Advertisements
Advertisements
Question
A uniform metre rule rests horizontally on a knife-edge at the 60 cm mark when a mass of 10 gram is suspended from one end. At which end must this mass be suspended? What is the mass of the rule?
Advertisements
Solution
Let M gram be the mass of the rule. Since the density of the rule is uniform, its weight Mg will act at its middle point i.e., at the 50 cm mark (Fig.).
The weight of the rule produces an anti-clockwise moment about the knife-edge O. In order to balance it, 10-gram mass is to be suspended at the end B, to produce a clockwise moment.
From the principle of moments,
Anti-clockwise moment = Clockwise moment
Mg × (60 – 50) = 10 g × (100-60)
or Mg × 10 = 10 g × 40
∴ M = 40 gram
RELATED QUESTIONS
The S.I. unit of force is kgf. (kilogram-force)
Thrust is the ............ force acting on a surface.
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 |
The moment of a force of 25 N about a point is 2.5 N m. Find the perpendicular distance of force from that point.
Classify the following amongst contact and non-contact force.
force of tension in a string
Give two examples of a couple's actions in our daily life.
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?
Write the expression for calculating the moment of force about a given point.
A door lock is opened by turning the lever (handle) of length 0.2 m. If the moment of force produced is 1 Nm, then the minimum force required is ______.
