Answer to Question #153015 in Mechanics | Relativity for Faniyi Abdulwaasi

Question #153015

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?


1
Expert's answer
2020-12-29T15:31:55-0500

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 ,

Tsinθmg=0Tsinθ=60kg×9.8=588N(1)\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,

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

  • By (1)/(2),

tanθ=588FF=588tanθ\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, tanθ\tan\theta can be found. Chain is 13m, horizontal is 5m hence the verticle side 12m (by pythagoras's)
  • Thereofre,

F=588125=245N\qquad\qquad\qquad \begin{aligned} \small F&= \small \frac{588}{\frac{12}{5}}=245N \end{aligned}


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