Question

A uniform wooden plank 5 m long and weighing 40 kg is resting on two supports 0.5 m from each end. A boy weighing 45 kg stands 1.5 m from one end of the wooden plank. Find the reactions at the supports.

Answer 1

FOR DIAGRAM SEE THE ATTACHMENT
Now from the laws of equilibrium
∑F=0⇒R1−40g−45g+R2=0⇒R1+R2=85g=850 N
and
∑τ=0⇒R1×0.5−40g×2.5−45g×3.5+R2×4.5=0⇒R1+9R2=5150
Solving for R1 and R2
R1=312.5 NR2=537.5 N
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Answer 2

Answer :-

Refer the image.

∑ F = 0 -- From laws of equilibrium

R₁ - 40g - 45g + R₂ = 0

Therefore, R₁ +  R₂ = 85 g

R₁ +  R₂ = 850 N - - - (1)

∑ ζ = 0

R₁ x 0.5 - 40 x 2.5 - 45 x 3.5 + R₂ x 4.5 = 0

Therefore, R₁ + 9 R₂ = 5150 - - - (2)

Solving we get,

R₁ = 312.5 N

R₂ = 537.5 N