Torque is a measure of the tendency of a force to cause an object to rotate about a specific axis or pivot point.
It depends not only on the magnitude of the force but also on the perpendicular distance from the axis of rotation, known as the lever arm, and the angle at which it is applied. Torque essentially quantifies how effectively a force can produce rotational motion, making it the rotational equivalent of linear force.
Examples: Using a wrench to loosen a bolt or pushing a door at its handle to open it.
The torque is given by the cross product between the force and the displacement vector from the pivot point. Thus, mathematically, torque can be written as
Torque = Force × Displacement Vector
OR
{\tau = \vec{r}\times\vec{F} = rF\sin\theta } Where θ is the angle between vector r and vector F.
The SI unit of torque is N·m (Newton-meter) or kg·m²·s⁻². Other units include dyne-cm, pound-feet, and pound-inches. The dimensional formula of torque is [ML²T⁻²], since it is the product of force [MLT⁻²] and distance [L].
Torque Calculation
As shown in the figure, N denotes the axis of rotation, F is the horizontal force applied at p to rotate, and d represents the moment arm (perpendicular distance between the line of action of the force and the axis of rotation).
Torque = Force × Perpendicular distance
τ = F d × sin90° [θ = 90°]
τ = F × d × 1 [sin 90° = 1]
τ = F × d
Or in other words,
\tau = F \times r
Therefore, Torque = Force × Moment of arm
Types of Torque
1. Static Torque
Static torque is the torque applied to maintain an object in rotational equilibrium without causing angular acceleration. When someone pushes on a closed door, the door receives a static torque because, despite the exerted force, it is not spinning about its hinges. Because they are not accelerating, someone riding a bicycle at a steady speed is also creating a static torque. Some other examples of static torque include tightening of the bolt with the help of a wrench, opening the caps of bottles using a bottle opener, turning the steering wheel of the vehicle, etc.
2. Dynamic Torque
The torque that results in angular acceleration is called dynamic torque. When a racing car accelerates off the line, the drive shaft must be creating an angular acceleration of the wheels, given that the vehicle is moving quickly around the track. Also, when you are riding a bicycle, you start pedaling, and the bicycle starts to move at varying speeds, which is also an example of dynamic torque. Some more examples of dynamic torque include spinning a top, the operations of a wind turbine, and the use of a power drill.
Applications of Torque
- Automotive Industry: Torque measures engine power and is transmitted to the wheels to move the vehicle.
- Construction: Used in tightening bolts and screws with tools like torque wrenches for precise application.
- Sports: Players generate torque through body rotation to increase power, e.g., in golf, baseball, and tennis.
- Robotics: Motors produce torque to move arms and joints precisely, enabling robots to perform tasks.
Solved Problems
Question 1: A mechanic applies a force of 400 N to a wrench for loosening a bolt. He applied the force that is perpendicular to the arm of the wrench. The distance between the bolt and the hand is 60 cm. Find out the torque applied.
Solution: As mentioned in the question, the applied force is perpendicular to the arm of the wrench, so the angle will be 90°.
F = 400N
r = 60cm = 60⁄100 = .60
\tau = F \cdot r \cdot \sin \theta Substitute the values:
\tau = 400 \times 0.6 \times \sin 90^\circ
\tau = 400 \times 0.6 \times 1
\boxed{\tau = 240 \, \text{Nm}}
Question 2: The width of a door is 50 cm. A force of 3 N is applied at its edge (which is away from the hinge). Calculate the torque produced that causes the door to open.
Solution: Given
F = 3 N
r = 50 cm = 50/100 = 0.5 m
As the most efficient way to open the door, if the force applied to the edge of the door is perpendicular to the plane of the door, θ = 90°
\tau = F \cdot r \cdot \sin \theta Substitute the values
\tau = 3 \times 0.5 \times \sin 90^\circ
\tau = 3 \times 0.5 \times 1
\boxed{\tau = 1.5 \, \text{Nm}}
Question 3: A motor exerts a torque of 100 N·m on a shaft that has a radius of 0.5 meters. What force is exerted by the motor on the shaft?
Solution: The force exerted by the motor can be calculated using the formula.
\tau = F \cdot r where τ is the torque, F is the force applied, and r is the radius of the shaft.
Given: τ = 100 N·m, r = 0.5 meters.
F = \frac{\tau}{r} Substitute the values:
F = \frac{100}{0.5}
F = 200 \, \text{N}
Question 4: A spanner of length 0.4 m is used to tighten a nut. If a force of 250 N is applied perpendicular to the spanner, calculate the torque exerted on the nut.
Solution: Given
F = 250 N, r = 0.4 m
The force is perpendicular to the spanner, so θ = 90 ° and sin90 ° = 1.
formula for torque
\tau = F \times r \times \sin\theta
\tau = F \times r \times 1
\tau = 250 \times 0.4
\tau = 100 \, \text{N}\cdot{m} The torque exerted on the nut is 100 N · m
Unsolved Problems
Question 1: A uniform solid cylinder of mass 10 kg and radius 0.2 m is placed on a horizontal surface. A horizontal force of 50 N is applied tangentially at the rim. Calculate the torque produced about the cylinder’s axis.
Question 2: A wrench 0.5 m long is used to loosen a bolt. The mechanic applies a force of 300 N at an angle of 60° to the wrench. Find the magnitude of the torque on the bolt.
Question 3: A horizontal rod of length 2 m is fixed at one end. A force of 150 N is applied at a point 1.2 m from the fixed end at an angle of 30° to the rod. Calculate the torque about the fixed end.
Question 4: A door of width 1 m is pushed with a force of 10 N at its handle, perpendicular to the door. Find the torque exerted on the door.
Question 5: A motor applies a torque of 120 N ·m to a shaft of radius 0.3 m. Determine the force applied by the motor on the shaft.