# Moment of force

1. Lever arm

Review an object that rotates, such as the door of a room. When the door is opened or closed, the door rotates. The hinges that connect the door to the wall act as the axis of rotation.

The door image is seen from above. Review an example where the door is pushed in the same two forces that have the same magnitude and direction, where the direction of the force is perpendicular to the door. At first, the door is pushed with a force of F_{1}, r_{1} from the axis of rotation. After that, the door is pushed with the force of F_{2}, r_{2} away from the axis of rotation. Although the magnitude and direction of the force F_{1 }= F_{2}, the force of F_{2} causes the door to rotates faster than the force of F_{1}. In other words, the force of F_{2 }causes a greater angular acceleration compared to the force of F_{1}. You can prove this.

The magnitude of the angular velocity of the moving object is not only influenced by force but is also influenced by the distance between the working points of the force and the rotary axis (r). If the direction of the force is perpendicular to the surface of the object as in the example above, then the lever arm (l) is equal to the distance between the points of work with the axis of rotation (r). What if the direction of the force is not perpendicular to the surface of the object?

Review two other examples, as shown in the figure on the side. Although the magnitude of the force is the same the direction of the force is different so the lever arm (l) is also different. In Figure 3, the direction of the work line of force coincides with the axis of rotation so that the lever arm is zero. The lever arm is known by describing the line from the axis of rotation to the line of the workforce, where the line of the axis of rotation must be perpendicular or form an angle of 90o with a line of force.

Observe figure 2 so that you better understand the equation of the lever arm.

Sin θ = l / r

l = r sin θ

*l = lever arm, r = distance of the point of the workforce with the axis of rotation.*

The equation above is used to calculate the lever arm. If F is perpendicular to r then the angle formed is 90^{o}.

l = r sin 90^{o }= r (1)

l = r

If F coincides with r then the angle formed is 0^{o}.

l = r sin 0^{o }= r (0)

l = 0

2.2 Moment of force (torque)

2.2.1 Magnitude of the moment of force

Mathematically, the magnitude of the moment force is the result of the multiplication of the force (F) and the lever arm (l).

τ = F l

*τ = moment of force (Newton meter), F = force (Newton), l = lever arm (meter)*

Equation 2 is used to calculate the magnitude of the moment of force. The international system of torque is the same as work, but torque is not energy so the unit does not need to be replaced with Joule. Physicists often use the term torque while engineers use the term the moment of force.

2.2.2 Direction of the moment of force

The moment of force is a vector quantity because, in addition to having a magnitude, the moment of force also has a direction. The direction of the moment of force is known easily using the right-hand rule. Rotate the four fingers of your right hand, while the thumb of the right hand is upheld. The direction of the four fingers is the direction of rotation of the object while the direction shown by the thumb is the direction of the moment of force.

If the direction of the moment of force is upward (in the direction of the y-axis) or rightward (in the direction of the x-axis) then the moment of force is positive. Conversely, if the direction of the moment of force is downward (in the direction of the y-axis) or to the leftward (in the direction of the –x-axis), the moment of force is negative. In other words, if the direction of rotation of objects is clockwise then the moment of force is negative. Conversely, if the direction of rotation of objects is opposite to the clockwise rotation then the moment of force is positive.

2.2.3 Sample problems of the moment of force

Sample problems 1.

A block has a length of 8 meters. On the beam works three forces, as in the figure. What is the magnitude of the moment of force that causes the beam to rotate about its center?

Solution:

The center of the beam is located in the middle. long of the beam is 8 meters, therefore the center of the beam is 4 meters from the end of the beam.

Torque 1 = F_{1} l_{1 }= (10 N)(4 m) = 40 N m

Torque 1 is positive because torque 1 causes the beams to rotate opposite clockwise.

Torque 2 = F_{2} l_{2} = (10 N)(2 m) = – 20 N m

Torque 2 is negative because torque 2 causes the beam to rotate clockwise.

Torque 3 = F_{3} l_{3 }= (15 N)(2 m) = – 30 N m

Torque 3 is negative because torque 3 causes the beam to rotate clockwise.

Net torque = 40 N m – 20 N m – 30 N m = – 10 N m

The resultant torque is negative, this indicates that the direction of rotation of the beam is clockwise.

Sample problems 2.

F_{1 }= 10 N, F_{2} = 15 N, F_{3} = 15 N and F_{4 }= 10 N, work on the rod of ABCD as shown in figure. The length of rod ABCD is 20 meters. If the rod mass is ignored, the axis of rotation located at point D determines the magnitude of the moment of force.

Solution:

The question of this problem is what is the net torque that causes the beam to rotate, where the axis of rotation is located at point D.

Torque 1 = F_{1} l_{1} = (10 N)(15 m) = 150 N m

Torque 1 is positive because torque 1 causes the beam to rotate

opposite to the direction of clockwise.

Torque 2 = F_{2 }l_{2 }= (15 N)(5 m) = -75 N m

Torque 2 is negative because torque 2 causes the beam to rotate in the direction clockwise.

Torque 3 = F_{3} l_{3 }= (15 N)(0 m) = 0 N m

Torque 3 is zero because F_{3} coincides with the axis of rotation.

Torque 4 = F_{4} l_{4 }= (10 N)(5 m) = 50 N m

Torque 4 is positive because torque 4 causes the beam to rotate in opposite directions clockwise.

Net torque = 150 N m – 75 N m + 50 N m = 125 N m

The net torque is positive, therefore the direction of rotation of the beam is opposite to the direction of clockwise.