Problem A particle of mass 4kg is suspended from a point A on a vertical wall by means of a light inextensible string of length 130cm.
a) A horizontal force, P is applied to the particle so that it is held in equilibrium a distance of 5ocm from the wall. Find the value of P and the tension in the string.
b) By drawing a triangle of forces, or otherwise, find the magnitude and direction of the minimum force that would hold the particle in this position, and the tension in the string which would result.
I have been trying to understand this problem for the past 4 days and I am struggling to get an intuition for forces.
In the problem described below, what does the magnitude of a vector, or length of a side of a triangle have to do with the force?
Is the force P different to the tension in the string?
Ultimately, what am I trying to find?
The first thing to do in these kind of problems is to draw a body-diagram in which you draw the situation and the forces involved in the problem. So here it is:
Now we use the information given in the problem, for instance we know that $$130\sin\theta=50 \Longrightarrow \theta\approx 22.6º.$$
We know that the particle doesn't move, so if we sum the forces acting on the object in the $y$-axis and in the $x$-axis we will obtain $0$ as the answer. So, taking $g=10$m/s (I won't write units from now on): $$ T_y-mg=0\Longrightarrow T_y=mg=4·10=40= \|\boldsymbol T\|\cos\theta\Longrightarrow \|\boldsymbol T\|\approx 43.2.$$ Now it's time to sum forces in the $x$ axis: $$T_x-P=0\Longrightarrow P=T_y=\|\boldsymbol T\|\sin\theta\approx16.6.$$
In a general case we have the following diagram:
where $b$ and $a$ are the known lengths.
Now for part b you know that: \begin{align} \text{$x$-axis: } P\cos\alpha=T\sin\theta, &&\text{$y$-axis: }P\sin\alpha+T\cos\theta=mg. \end{align} Here you need to solve for $P$ and you will obtain a function $P(\alpha)$. You then can minimize it and obtain your answer.