(a) 40.3 Nm
When a force is applied perpendicular to the door, the magnitude of the torque exerted by the person is given by
[tex]\tau = F d[/tex]
where
F = 42 N is the magnitude of the force
d = 96 cm = 0.96 m is the distance between the point of application of the force and the centre of rotation of the door
Substituting numbers into the formula, we find
[tex]\tau = (42 N)(0.96 m)=40.3 Nm[/tex]
(b) 34.9 Nm
In this case, the force is no longer perpendicular to the door, so we must use the formula for the general case:
[tex]\tau = Fd sin \theta[/tex]
where
[tex]\theta = 60^{\circ}[/tex] is the angle between the direction of the force and the face of the door
Substituting numbers into the formula, we find
[tex]\tau=(42 N)(0.96 m)(sin 60^{\circ})=34.9 Nm[/tex]
(a) The magnitude of the torque if the force is exerted perpendicular to the door is 40.32 Nm.
(b) The magnitude of the torque if the force is exerted at a 60 degrees angle to the face of the door is 33.46 Nm.
The torque exerted by the applied force is determined from the product of the applied force and the radius through which the force is applied.
τ = Fr x sin(90)
τ = 42 x 0.96
τ = 40.32 Nm
τ = Fr x sin(60)
τ = 42 x 0.92 x 0.886
τ = 33.46 Nm
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