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 force always produces both the linear and turning motions.
To obtain a given moment of force for turning a body, the force needed can be decreased by
Write the expression for the moment of force about a given axis of rotation.
On what factors does the effect of thrust on a surface depend?
A wheel of diameter 2 m can be rotated about an axis passing through its center by a moment of force equal to 2.0 Nm. What minimum force must be applied on its rim?
State Newton's third law of motion.
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.
Classify the following amongst contact and non-contact forces.
Frictional force
Calculate the resultant moment of forces about O and state its direction in fig.

The diagram in Fig shows two forces F1 = 5 N and F2 = 3 N acting at points A and B respectively of a rod pivoted at a point O, such that OA = 2 m and OB = 4 m.

Calculate:
- the moment of force F1 about O.
- the moment of force F2 about O.
- total moment of the two forces about O.
