The classroom door is of width 50cm. The force of 5N is applied on the handle which is situated in the middle of the smaller plate as shown below. Compute the torque about the hinge (lever arm) of the door.
A
3.5Nm
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B
1.5Nm
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C
2.5Nm
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D
2.0Nm
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Solution
The correct option is D2.0Nm
The handle of the door is located at 10cm as shown in the figure.
Thus, the line of action is 202=10cm from the outer edge.
Distance from axis of rotation or lever arm =d=50−10=40cm=0.4m
Force exerted =5N
Thus, torque (τ)=5×0.4 τ=2.0Nm