The hinge of a door is 1.5 m away from its handle. If a force of 80 N is applied on the handle, find the moment of force.

Open in App

Solution

Step 1: Given data
Given data-
Perpendicular distance d= 1.5 m
Force applied F= 80 N
Step 2: To find
To find: Moment of force
Step 3: Formula
Formula- $M o m e n t o f f o r c e (M) = F o r c e (F) \times P e r p e n d i c u l a r d i s \tan c e b e t w e e n h i n g e a n d h a n d l e (d)$
Step 4: Calculating the moment of force
Calculation-
The moment of force is given by,
$M o m e n t o f f o r c e (M) = F o r c e (F) \times P e r p e n d i c u l a r d i s \tan c e b e t w e e n h i n g e a n d h a n d l e (d) M = 80 N \times 1.5 m = 120 N m$
Hence, The moment of force applied on the door is 120 Nm

Suggest Corrections

Similar questions

If a force is applied at the hinge of a door instead of the handle, then

Q. A door can just be opened with

10 N

force on the handle of the door. The handle is at a distance of

50 c m

from the hinges. Then the torque applied on the door is ( in

N m

Q. Force applied normally on the handle of a door is an example of:

What is the moment of force developed about the hinge of a door if a person tries to open it with a force of 100 N, the force being applied normal to the door at handle which is at a distance of 60 cm from the hinge?

Q. The handle of a door is at a distance

40 c m

from axis of rotation. If a force

5 N

is applied on the handle in a direction

30^{0}

with plane of door, then the torque is:

Join BYJU'S Learning Program