Question

Two builders carry a sheet of drywall up a ramp. Assume that W = 2.00 m, L = 3.10 m, θ = 20.0°, and that the lead builder carries a (vertical) weight of 197.0 N (44.3 lb). What is the (vertical) weight carried by the builder at the rear?

Answer 1

Answer:

Force carried by builder at rear end is F = 308.1 N

Explanation:

As we know that the weight is carried up along the plane in rotational equilibrium condition

So here we can use the torque equilibrium condition

Torque due to force of rear person about the position of front person = Torque due to weight of the block about the position of front person

so we will have

$F(W cos\theta) = mg sin\theta(L/2) + mgcos\theta(W/2)$

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

$F cos20 = 289.55$

$F = 308.1 N$