Question

A door of width 1m can rotate if a
moment of of 10 Nm is applied. The
minimum force that can be applied to
open it is
O A) 8.66 N
O B)10 N
O C5 N
O D)None of the above​

Answer 1

Answer:

10 N

Explanation:

According to this problem, M is the moment of force, F is the force applied and d is the distance from the hinges.

Given :

$M = 10 Nm$

$d = 1 m$

The torque depends only on the magnitude of the force applied and the perpendicular distance of the force’s application from the hinges.

The SI unit of torque is written as Nm and the SI unit of force is N.

$M = F.d$

$10 = F .1\\F = 10 N$

Therefore, the minimum force that can be applied to open is 10 N