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DedPeter [7]
2 years ago
15

(15 Points)

Physics
1 answer:
oksian1 [2.3K]2 years ago
5 0

The vertical weight carried by the builder at the rear end is F = 308.1 N

<h3>Calculations and Parameters</h3>

Given that:

The weight is carried up along the plane in rotational equilibrium condition

The torque equilibrium condition can be used to solve

We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person

This would lead to:

F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)

F(1cos20)= 197/2(3.10sin20 + 2 cos 20)

Fcos20= 289.55

F= 308.1N

Read more about vertical weight here:

brainly.com/question/15244771

#SPJ1

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This question discusses about the increased range. So, we shall assume that the angle of jumping will be  as the horizontal range is maximum at this angle.

Step 1 of 3<

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Step 2 of 3<

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