Question
3. A person exerts a force of 55 N on the end of a door 74 cm wide. What is the magnitude of the torque if the force is exerted a)perpendicular to the door b)at a 45 degree angle to the door, c) at a 125 degree angle to the door?
Answers
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Torque= (F)sinTheta(r)
Given:
F=55N
d= 74cm (converted to M=0.74M)
Theta=90degrees.
Since,
Perpendicular is equal to 90 degrees, sin90 = 1.
T=55N(1)(0.74m)
Tperpendicular= 41N.
45degrees.
T=55(sin45)0.74
T=29N.
Hope this will help :)
Given:
F=55N
d= 74cm (converted to M=0.74M)
Theta=90degrees.
Since,
Perpendicular is equal to 90 degrees, sin90 = 1.
T=55N(1)(0.74m)
Tperpendicular= 41N.
45degrees.
T=55(sin45)0.74
T=29N.
Hope this will help :)