Advertisements
Advertisements
प्रश्न
The perpendicular distance between the point of application of force and turning point is 1.75 m when a force of 80 N acts on a rigid body. Calculate the moment of force.
Advertisements
उत्तर
Force (F) = 80 N
Perpendicular distance = 1.75 m
∴ Moment of force = F × perpendicular distance
= 80 × 1.75 = 140 Nm.
संबंधित प्रश्न
A half metre rod is pivoted at the centre with two weights of 20gf and 12gf suspended at a perpendicular distance of 6 cm and 10 cm from the pivot respectively as shown below.
1) Which of the two forces acting on the rigid rod causes clockwise moment?
2) Is the rod in equilibrium?
3) The direction of 20 kgf force is reversed. What is the magnitude of the resultant moment of the forces on the rod?
What are contact forces? Give two Example
Name of S.I. unit of momentum.
A body X of mass 5 kg is moving with velocity 20 m s-2 while another body Y of mass 20 kg is moving with velocity 5 m S-1. Compare the momentum of the two bodies.
The stone of hand flour grinder is provided with a handle near its rim. Give reason.
Two forces each of magnitude 10 N act vertically upwards and downwards respectively at the two ends A and B of a uniform rod of length 4 m which is pivoted at its midpoint O as shown in the figure. Determine the magnitude of the resultant moment of forces about the pivot O.
In a beam balance when the beam is balanced in a horizontal position, it is in ________ equilibrium.
State one factor on which the magnitude of a non-contact force depends. How does the magnitude of a non-contact force change along with distance?
On a see-saw, two children of masses 30 kg and 50 kg are sitting on one side of it at distance 2 m and 2.5 m respectively, from its middle, where should a man of mass 74 kg sit to balance it?
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 ______.
