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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.
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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.
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This results in the translation of the center of mass of the body in that direction.
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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.
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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.
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We will begin by considering what this looks like in practice.
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In the diagram drawn, we have a thin rod suspended vertically at a point π.
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A force vector π
acts on the road horizontally.
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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 π.
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We can go one stage further and say that the rod rotates in a clockwise direction due to the moment of vector π
about π.
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We will see later in this video that moment can act in both a clockwise and counterclockwise direction about a point.
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Letβs now consider what happens if the line of action of force π
changes so that the line passes through point π as shown in the diagram.
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This time, the rod does not rotate about π due to the moment of π
.
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And we can conclude that for π
to have a nonzero moment about π, there must be a nonzero distance between π and the line of action of π
.
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When the line of action of the force π
and the line connecting a point π and a point where the force π
acts are perpendicular, then the magnitude of the moment of the force about point π is the product of the magnitude of the force π
and the distance π between the point and the line of action of the force.
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This can be written as π is equal to π
multiplied by π.
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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.
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We will now look at an example where we need to find the moment of a force about a point.
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If a force of magnitude 498 newtons is eight centimeters away from a point π΄, find the norm of the moment of the force about point π΄, giving your answer in newton-meters.
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In this question, we are told that there is a force of magnitude 498 newtons acting at point eight centimeters from a point π΄.
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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.
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We can therefore add the line of action of our force to the diagram.
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We are asked to find the norm of the moment of the force, which is its magnitude.
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And this means it doesnβt matter if the force is acting in a clockwise or counterclockwise direction.
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We know that the moment can be calculated using the formula π is equal to π
multiplied by π, where π
is the magnitude of the force in newtons and π is the perpendicular distance from the line of action of the force to the point at which weβre taking moments.
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In order for the moment to be in newton-meters, we need this distance to be in meters.
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And since there are 100 centimeters in a meter, the perpendicular distance is equal to 0.08 meters.
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We can therefore calculate the moment by multiplying 498 by 0.08.
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This is equal to 39.84.
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And we can therefore conclude that the norm of the moment of the force is 39.84 newton-meters.
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We mentioned earlier that a moment of a force can produce rotation of a body either clockwise or counterclockwise.
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We will briefly consider this before moving on to our next example.
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Letβs consider a thin rod suspended vertically at point π once again.
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This time, there are two forces acting on it, π
sub one and π
sub two.
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The force π
sub one produces a clockwise moment π sub one about point π.
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And if we let the perpendicular distance between the line of action of this force and point π be π sub one, then π sub one is equal to π
sub one multiplied by π sub one.
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The force π
sub two, on the other hand, produces a counterclockwise moment π sub two about point π, where π sub two is equal to π
sub two multiplied by π sub two.
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The sum of the moments π sub one and π sub two is Ξ£π, where Ξ£π is equal to π sub two minus π sub one.
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This can also be written π net.
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We subtract π sub one from π sub two as counterclockwise moments are usually taken as positive.
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We will now look at an example where we need to calculate the net moment due to multiple forces.
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π΄π΅ is a rod of length 114 centimeters and negligible weight.
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Forces of magnitudes 83 newtons, 225 newtons, 163 newtons, and 136 newtons are acting on the rod as shown in the following figure.
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πΆ and π· are the points of trisection of π΄π΅, and point π is the midpoint of the rod.
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Find the algebraic sum of the moments of these forces about point π.
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We are asked to find the sum of the moments of the four forces about point π.
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In order to do this, we firstly need to calculate the distances between each point on the rod.
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We are told that π΄π΅ has length 114 centimeters.
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πΆ and π· are the points of trisection of π΄π΅.
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And this means that π΅π· is equal to π·πΆ, which is equal to πΆπ΄.
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All three of these will be equal to 114 centimeters divided by three.
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This is equal to 38 centimeters.
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The lengths π΅π·, π·πΆ, and πΆπ΄ are all equal to 38 centimeters.
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We are told that π is the midpoint of the rod, which means that π΅π is equal to ππ΄.
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These lengths are equal to 114 divided by two, which is equal to 57.
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The lengths π΅π and ππ΄ are equal to 57 centimeters.
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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 is equal to π
multiplied by π·, where π
is the magnitude of the force measured in newtons and π· is the perpendicular distance to the point at which we are taking moments, in this question, measured in centimeters.
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This means that the units for each moment will be newton-centimeters.
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We can take moments about any point on the rod.
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However, in this question, we are asked to find the moment of the forces about point π.
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We can find the sum of the moments by firstly finding the moment of each force at points π΄, π΅, πΆ, and π·.
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We are told that moments acting in the counterclockwise direction will be positive.
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From the diagram, it is clear that this is true for the 83- and 163-newton forces at points π΄ and π·, respectively.
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As these produce counterclockwise moments about π, the moment will be positive.
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The forces acting at point π΅ and πΆ produce clockwise moments about π.
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This means that the moments of the 136- and 225-newton forces will be negative.
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Letβs begin by calculating the moment of force π΄.
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This is equal to 83 newtons multiplied by a perpendicular distance of 57 centimeters, which is equal to 4731 newton-centimeters.
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As already mentioned, the moment of the force acting at π΅ will be negative.
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This is equal to negative 136 multiplied by 57, which is equal to negative 7752 newton-centimeters.
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The moment of the force acting at point πΆ is also negative.
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It is equal to negative 225 multiplied by 19, as the distance ππΆ is 19 centimeters.
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The moment here is equal to negative 4275 newton-centimeters.
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Finally, we have the moment of the force acting at π· which is equal to 163 multiplied by 19.
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This is equal to 3097 newton-centimeters.
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The sum of the moments is therefore equal to 4731 minus 7752 minus 4275 plus 3097.
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This is equal to negative 4199.
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The algebraic sum of the moments of the four forces about point π is negative 4199 newton-centimeters.
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In the examples we have seen so far, the line of action of the forces have always been perpendicular to the rods.
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Letβs now consider what happens when this is not the case.
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We once again have a thin rod suspended vertically at a point π.
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This time, however, the line of action of force π
makes an angle of π degrees with the vertical.
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We know that if π is zero, the line of force π
must pass through π.
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And this would produce zero moment about π.
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If π is equal to 90 degrees, which means the line of action is perpendicular to the rod, then π
produces its maximum moment about π.
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This means that the calculation of the moment of a force must therefore include the angle at which the force acts.
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Using our knowledge of right angle trigonometry, the component of the force acting perpendicular to the rod is therefore equal to π
multiplied by sin π.
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And we can therefore calculate the moment of a force about a point using the formula π is equal to π
multiplied by π multiplied by sin π, where π
is the magnitude of the force and π is the angle between the direction of the force and the direction of the line intersecting π and the point where the force acts.
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It is worth noting that in our previous examples, π was equal to 90 degrees.
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And we know that sin of 90 degrees is equal to one.
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This allowed us to use the formula π is equal to π
π.
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We can use this simplified version of the formula when the line of action of our force is perpendicular to the rod.
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We will now look at one final example where the angles at which the forces act must be considered.
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Three forces, measured in newtons, are acting along the sides of an equilateral triangle π΄π΅πΆ as shown in the figure.
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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 π΄π΅ rounded to two decimal places.
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In this question, we are asked to find the sum of the moments of the three forces 300, 100, and 150 newtons.
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Whilst we could take moments about any point, in this question, we are asked to take them about the midpoint of π΄π΅, labeled π on the diagram.
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Since π is the midpoint of π΄π΅, we know that π΄π is equal in length to ππ΅.
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And since the side length of the equilateral is seven centimeters, then both π΄π and ππ΅ are equal to 3.5 centimeters.
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The line of action of the 300-newton force is along π΄π΅.
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And it therefore passes through point π.
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This means that it produces zero moment about π and can therefore be ignored.
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We have two other forces of magnitude 100 and 150 newtons that we need to consider.
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The 100-newton force acts at point π΅ and the 150-newton force acts at point πΆ.
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Recalling that we can calculate the moment of a force using the formula π
multiplied by π multiplied by sin π, where π 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 ππΆ.
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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.
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The tangent ratio tells us that tan π is equal to the opposite over the adjacent.
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This means that in triangle ππ΅πΆ, ππΆ over 3.5 is equal to the tangent of 60 degrees.
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60 degrees is one of our special angles.
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And the tangent of 60 degrees is root three.
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Multiplying through by 3.5, we have ππΆ is equal to 3.5 root three centimeters.
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We can now sketch a diagram showing the magnitude of the forces acting, the angles at which they act, and the distances from π.
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Convention dictates that moments acting in a counterclockwise direction are positive.
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In this question, both the 150-newton force and the 100-newton force act clockwise.
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This means that the moment of both forces will be negative.
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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.
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We know that sin of 30 degrees is one-half.
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The moment of this force is therefore equal to negative 525 root three over two newton-centimeters.
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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.
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As the sin of 60 degrees is root three over two, this is equal to negative 175 root three newton-centimeters.
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The sum of the moments about π is therefore equal to the sum of these two moments.
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This can be written Ξ£π or π net and is equal to negative 757.7722 and so on.
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Rounding to two decimal places as required, the sum of the moments about the midpoint of π΄π΅ is negative 757.77 newton-centimeters.
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We will now finish this video by summarizing the key points.
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The moment of a force π
about a point π is the distance π from π to the point where the force acts multiplied by the component of the force perpendicular to the direction of the line intersecting π and the point where the force acts.
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This can be written as π is equal to π
multiplied by π multiplied by sin π, where π
is the magnitude of the force and π is the angle between the direction of the force and direction of the line intersecting π and the point where the force acts.
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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.