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Julli [10]
3 years ago
5

A uniform 40-N board supports two children weighing 500 N and 350 N. If the support is at the center of the board and the 500-N

child is 1.5 m from its center, How far is the 350-N child from the center?
Physics
1 answer:
serg [7]3 years ago
5 0

Answer:

b= 2.14 m

Explanation:

Given that

Weight of the board ,wt = 40 N

Wight of the first children , wt₁=500 N

Weight of the second children ,wt₂ = 350 N

The distance of the 500 N child from center ,a= 1.5 m

lets take distance of the 350 N child from center = b m

Now by taking the moment about the center of the board

We know that moment = Force x Perpendicular distance from the force

wt₁ x a = wt₂ x b

500 x 1.5 = 350 x b

b= 2.14 m

Therefore the distance of the 350 N weight child from the center is 2.14 m.

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(2,1), (2,2) (2,3), (2,4) (2,5) (2,6)

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(5,1), (5,2) (5,3), (5,4) (5,5) (5,6)

(6,1), (6,2) (6,3), (6,4) (6,5) (6,6)

From here it can be seen that there are only two cases where the sum on the two die is 11 i.e., (6,5) and (5,6)

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