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A door 1.2 mwide requires a force of $1 \mathrm{~N}$ to be applied perpendicular at the free end to open or close it. The perpendicular force required at a point 0.2 mdistant from the hinges for opening or closing the door is
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$6.0 \mathrm{~N}$
$\tau$ torque required to open/close the door will remain constant.
$\begin{aligned} & \tau=r_1 \times F_1=r_2 \times F_2 \\ & 1.2 \times 1=0.2 \times F_2 \\ & \Rightarrow F_2=6 \mathrm{~N}\end{aligned}$
$\begin{aligned} & \tau=r_1 \times F_1=r_2 \times F_2 \\ & 1.2 \times 1=0.2 \times F_2 \\ & \Rightarrow F_2=6 \mathrm{~N}\end{aligned}$
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