Question

a uniform wooden plank of mass 30kg is kept on two supports X and Y the length of plank is L and reaction force applied by Y is
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Answer 1

Given:

A uniform wooden plank of mass 30kg is kept on two supports X and Y the length of plank is L.

To find:

Reaction force applied by Y.

Calculation:

First refer to the attached diagram to understand the position of X and Y.

Since the plank is in Rotational Equilibrium, we can consider the position of X as the instantaneous axis of rotation ;

$\therefore \: \sum{( \tau)} = 0$

$= >( N_{X} \times 0) + (N_{Y} \times L) - (30g \times \dfrac{L}{2} ) = 0$

$= > (N_{Y} \times L) - (30g \times \dfrac{L}{2} ) = 0$

Putting g = 10 m/s²:

$= > (N_{Y} \times L) - (300 \times \dfrac{L}{2} ) = 0$

$= > (N_{Y} \times L) = (300 \times \dfrac{L}{2} )$

$= > N_{Y} = 300 \times \dfrac{1}{2}$

$= > N_{Y} = 150 \: N$

So, final answer is:

$\boxed{ \red{ \bold{ \large{N_{Y} = 150 \: N }}}}$

Answer 2

Answer:

200N

Explanation:

300×L/2=Ry×3L/4

Ry=200N