A particle of mass $m$ is suspended from a point A on a vertical wall by means of a light inextensible string of length $b$. Find the magnitude and direction of the minumum force $\boldsymbol P$ that would hold the particle in equilibrium at a distance $a$ from the wall.

I found this problem on MathSE and tried to help the OP with it but with no success. This is my sketch of the situation:

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So, for $b,a$ fixed lengths we know that $\theta = \arctan\frac{a}{\sqrt{b^2-a^2}}$. Since the sum of the forces must be $0$, the magnitude of $\boldsymbol P$ can be expressed by:

$$\|\boldsymbol P\|=\frac{mg-T\cos\theta}{\sin\alpha}=\frac{T\sin\theta}{\cos\alpha}.$$

I really don't know how to get another equation that relates $\alpha,\boldsymbol T$ and $\boldsymbol P$. Also I'm struggling with expressing equations that show a path leading towards a raw maximization-minimization calculus problem. I appreciate your thoughts on the problem and will be very thankful if you could give me a hint.


closed as off-topic by ACuriousMind, Kyle Kanos, Brandon Enright, JamalS, Pranav Hosangadi Dec 8 '14 at 22:28

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  • $\begingroup$ Is the hypotenuse, b, really involved in the tangent function? $\endgroup$ – DJohnM Dec 8 '14 at 4:53
  • $\begingroup$ Could you provide the link to Math.SE post/chat transcript for reference? $\endgroup$ – Kyle Kanos Dec 8 '14 at 15:33
  • $\begingroup$ @KyleKanos, Click here. $\endgroup$ – Vladimir Vargas Dec 8 '14 at 22:53

There's no such thing as minimum force for equilibrium. There is no range. Just a single value.

For equilibrium, there should be no net external forces on the particle, so $\vec{F}_{net} = \vec{P} + \vec{T} + m\vec{g} = 0$.

This would mean, in the horizontal direction: $P \cos \alpha = T \sin \theta$, or $P = T \sin \theta / \cos \alpha$

In the vertical direction, $P \sin \alpha + T \cos \theta = m g$

Solve these two, and you should get an equation that relates all the variables in the problem.

  • $\begingroup$ Oh my... The answer was in my face, I must have been so tired. Thanks. $\endgroup$ – Vladimir Vargas Dec 8 '14 at 13:47

In this problem you get two equations. One for forces in equilibrium along the horizontal direction and the other for the vertical direction. With the help of horizontal equation

$T \sin \theta = P \cos \alpha$

You can insert $T$ from above equation in the vertical force equation thus eliminating $T$. Then your answer gives $P$ as a function of $\alpha$ and $\theta$. Differentiate $P$ with respect to $\alpha$ to obtain the min-max or get a graphical solution. You have lumped up two equations in one.

  • $\begingroup$ Oh my... The answer was in my face, I must have been so tired. Thanks. $\endgroup$ – Vladimir Vargas Dec 8 '14 at 13:46

What you have written down is just the force balance. For any body to be in equilibrium, the net moment acting on it must also be zero. So use the moment balance relation also:

Moment about $A$ should be zero for equilibrium, $$ |\vec{P}|\times \cos (\theta-\alpha)=mg\times a$$

Now solve the above equation along with the force balance equation you have to get $\vec{P}$


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