F=ma
Therefore the net force = 1000kg × 2 metres per second per second
So F=2000 N
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:
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The buoyant force on the branch is given by:
F = pVg
F = buoyant force, p = water density, V = volume of branch submerged, g = gravitational acceleration
Given values:
p = 1000kg/m^3
V = 0.75x1.12m^3 (75% of total volume)
g = 9.81m/s^2
Plug in and solve for F:
F = 1000(0.75x1.12)(9.81)
F = 8240N
