Question

A mass of 60kg supported by a chain 13m long, is pulled a perpendicular distance of 5m from the original line of support. Calculate the horizontal force required?

Answer 1

Explanations & Calculations

• Assume the mass is hung by the chain of 13m & pulled perpendicularly.
• Considering the equilibrium of the system at a distance of 5m from the starting point, the force needed can be calculated.
• In the beginning, the needed force is zero as there is no force acting opposite to what we apply, and gradually a component of its own weight starts acting as an opposite force to what we apply as it moves in a circular arch from the start to the final position.

• At the equilibrium,

Applying Newton's second for the verticle equilibrium $\uparrow$ ,

$\qquad\qquad\qquad \begin{aligned} \small T\sin\theta-mg&=\small 0\\ \small T\sin \theta &= \small 60kg\times9.8=588N\cdots(1) \end{aligned}$

From the horizontal equilibrium,

$\qquad\qquad\qquad \begin{aligned} \small T\cos\theta &= \small F\cdots(2) \end{aligned}$

• By (1)/(2),

$\qquad\qquad\qquad \begin{aligned} \small \tan\theta&= \small \frac{588}{F}\\ \small F&= \small \frac{588}{\tan\theta} \end{aligned}$

• By simple geometry at the equilibrium position, $\tan\theta$ can be found. Chain is 13m, horizontal is 5m hence the verticle side 12m (by pythagoras's)
• Thereofre,

$\qquad\qquad\qquad \begin{aligned} \small F&= \small \frac{588}{\frac{12}{5}}=245N \end{aligned}$