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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
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Answer:
The spring constant = 104.82 N/m
The angular velocity of the bar when θ = 32° is 1.70 rad/s
Explanation:
From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:


Also;

Thus;

where;
= deflection in the spring
k = spring constant
b = remaining length in the rod
m = mass of the slender bar
g = acceleration due to gravity


Thus; the spring constant = 104.82 N/m
b
The angular velocity can be calculated by also using the conservation of energy;






Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s
The speed of car is 100.8km/h









v car= 28×3.6
=100.8km/h
Hence, the speed of the car is 100.8km/h
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