# TORQUE AND ITS CALCULATIONS

## The Physics lesson. Subject of toque. The turning effect of a force. The applied torque to the objects that rod-shaped. The definition of the torque and its equation. The subject expression and examples.

What is the Tork

If a force perpendicular to the axis of rotation of an object that can rotational motion is applied, this object rotates.

The turning effect of a force is called the torque.

Mathematical Expression of The Torque

Tork is a vectoral magnitude. Its unit according of SI is Newton-meter. The torque is represented by

The torque is equal to the multiplying of force and distance.

The Axis Of Rotation

A line around which an object rotates is called the axis of rotation.

The axis of rotation of a door is a line that passing on the hingers.

The axis of rotation of the Earth is a line assumed to pass through pole points.

The axis of rotation of a car wheel is the center point of the wheel.

1- Torque Applied To A Rod.

The rod shown in the figure can be rotated freely around on point O. The force F which applied to this rod was apply be perpendicular to the position vector of the rod. That is, force F is perpendicular to the line between the point of rotation and the point that the force is applied.

The d in the figure is the position vector.

If a force is applied to a ℓ length rod which can be rotate around on the point O, the torque of the force F is T.

τ = F. ℓ

τ : torque

F: force

ℓ: length

A force that does angle of 37° with the ℓ rod is seen in the figure in the following.

If we want to find torque of the force F, we must find the perpendicular component of the force F. This perpendicular component must be perpendicular to the rod.

Perpendicular component of the force F is the Fy as the following. This component is perpendicular to the rod.

Fy = F.sin37°

Fx = F.cos37°

τ = Fy. ℓ

τ = F.sin37°. ℓ

Example:

in the figure is seen a rod that can be rotate around on the point O and a force F is seen.

Length of the rod is 5 m, and magnitude of the force F is 12 N.

Calculate the torque applied to the rod.

(sin37° = 0.6)

Solution:

First, we must find the perpendicular component of the force F. This component must be perpendiculat to the rod.

The perpendicular component to the rod is found as follows.

Fy = F.sin37°

Fy = 12*0.6

Fy = 7.2 N

Now, let we calculate the torque.

τ = 7.2*5

τ = 36 N.m

Example:

In the figure a rod which can be rotate around of the point O and the forces that applied to that is shown.

Magnitude of the force F1 is 8 N, and magnitude of the force F2 is 5 N.

Angle between the force F1 and the rod is 37°. The distance between the point O and the point that the force F1 is applied is 3 meters

Angle between the force F2 and the rod is 30°. The distance between the point O and the point that the force F2 is applied is 5 meters.

1- Find the total torque.

2- Find the rotate direction of the rod.

(sin37° = 0,6, sin30° = 0,5, cos37° = 0,8, cos30° = 0,87)

First, we finds the perpendicular components of two forces as to the rod.

Then we multiply the perpendicular components by the distances that between the point O and the applied point of the forces F1 and F2 .

The Fx component of the force F1 do not affect rotation of the rod. The Fy component of the force F1 tries to rotate the rod in the direction of the 1.

Component on y-axis of the force F1 is

F1y = F1.sin37

F1y = 8*0.6

F1y = 4.8 N

The torque of the component (F2x) on the x-axis of the force F2 is 0. The component on y-axis of the force F2 tries to rotate the rod in the direction 2.

Component on yaxis of the force F2 is

F2y = F2.sin30

F2y = 5.0,5

F2y = 2.5 N

τ1 = F1y .d1

τ1 = 3*4.8

τ1 = 14.4 N.m

τ2 = F2y.d2

τ2 = 2,5.5

τ2 = 12.5 N.m

Let we suppose that direction 1 as + and direction 2 as –.

τ = τ1 - τ2

τ = 14.4 – 12.5

τ = 1.9 N.m

We had taken the direction 1 as + ans the direction 2 as –.

Because of the sign of the result is +, the rod rotates in the direction 1.

RISE KNOWLEDGE

December 21 2018

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