(a) 26.1 Nm
The magnitude of the torque exerted by a force acting perpendicularly to a surface is given by:
[tex]\tau = Fr[/tex]
where
F is the magnitude of the force
r is the distance from the pivot
In this situation,
F = 47.5 N is the force applied
[tex]r=\frac{1.10 m}{2}=0.55 m[/tex] is the distance from the hinges (the force is applied at the center of the door)
So, the magnitude of the torque is
[tex]\tau = (47.5 N)(0.55 m)=26.1 Nm[/tex]
(b) 52.3 Nm
In this case, the force is applied at the edge of the door farthest from the hinges. This means that the distance from the hinges is
r = 1.10 m
So, the magnitude of the torque is
[tex]\tau =(47.5 N)(1.10 m)=52.3 Nm[/tex]
The magnitude of torque applied about the axis of the door when the force is applied at the center is 26.125 N.m
The magnitude of the torque applied at the edge farthest from the center is 52.25 N.m.
The given parameters;
The torque applied by the janitor is the product of applied force and perpendicular distance.
τ = F.r
The torque applied about the axis of the door when the force is applied at the center.
[tex]\tau = F \times \frac{r}{2} \\\\\tau = 47.5 \times \frac{1.1}{2} \\\\\tau = 26.125 \ N.m[/tex]
The magnitude of the torque applied at the edge farthest from the center ;
[tex]\tau = F\times r\\\\\tau = 47.5 \times 1.1\\\\\tau = 52.25 \ N.m[/tex]
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