Advertisements
Advertisements
Question
Calculate the minimum distance between point of application of force and axis of rotation of a rigid body, if a force 125 N produces a moment of force of 25.75 Nm.
Advertisements
Solution
Moment of force = Force × Perpendicular distance
25.75 Nm = 125 N × Perpendicular distance
Perpendicular distance =`25.75/125` = 0.206 m.
RELATED QUESTIONS
Where can a body execute free vibrations?
In a door, handle is provided ......... from the hinges.
Match the column
| Column A | Column B |
| (a) Camel | (1) broad and deep foundation |
| (b) Truck | (2) broad feet |
| (c) Knife | (3) six or eight tyres |
| (d) High building | (4) sharp cutting edge |
| (e) Thrust | (5) atm |
| (f) Moment of force | (6) N |
| (g) Atmospheric pressure | (7) N m |
State the effect of force F in of the following diagram.

The steering wheel of a vehicle is of large diameter.
A spanner of length 10 cm is used to open a nut by applying a minimum force of 5.0 N. Calculate the moment of force required.
What are non-contact forces? Give two example
A car of mass 600 kg is moving with a speed of 10 ms-1 while a scooter of mass 80 kg us moving with a speed of 50 ms-1. Compare their momentum.
The diagram shows two forces F1 = 5 N and F2 = 3N acting at point A and B of a rod pivoted at a point O, such that OA = 2m and OB = 4m

Calculate:
- Moment of force F1 about O
- Moment of force F2 about O
- Total moment of the two forces about O.
How does the distance of separation between two bodies affect the magnitude of the non-contact force between them?
