Question

a torque of 100Nm is required to open a door. WHAT IS the minimum distance of the handle fromt he hinge. if the door is to be pulled open wth a force at handle not greater than 50N?

Answer 1

At least
`2\; \rm m`.

Step-by-step explanation:

The torque
`\tau` that a force exerts on a lever is equal the product of the following:

• 
  `F`, the size of that force,
• 
  `r`, the distance between the fulcrum and the point where that force is applied, and
• 
  `\sin\theta`, the sine of the angle between the force and the lever.

`\tau = F\cdot r \cdot \sin\theta`.

The force in this question is (at most)
`50\; \rm N`. That is:
`F = 50\; \rm N`.

`\sin \theta` is maximized when
`\theta = 90^\circ`. In other words, the force on the door gives the largest-possible torque when that force is applied perpendicular to the door. When
`\theta = 90^\circ\!`,
`\sin \theta =1`.

If the force here is applied at a distance of
`r` meters away from the hinge (the fulcrum of this door,) the torque generated would be:

`\begin{aligned}\tau &= F \cdot r \cdot \sin \theta \\ &= (50\, r)\; \rm N \cdot m\end{aligned}`.

That torque is supposed to be at least
`100\; \rm N\cdot m`. That is:

`50\, r \ge 100`.

`r \ge 2`.

In other words, the force needs to be applied at a point a minimum distance of
`2\; \rm m` away from the hinge of this door.