Home
Class 8
PHYSICS
The amount of force to be applied on the...

The amount of force to be applied on the spanner at a length of 0.25 m from the bolt is 400 N. Find the torque to open a bolt.

Text Solution

AI Generated Solution

The correct Answer is:
To find the torque required to open the bolt using a spanner, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Formula for Torque**: The torque (τ) is calculated using the formula: \[ \tau = \text{Force} \times \text{Distance} \] where: - τ is the torque, - Force is the applied force, - Distance is the length from the pivot point (in this case, the bolt) to the point where the force is applied. 2. **Identify the Given Values**: From the question, we have: - Force (F) = 400 N - Distance (d) = 0.25 m 3. **Substitute the Values into the Formula**: Now, we can substitute the given values into the torque formula: \[ \tau = 400 \, \text{N} \times 0.25 \, \text{m} \] 4. **Calculate the Torque**: Performing the multiplication: \[ \tau = 400 \times 0.25 = 100 \, \text{N m} \] 5. **State the Final Answer**: Therefore, the torque required to open the bolt is: \[ \tau = 100 \, \text{N m} \]
Promotional Banner

Topper's Solved these Questions

  • FORCE AND PRESSURE

    ICSE|Exercise Picture Study |4 Videos
  • FORCE AND PRESSURE

    ICSE|Exercise Exercises (Section II) Long answer questions |5 Videos
  • ENERGY

    ICSE|Exercise THEME ASSIGNMENT |4 Videos
  • HEAT TRANSFER

    ICSE|Exercise Picture Study |2 Videos

Similar Questions

Explore conceptually related problems

A spanner is used to unscrew a nut. A force of 30 N is applied to the end of the spanner, which is 10 cm away from the centre of the nut. Calculate the moment of force when the spanner is horizontal.

A spanner is used to unscrew a nut. A force of 40 N is applied to the end of the spanner, which is 10 cm away from the centre of the nut. Calculate the moment of force when the spanner is horizontal.

A force is applied on a body of mass 0.9kg that is at rest. The force is applied for a duration of 5s and as a result the body covers a distance of 250 m. Find the magnitude of the force.

A rod is pivoted about its centre. A 10 N force is applied 4 m from the pivot and another force of 5 N is applied 2 m from the pivot, as shown. The magnitude of the total torque about the pivot is

The length of a spring is alpha when a force of 4N is applied on it and the length is beta when 5N is applied. Then the length of spring when 9 N force is applied is-

A cubical block of side 1 m and mass 1 kg is pulled by a force F applied to central line so that it slides with an acceleration of 1m//s^(2) . What is the net torque on it about O. (in N-m).

A uniform string of length 10m and mass 20 kg lies on a smooth frictionless inclined plane. A force of 200N is applied as shown in the figure. (a) Find the acceleration of the string. (b) Find the tension in the string at 2m from end A.

A door 1.6 m wide requires a force of 1 N to be applied at the free end to open or close it. The force that is required at a point 0.4 m distant from the hinges for opening or closing the door is