Answer to Question #306968 in Mechanics | Relativity for Aria

Question #306968

A uniform 10 kg ladder 2.5 m long is placed against a frictionless wall with its base on the ground 80 cm from the wall.




Find the magnitude of the force exerted on the wall and on the ground.





1
Expert's answer
2022-03-10T10:18:32-0500

Explanations & Calculations


  • Since the wall is frictionless, the only force acting on the ladder by the wall is the normal reaction force say R\small R.
  • And the floor should have friction to keep the ladder sliding on the floor so there are 2 forces acting on the ladder from the floor: the normal force and the friction force say S\small S and f\small f respectively.
  • With the horizontal equilibrium, it is obvious that f=R\small f = R and from the vertical equilibrium S=mg\small S= mg therefore, only R\small R or f\small f needs to be found.
  • Take moment around the point where the ladder touches the floor.

R×(2.5m)2(0.8m)210kg×9.8ms2×0.4m=0R=16.6N\qquad\qquad \begin{aligned} \small R\times\sqrt{(2.5m)^2-(0.8m)^2}-10kg\times9.8\,ms^{-2}\times0.4\,m&=\small 0\\ \small R&=\small 16.6\,N \end{aligned}

  • Then, the force at the wall is R=16.6N\small R=16.6\,N
  • Force on the ground point is

F=S2+f2=(10×9.8)2+16.62=\qquad\qquad \begin{aligned} \small F&=\small \sqrt{S^2+f^2}=\sqrt{(10\times9.8)^2+16.6^2}\\ &=\small \end{aligned}


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