Advertisements
Advertisements
प्रश्न
Find the moment of force of 20 N about an axis of rotation at distance 0.5 m from the force.
Advertisements
उत्तर
F = 20 N ⊥ distance = 0.5 m
Moment of force = Force x perpendicular distance of force from the point of rotation
= F x ⊥ distance
= 20 N × 0.5 m
= lONm
APPEARS IN
संबंधित प्रश्न
A force is applied on (i) a non-rigid body and (ii) a rigid body. How does the effect of the force differ in the above two cases?
How does the effect of force differ when it is applied on a rigid body?
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.
State Newton's third law of motion.
Name and define the S.I. and C.G.S. units of force. How are they related?
A uniform metre scale can be balanced at the 70.0 cm mark when a mark when a mass 0.05 kg is hung from the 94.0 cm mark
- draw a diagram of the arrangement
- Find the mass of the metre scale
What is the work done when no net force is applied on the body?
Three forces A, B and C are acting on a rigid body which can turn about O in fig.9. If all the three forces are applied simultaneously, in which direction will the body move? Explain.

The diagram shows a uniform metre rule weighing 100gf, pivoted at its centre O. Two weights 150gf and 250gf hang from the point A and B respectively of the metre rule such that OA = 40 cm and OB = 20 cm. Calculate :
the difference of anticlockwise and clockwise moment

