Lesson Video: Moment of a Force about a Point in 2D: Scalar Mathematics

In this video, we will learn how to find the sum of moments of a group of forces acting on a body about a point in 2D.

20:39

Video Transcript

In this video, we will learn how to find the sum of moments of a group of forces acting on a body about a point in two dimensions.

We know that a nonzero net force acting on a rigid body produces a linear acceleration of the body in the direction that the net force acts in. This results in the translation of the center of mass of the body in that direction. However, a force acting on a body may also result in the angular acceleration of the body about a point, producing rotation of the body. The magnitude of this angular acceleration of the body due to the action of the force is proportional to the moment of the force about the point.

We will begin by considering what this looks like in practice. In the diagram drawn, we have a thin rod suspended vertically at a point 𝑃. A force vector 𝐅 acts on the road horizontally. Since a moment is the turning effect of a force about a point, the moment of the force here acts to rotate the rod about 𝑃. We can go one stage further and say that the rod rotates in a clockwise direction due to the moment of vector 𝐅 about 𝑃.

We will see later in this video that moment can act in both a clockwise and counterclockwise direction about a point. Let’s now consider what happens if the line of action of force 𝐅 changes so that the line passes through point 𝑃 as shown in the diagram. This time, the rod does not rotate about 𝑃 due to the moment of 𝐅. And we can conclude that for 𝐅 to have a nonzero moment about 𝑃, there must be a nonzero distance between 𝑃 and the line of action of 𝐅.

When the line of action of the force 𝐅 and the line connecting a point 𝑃 and a point where the force 𝐅 acts are perpendicular, then the magnitude of the moment of the force about point 𝑃 is the product of the magnitude of the force 𝐅 and the distance 𝑑 between the point and the line of action of the force. This can be written as π‘š is equal to 𝐅 multiplied by 𝑑. If the force is measured in newtons and the distance in meters, then the unit of the moment of a force is in newton-meter. We will now look at an example where we need to find the moment of a force about a point.

If a force of magnitude 498 newtons is eight centimeters away from a point 𝐴, find the norm of the moment of the force about point 𝐴, giving your answer in newton-meters.

In this question, we are told that there is a force of magnitude 498 newtons acting at point eight centimeters from a point 𝐴. Whilst the question does not specify that the eight-centimeter distance and the line of action of the force are perpendicular, we can assume this if nothing indicates otherwise. We can therefore add the line of action of our force to the diagram. We are asked to find the norm of the moment of the force, which is its magnitude. And this means it doesn’t matter if the force is acting in a clockwise or counterclockwise direction.

We know that the moment can be calculated using the formula π‘š is equal to 𝐅 multiplied by 𝑑, where 𝐅 is the magnitude of the force in newtons and 𝑑 is the perpendicular distance from the line of action of the force to the point at which we’re taking moments. In order for the moment to be in newton-meters, we need this distance to be in meters. And since there are 100 centimeters in a meter, the perpendicular distance is equal to 0.08 meters. We can therefore calculate the moment by multiplying 498 by 0.08. This is equal to 39.84. And we can therefore conclude that the norm of the moment of the force is 39.84 newton-meters.

We mentioned earlier that a moment of a force can produce rotation of a body either clockwise or counterclockwise. We will briefly consider this before moving on to our next example.

Let’s consider a thin rod suspended vertically at point 𝑃 once again. This time, there are two forces acting on it, 𝐅 sub one and 𝐅 sub two. The force 𝐅 sub one produces a clockwise moment π‘š sub one about point 𝑃. And if we let the perpendicular distance between the line of action of this force and point 𝑃 be 𝑑 sub one, then π‘š sub one is equal to 𝐅 sub one multiplied by 𝑑 sub one. The force 𝐅 sub two, on the other hand, produces a counterclockwise moment π‘š sub two about point 𝑃, where π‘š sub two is equal to 𝐅 sub two multiplied by 𝑑 sub two.

The sum of the moments π‘š sub one and π‘š sub two is Ξ£π‘š, where Ξ£π‘š is equal to π‘š sub two minus π‘š sub one. This can also be written π‘š net. We subtract π‘š sub one from π‘š sub two as counterclockwise moments are usually taken as positive. We will now look at an example where we need to calculate the net moment due to multiple forces.

𝐴𝐡 is a rod of length 114 centimeters and negligible weight. Forces of magnitudes 83 newtons, 225 newtons, 163 newtons, and 136 newtons are acting on the rod as shown in the following figure. 𝐢 and 𝐷 are the points of trisection of 𝐴𝐡, and point 𝑂 is the midpoint of the rod. Find the algebraic sum of the moments of these forces about point 𝑂.

We are asked to find the sum of the moments of the four forces about point 𝑂. In order to do this, we firstly need to calculate the distances between each point on the rod. We are told that 𝐴𝐡 has length 114 centimeters. 𝐢 and 𝐷 are the points of trisection of 𝐴𝐡. And this means that 𝐡𝐷 is equal to 𝐷𝐢, which is equal to 𝐢𝐴. All three of these will be equal to 114 centimeters divided by three. This is equal to 38 centimeters. The lengths 𝐡𝐷, 𝐷𝐢, and 𝐢𝐴 are all equal to 38 centimeters.

We are told that 𝑂 is the midpoint of the rod, which means that 𝐡𝑂 is equal to 𝑂𝐴. These lengths are equal to 114 divided by two, which is equal to 57. The lengths 𝐡𝑂 and 𝑂𝐴 are equal to 57 centimeters. As the lines of action of all four forces are acting perpendicular to the rod, we can calculate the moment of each force using the equation π‘š is equal to 𝐅 multiplied by 𝐷, where 𝐅 is the magnitude of the force measured in newtons and 𝐷 is the perpendicular distance to the point at which we are taking moments, in this question, measured in centimeters. This means that the units for each moment will be newton-centimeters.

We can take moments about any point on the rod. However, in this question, we are asked to find the moment of the forces about point 𝑂. We can find the sum of the moments by firstly finding the moment of each force at points 𝐴, 𝐡, 𝐢, and 𝐷. We are told that moments acting in the counterclockwise direction will be positive. From the diagram, it is clear that this is true for the 83- and 163-newton forces at points 𝐴 and 𝐷, respectively. As these produce counterclockwise moments about 𝑂, the moment will be positive. The forces acting at point 𝐡 and 𝐢 produce clockwise moments about 𝑂. This means that the moments of the 136- and 225-newton forces will be negative.

Let’s begin by calculating the moment of force 𝐴. This is equal to 83 newtons multiplied by a perpendicular distance of 57 centimeters, which is equal to 4731 newton-centimeters. As already mentioned, the moment of the force acting at 𝐡 will be negative. This is equal to negative 136 multiplied by 57, which is equal to negative 7752 newton-centimeters. The moment of the force acting at point 𝐢 is also negative. It is equal to negative 225 multiplied by 19, as the distance 𝑂𝐢 is 19 centimeters. The moment here is equal to negative 4275 newton-centimeters.

Finally, we have the moment of the force acting at 𝐷 which is equal to 163 multiplied by 19. This is equal to 3097 newton-centimeters. The sum of the moments is therefore equal to 4731 minus 7752 minus 4275 plus 3097. This is equal to negative 4199. The algebraic sum of the moments of the four forces about point 𝑂 is negative 4199 newton-centimeters.

In the examples we have seen so far, the line of action of the forces have always been perpendicular to the rods. Let’s now consider what happens when this is not the case. We once again have a thin rod suspended vertically at a point 𝑃. This time, however, the line of action of force 𝐅 makes an angle of πœƒ degrees with the vertical. We know that if πœƒ is zero, the line of force 𝐅 must pass through 𝑃. And this would produce zero moment about 𝑃.

If πœƒ is equal to 90 degrees, which means the line of action is perpendicular to the rod, then 𝐅 produces its maximum moment about 𝑃. This means that the calculation of the moment of a force must therefore include the angle at which the force acts. Using our knowledge of right angle trigonometry, the component of the force acting perpendicular to the rod is therefore equal to 𝐅 multiplied by sin πœƒ. And we can therefore calculate the moment of a force about a point using the formula π‘š is equal to 𝐅 multiplied by 𝑑 multiplied by sin πœƒ, where 𝐅 is the magnitude of the force and πœƒ is the angle between the direction of the force and the direction of the line intersecting 𝑃 and the point where the force acts.

It is worth noting that in our previous examples, πœƒ was equal to 90 degrees. And we know that sin of 90 degrees is equal to one. This allowed us to use the formula π‘š is equal to 𝐅𝑑. We can use this simplified version of the formula when the line of action of our force is perpendicular to the rod. We will now look at one final example where the angles at which the forces act must be considered.

Three forces, measured in newtons, are acting along the sides of an equilateral triangle 𝐴𝐡𝐢 as shown in the figure. Given that the triangle has a side length of seven centimeters, determine the algebraic sum of the moments of the forces about the midpoint of 𝐴𝐡 rounded to two decimal places.

In this question, we are asked to find the sum of the moments of the three forces 300, 100, and 150 newtons. Whilst we could take moments about any point, in this question, we are asked to take them about the midpoint of 𝐴𝐡, labeled 𝑃 on the diagram. Since 𝑃 is the midpoint of 𝐴𝐡, we know that 𝐴𝑃 is equal in length to 𝑃𝐡. And since the side length of the equilateral is seven centimeters, then both 𝐴𝑃 and 𝑃𝐡 are equal to 3.5 centimeters. The line of action of the 300-newton force is along 𝐴𝐡. And it therefore passes through point 𝑃. This means that it produces zero moment about 𝑃 and can therefore be ignored.

We have two other forces of magnitude 100 and 150 newtons that we need to consider. The 100-newton force acts at point 𝐡 and the 150-newton force acts at point 𝐢. Recalling that we can calculate the moment of a force using the formula 𝐅 multiplied by 𝑑 multiplied by sin πœƒ, where 𝑑 is the distance between the point at which the force acts and the point at which we are taking moments, then we need to calculate the length 𝑃𝐢. We can do this using our knowledge of right angle trigonometry and the fact that an equilateral triangle has equal angles equal to 60 degrees.

The tangent ratio tells us that tan πœƒ is equal to the opposite over the adjacent. This means that in triangle 𝑃𝐡𝐢, 𝑃𝐢 over 3.5 is equal to the tangent of 60 degrees. 60 degrees is one of our special angles. And the tangent of 60 degrees is root three. Multiplying through by 3.5, we have 𝑃𝐢 is equal to 3.5 root three centimeters. We can now sketch a diagram showing the magnitude of the forces acting, the angles at which they act, and the distances from 𝑃. Convention dictates that moments acting in a counterclockwise direction are positive. In this question, both the 150-newton force and the 100-newton force act clockwise. This means that the moment of both forces will be negative.

The moment of the 150-newton force is therefore equal to negative 150 multiplied by 3.5 root three multiplied by sin of 30 degrees. We know that sin of 30 degrees is one-half. The moment of this force is therefore equal to negative 525 root three over two newton-centimeters. Repeating this for the 100-newton force, we have a moment equal to negative 100 multiplied by 3.5 multiplied by the sin of 60 degrees. As the sin of 60 degrees is root three over two, this is equal to negative 175 root three newton-centimeters. The sum of the moments about 𝑃 is therefore equal to the sum of these two moments. This can be written Ξ£π‘š or π‘š net and is equal to negative 757.7722 and so on. Rounding to two decimal places as required, the sum of the moments about the midpoint of 𝐴𝐡 is negative 757.77 newton-centimeters.

We will now finish this video by summarizing the key points. The moment of a force 𝐅 about a point 𝑃 is the distance 𝑑 from 𝑃 to the point where the force acts multiplied by the component of the force perpendicular to the direction of the line intersecting 𝑃 and the point where the force acts. This can be written as π‘š is equal to 𝐅 multiplied by 𝑑 multiplied by sin πœƒ, where 𝐅 is the magnitude of the force and πœƒ is the angle between the direction of the force and direction of the line intersecting 𝑃 and the point where the force acts. The net moment due to a set of moments about a point is the sum of the clockwise and counterclockwise moments about the point, where the counterclockwise moments are positive.

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