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Grade 12Mechanics

A painter of mass 60 kg paints the wall of a house standing on a criss cross stand and a long wooden board placed on it of mass 15kg. How close to its end can he stand without risking to topple board?

Vaishnavi dagabaj

7 Years agoGrade 12

2 Answers

Arun

7 Years ago

The center of gravity of the board is at it's center, i.e. 1.25 m from the support (pivot) point. The weight of the painter (W_p) is 60*g, the weight of the board (W_b) is 15*g. Take the moments about the support point closest to the painter, using "x" as the painter's distance from the end of the board. The moment arm of the painter to the support point is ( 1.5 - x ).

(W_p)*(1.5 - x) - (W_b)*(1.25) = 0

(W_p)*1.5 - (W_b)*1.25 = (W_p)*x

x = [ (W_p)*1.5 - (W_b)*1.25 ] / (W_p)

x = [ 60*9.9*1.5 – 15*9.8*1.25 ] / (60*9.8)

(Notice that "g" could have been factored out and canceled here.)

Now calculate x

Karan Joshi

5 Years ago

Given,

Mp=60kg

Mb=15kg

To not topple down it must be in mechanical equillibrium.

So, Total torque must be zero.

i.e. Torque by painter= Torque by board

Mp×g×(1.5-x)=Mb×g×1.25

60×(1.5-x)=15×1.25

x=1.1875m or 1.2m

So the painter can stand at 1.5-x distance from the sypport

So, 1.5-1.2m = 0.3 m is the max. distance from the support.

If you liked the explanation the much better content will be provided by me on youtube with name STUDY WITH ME by karan joshi.

Thank you. Waiting for your comments

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