Q. A door $1.6 \, m$ wide requires a minimum force of $1 \, N$ to be applied at the free end to open or close it. The minimum force that is required at a point $0.4 \, m$ away from the hinges for opening or closing the door is
NTA AbhyasNTA Abhyas 2022
Solution:
The torque required in the initial case is,
$\tau=1.6\times 1=1.6Nm$
Now when the point of application of force is changed to,
$d=0.4m$
Hence, the force required will be,
$F=\frac{\tau}{d}=\frac{1 . 6}{0 . 4}=4N$ will be the magnitude of force required to open or close the door.
$\tau=1.6\times 1=1.6Nm$
Now when the point of application of force is changed to,
$d=0.4m$
Hence, the force required will be,
$F=\frac{\tau}{d}=\frac{1 . 6}{0 . 4}=4N$ will be the magnitude of force required to open or close the door.