milli newton since we took mass in grams)

We cannot assume that the force exerted is horizontal because if the force is angled upward, the normal force is decreased and thus the frictional force is decreased. So, we can find the exerted force as a function of its angle with respect to the horizontal, and optimize this function.

Using a free body diagram ( here is an example), we put the exerted force at an angle $θ$ . We know that this angle is above the horizontal because putting it below the horizontal would increase the normal force and thus increase the frictional force. Using Newton's second law for $x$ and $y$ components, we get:

1. $N+F\sinθ-mg=0\Rightarrow N=mg-F\sinθ$
2. $F\cosθ-f=0\Rightarrow F\cosθ=f$

Since we know $f=μ_{k}N$ , we can plug in the value of $N$ from the first equation. We can also substitute the value of $f$ from the second equation. Now, we have:

$F\cosθ=μ_{k}(mg-F\sinθ)$

Solving for F, we get:

$F=\frac{μ_{k}mg}{\cosθ+μ_{k}\sinθ}$

F is minimized when $\frac{dF}{dθ}=0$ , so we find the derivative of F using the chain rule:

$\frac{dF}{dθ}=μ_{k}mg\frac{\sinθ-μ_{k}\cosθ}{(\cosθ+μ_{k}\sinθ)^{2}}=0$

For this to be equal to 0, the numerator needs to be 0:

$\sinθ-μ_{k}\cosθ=0\Rightarrow θ=\tan^{-1}(μ_{k})=\tan^{-1}(0.5)=26.565°$

Plugging in this value of $θ$ into $F(θ)$ yields $F=0.0219 N=\fbox{21.9 mN}$ .

For minimum force , F cos a = f [eqn 1] Where F is the force required , a is the optimum angle for min.force f is the friction force

f is related to F and mg by : f = ( mg - F sin a ) k [eqn 2 ] Where k represents the coefficient of friction

Equiting the two equations F cos a = ( mg - F sin a ) k

Solving for F , F = (mgk)/( ( cos a ) + k sin a ) [eqn 3]

For min.force , dF/da = 0 Differentiating equation 3 with respect to a, dF/da = -mgk*([k cos a ] - sin a)/(cos a + k sin a )^2

Solving for a when dF/da equal to 0 , a = 26.5650512° Substituting a into eqn 3 , F = 0.0219 N or 21.9 mN

Perhaps the hardest part of the problem is enumerating the forces.

Say the mass of the cube is $W$ and we have our normal force $F_N$ and our applied force $F_A$ . It is important to note that $F_N$ does, indeed, depend on $F_A$ . Since there is no net acceleration, the forces must balance, giving us: $\sum_FF=0$ That is: $\begin{aligned} W-F_N-F_A\sin\theta&=0\\ \mu F_N-F_A\cos \theta&=0 \end{aligned}$ Where $\theta$ is the angle from the horizontal.

We solve for $F_N$ , giving us: $F_N=F_A\frac{\cos\theta}{\mu}$ And put it back into (1): $W-F_A\frac{\cos\theta}{\mu}-F_A\sin\theta=0$ A quick rearrangement shows: $F_A=\frac{W}{\frac{\cos\theta}{\mu}+\sin\theta}$ We derive to find our minimum: $\frac{\partial}{\partial\theta}F_A=\frac{\mu W(\mu \cos\theta-\sin\theta)}{(\cos\theta+\mu\sin\theta)^2}=0=\mu\cos\theta-\sin\theta$

Solving, we receive $\theta \approx .46365$ , hence: $F_A\approx .0219135\text{ N}=21.9135\text{ mN}$

To have the minimum force, u need the least friction. To decrease friction, u can have a vertical component of force, so that the normal force decreases. However, we too much vertical component means too little horizontal component, and thus it won't be able to overcome the force of friction. We want to find the point where the horizontal force = friction force (that is the minimum force required), and we also want at the same time as much vertical force as possible. Then we can differentiate to find a maximum.

$\Sigma F_x = F\cos\theta - \mu n = 0 \Rightarrow F\cos\theta = \mu n$ $[1]$

$\Sigma F_y = n + F\sin\theta - mg = 0 \Rightarrow n = mg - F\sin\theta$ $[2]$

$[2]\rightarrow [1] : F\cos\theta = \mu (mg - F\sin\theta)$

$F(\mu\sin\theta + \cos\theta) = \mu mg$

$F = \frac{\mu mg}{\mu\sin\theta + \cos\theta}$ $[3]$

In order to find the minimum force, we must find the derivative and set it to 0

$\frac{dF}{d\theta} = \frac{-\mu mg(\mu\cos\theta - \sin\theta)}{(\mu\sin\theta + \cos\theta)^{2}} = 0$

$\mu\cos\theta - \sin\theta = 0 \Rightarrow \tan\theta = \mu \Rightarrow \theta = \tan^{-1}\mu$ $[4]$

$[4]\rightarrow [3] : F_{min} = \frac{\mu mg}{\mu\sin(\tan^{-1}\mu) + \cos(\tan^{-1}\mu)} = 0.02191$ $N$ $= 21.91$ $mN$