Answer:
Explanation:
From the given information:
From the rotational axis, the distance of the force of gravity is:
d_g = 25+5.0 cm
d_g = 30.0 cm
d_g = 30.0 × 10⁻² m
However, the relative distance of FB cos 75.9° from the axis is computed as:
d_B = 5.0 cm
d_B = 5.0 × 10⁻² m
The net torque rotational equilibrium = zero (0)
i.e.
[tex]\tau_g -\tau_B = 0 \\ \\ F_gd_g -F_gcos 75.9^0 d_B = 0 \\ \\ F_B = \dfrac{F_g d_g}{F_g cos 65.6} \\ \\ F_B = \dfrac{(3.2)(9.8)(30*10^{-2})}{(5.0*10^{-2} * cos 75.9)} \\ \\ \mathbf{F_B = 772.4 N}[/tex]
= 772.4 N
Thus, the force exerted = 1772.4 N
