Sep 2019 CSAT mock test, Physics I #19

Classical Mechanics Level 3

As shown in the picture, a rigid bar with length 8 L 8L is in equilibrium on two fulcrums A , B , \rm A,~B, horizontally. A \rm A and the weight are respectively 2 L 2L and L L away from the left end of the bar, and the ball at rest on the horizontal plane is connected with the bar with a string. The mass of the weight, the bar, and the ball are 2 m , 5 m , 2m,~5m, and 3 m 3m respectively, and the distance between A A and B \rm B is x . x. The force applied to the bar by A \rm A is equal to the force applied to the bar by B , \rm B, and is twice the force applied to the ball by the horizontal plane.

Find 6 x L . \dfrac{6x}{L}.

Assumptions

  • The density of the bar is uniform.
  • The width and thickness of the bar, the mass of the string, and the size of the weight are negligible.


The answer is 21.

Boi (보이) by Boi (보이) , 19, South Korea

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1 solution

Let the force applied to the ball by the horizontal plane be N N . Then the force applied to the bar by the fulcrums A A and B B is 2 N 2N each. Applying force balance equation to the system, we get 5 N = 10 m g 5N=10mg or N = 2 m g N=2mg . Applying the moment balance equation about the left end of the bar, we get N L ( 4 + 4 + 2 x L ) = m g L ( 2 + 20 + 24 ) NL(4+4+\dfrac{2x}{L})=mgL(2+20+24) or 32 + 4 m g x L = 46 32+4mg\dfrac{x}{L}=46 , or x L = 7 2 \dfrac{x}{L}=\dfrac{7}{2} or 6 x L = 21 \dfrac{6x}{L}=\boxed{21} .

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