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
A brass ball is hanging from a stiff cotton thread. Draw a neat labeled diagram showing the forces acting on the brass ball and cotton thread
Where can a body execute free vibrations?
The S.I. unit of force is kgf. (kilogram-force)
How does the effect of force differ when it is applied on a rigid body?
State the relationship between force, mass and acceleration. Draw graphs showing the relationship between Acceleration and force for a constant mass.
State one way to reduce the moment of given force about a given axis of rotation.
The moment of a force of 10 N about a fixed point O is 5 N m. Calculate the distance of the point O from the line of action of the force.
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.
The S.I. unit of moment of force is ______.
