Moments
In this article, we will be learning about the concept of moments as per the Secondary 3 Physics Syllabus. We will be covering the following subtopics:
- The moment of a force is a measure of the turning effect.
- The formula of the moment of force.
- Moments due to oblique forces.
- Calculation of resultant moment.
Moment of a Force
When a force is applied on a pivoted system, it will cause a turning effect on the system.
This turning effect is known as a moment or torque.
The moment of a force is the product of the force and the perpendicular distance from the pivot to the line of action of the force.
Moment of a force = F × d
where F = force (in N)
d = perpendicular distance from pivot (in m)
The SI unit for the moment is Newton metre (N m).
The moment is a vector quantity and has both the magnitude and direction.
The direction is either clockwise or anti-clockwise.
Practice Questions
Question 1:
A vertical force is applied on a uniform beam as shown below. What is the direction of the moment caused by the force?
- Clockwise
- Anti-clockwise
Solution:
Option B is the correct answer.
Explanation:
Since a downward force is applied on the left side of the beam it will cause the beam to turn in an anti-clockwise direction.
Question 2:
A vertical force is applied on a uniform beam as shown below. What is the direction of the moment caused by the force?
- Clockwise
- Anti-clockwise
Solution:
Option A is the correct answer.
Explanation:
An upward force is applied on the left side of the beam. Hence, this will create a clockwise moment.
Line of Action
The line of action of a force is a line along which a force acts.
In order to find the moments, we need to find the perpendicular distance from the pivot to the line of action of the force.
Question 3:
A uniform 1 m beam is pivoted in the middle. A girl applied a downward force of 50 N at the extreme right end of the beam.
Determine the moment she produces on the beam and state its direction.
Solution:
The beam is 1 m long and has the pivot in the centre.
Moment = F × d
= 50 × 0.5
= 25 Nm clockwise
Since a downward force acts on the right side of the beam, the beam will turn in a clockwise direction.
Question 4:
A uniform 1 m beam is pivoted in the middle. A girl applied a force of 50 N at an angle to the beam as shown below. Determine the magnitude of the moment.
- 0 Nm
- 15 Nm
- 20 Nm
- 25 Nm
Solution:
Option C is the correct answer
Explanation:
Moment = force × distance
Since the angle between the line of action of force and the distance from the pivot should be perpendicular, we have to use the distance of 0.4 m.
Moment = 50 × 0.4
= 20 Nm
Moments due to Oblique Forces
If a force is applied at an angle on the beam, we will need to find the perpendicular distance between the line of action of the force and the pivot.
The perpendicular distance between the line of action of the force to the pivot is d. Hence,
sin 30⁰ = \(\frac {\;d\;}{\;0.5\;}\)
d = 0.5 × sin 30º
= 0.25 m
Moment = force × distance
= 10 × 0.25
= 2.5 N m
Question 5:
A uniform 1 m beam is pivoted in the middle. A girl applied a force of 50 N at an angle to the beam as shown below. Determine the magnitude of the moment.
- 10.2 Nm
- 17.7 Nm
- 25.0 Nm
- 50.0 Nm
Solution:
Option B is the correct answer.
Explanation:
sin 45º = \(\frac {\;d\;}{\;0.5\;}\)
d = sin 45º × 0.5
= 0.3535 m
Moment = 50 0.3535
= 17.7 Nm
Question 6:
Two forces act on a pivoted door as shown below. Calculate the resultant moment due to the forces.
- 50 Nm clockwise
- 50 Nm anti-clockwise
- 75 Nm clockwise
- 75 Nm anti-clockwise
Solution:
Option D is the correct answer.
Explanation:
Clockwise Moment = 50 × 0.9
= 45 Nm
Anticlockwise Moment = 150 × 0.8
= 120 Nm
Resultant Moment = Larger Moment − Smaller Moment
= 120 − 45
= 75 Nm in the Anticlockwise direction
Conclusion
In this article, we have learnt that the moment of a force is a measure of the turning effect. We discussed the formula for calculating the moment of a force. We understood the line of action of a force, the moments due to oblique forces and how to calculate the resultant moment. The details covered are as per the syllabus defined for the Secondary 3 Physics.
Bookmark the article for coming back for a quick refresher and ensure that you understand these concepts thoroughly as these will be critical to solving more complex problems.