A child pushes a door of width $0.8$ m perpendicular to the door at the handle with $20$ N force. The torque about the hinge is:
A$10$ N m, the simple half-width product value here
B$16$ N m, since $\tau = rF = 0.8\cdot 20 = 16$
C$25$ N m, dividing force by width on the door
D$0.04$ N m, the inverse ratio of force to width
Answer & Solution
Correct answer: B. $16$ N m, since $\tau = rF = 0.8\cdot 20 = 16$
$\tau = rF\sin 90^\circ = 0.8\cdot 20 = 16$ N m.
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