Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Moment of force = 2.5 N m
Force applied = 25
⊥ distance from the point of rotation?
Moment of force = Force × ⊥ distance
2.5 = 25 × χ
∴ x = ⊥ distance = `2.5/25 = 1/10` m
= `1/10` × 100
= 10 cm
APPEARS IN
संबंधित प्रश्न
A potter turns his wheel by applying a force through the stick near the rim of wheel.
State Newton's Second law of motion. Under what conditon does it take the form F = ma?
Name of S.I. unit of the rate of change in momentum.
A car of mass 600 kg is moving with a speed of 10 m-1 while a scooter of mass 80 kg is moving with speed of 50 m s-1. which vehicle will require more force to stop it in the (1) same interval of time (2) same distance
What do you understand by the clockwise and anticlockwise moment of force? When is it taken positive?
The figure shows two forces each of magnitude 10 N acting at the point A and B at a separation of 50 cm, in opposite directions. Calculate the resultant moment of two forces about the point.
- A,
- B and
- O situated exactly at the middle of the two forces.
A uniform metre rule of weight 10gf is pivoted at its 0 mark.
What moment of force depresses the rule?
Define the moment of force.
Fig. 5 shows a uniform meter scale weighing 200 gf. Provided at its centre. Two weights 300 gf and 500 gf are suspended from the ruler as shown in the diagram. Calculate the resultant torque of the ruler and hence calculate the distance from mid-point where a 100 gf should be suspended to balance the meter scale.
Give any two effects of a force on a non-rigid body.
