Respuesta :
Explanation:
It is given that,
Force acting on the door, [tex]F_1=555\ N[/tex]
Perpendicular distance, [tex]r_1=0.594\ m[/tex]
The torque acting on the business man is given by :
[tex]\tau_1=F_1\times r_1[/tex]
[tex]\tau_1=555\ N\times 0.594\ m[/tex]
[tex]\tau_1=329.67\ N[/tex].........(1)
Force applied by the boy, [tex]F_2=376\ N[/tex]
Perpendicular distance, [tex]r_2=1.45\ m[/tex]
The torque acting on the boy is given by :
[tex]\tau_2=F_2\times r_2[/tex]
[tex]\tau_2=376\ N\times 1.45\ m[/tex]
[tex]\tau_2=545.2\ N[/tex].........(2)
From equation (1) and (2) it is clear that the force applied by the business man is less than the boy.
Let clockwise direction is taken as positive while counterclockwise direction is negative. As the boy is pushing to the left of the hub. So, the direction of the torque exerted by the boy is counterclockwise. Hence, this is the required solution.
The direction of the torque exerted by the boy is; counterclockwise.
What is the direction of the Torque?
We are given;
Force acting on door; F₁ = 555 N
Perpendicular distance; r₁ = 0.594 m
Force applied by boy; F₂ = 376 N
Perpendicular distance; r₂ = 1.45 m
The torque acting on the businessman is gotten from;
τ₁ = F₁ * r₁
τ₁ = 555 * 0.594
τ₁ = 329.67 N
The torque acting on the boy is gotten from;
τ₂ = F₂ * r₂
τ₂ = 376 * 1.45
τ₂ = 545.2 N
Since the torque acting on the boy is more than that acting on the business man, it means that the boy is pushing to the left of the hub. Thus, the direction of the torque exerted by the boy will be counterclockwise.
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