A door can just be opened with $10 N$ force on the handle of the door. The handle is at a distance of $50 c m$ from the hinges. Then the torque applied on the door is ( in $N m$ ):

5

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Solution

The correct option is A $5$
The torque would simply be the cross product of the distance and the force applied.
Thus
r = 50 cm = 0.5 m
$¯ τ = \to r \times \to F$
$= (0.5) \times (10)$
$= 5 N m$

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A door can just be opened with 10N force on the handle of the door. The handle is at a distance of 50cm from the hinges. Then the torque applied on the door is ( in Nm ):

The correct option is A 5The torque would simply be the cross product of the distance and the force applied.Thus r = 50 cm = 0.5 m¯τ=→r×→F=(0.5)×(10)=5Nm

A door can just be opened with $10 N$ force on the handle of the door. The handle is at a distance of $50 c m$ from the hinges. Then the torque applied on the door is ( in $N m$ ):

The correct option is A $5$
The torque would simply be the cross product of the distance and the force applied.
Thus
r = 50 cm = 0.5 m
$¯ τ = \to r \times \to F$
$= (0.5) \times (10)$
$= 5 N m$