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

Answer:

At least $2\; \rm m$.

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.