Moment of Forces Acting on a Non-Uniform Rod
Neil Trivedi
Teacher
Moment of Forces Acting on a Non-Uniform Rod
Moment of Force Force Perpendicular Distance
The perpendicular distance is the shortest distance from a point to the force.
In this note, we’ll explore how moments act when forces are applied to a non-uniform rod.
Because the rod is non-uniform, its weight is not evenly distributed, so it doesn’t necessarily act vertically downward through the midpoint. Instead, the centre of mass of the rod can be at any point along the rod.
The general steps for calculating the moments acting on a non-uniform rod about a fixed pivot are essentially the same as for a uniform rod. The only difference is that the centre of mass of the rod is not necessarily at the midpoint.
General Steps for Calculating the Moment about a Fixed Pivot on a Non-Uniform Rod
1) Identify and label all forces acting on the rod (e.g. tensions, weights and reactions), drawing a diagram if necessary.
2) Find the perpendicular distances from the pivot to each force’s line of action. If you cannot draw a perpendicular line from the pivot point to the force, extend the force line on your diagram.
3) Determine whether each force would cause the rod to rotate clockwise or anti-clockwise about the pivot.
4) Calculate the clockwise and anti-clockwise moments separately.
5) Find the resultant (net) moment by subtracting the smaller moment from the larger moment.
6) If we have a question where the system is in equilibrium with unknown forces and distances, we will set all upward forces to equal all downward forces, and that the clockwise moment is equal to the anti-clockwise moment.
Example 1:
The diagram shows a non-uniform rod with a length of m and a mass of kg. The centre of mass of the rod is at a point such that m. The rod is supported by a smooth
pivot , where m. A box of mass kg has been placed at . The box is modelled as a particle. Find the resultant moment about .
Step 1: Identify and label the forces on the diagram.
In this example, the rod exerts a downward weight force of at point , while the box exerts a downward weight force of (we do not consider the reaction force of exerted by the rod on the box as it does not act on the rod). There is also a reaction force exerted by on the rod.
Step 2: Find the perpendicular distances between and all forces’ lines of action.
• The force acts at a perpendicular distance of m.
• The force acts at a perpendicular distance of m.
• The reaction force at acts directly on the pivot, so the perpendicular distance is . Therefore, we can exclude this from our calculation.
Step 3: Determine whether each force would cause the rod to rotate clockwise or
anti-clockwise about .
• The force would cause the rod to rotate anti-clockwise.
• The force would cause the rod to rotate clockwise.
All of this is illustrated in the annotated diagram below.
Step 4: Calculate the clockwise and anti-clockwise moments.
Clockwise moment: Nm
Anti-clockwise moment: Nm
Step 5: Find the resultant moment.
Nm anti-clockwise.
No answer provided.
Example 2:
A wooden plank has a length of m and a mass of kg. The plank is supported by two smooth pivots at its two ends, and . Maria has a mass of kg and stands on the plank at point , where m. The plank is in horizontal equilibrium and the magnitudes of the reactions on the plank at and are each equal to Newtons. The plank is modelled as a non-uniform rod, and the woman is modelled as a particle.
Find the value of and the distance of the centre of mass of the plank from .
Step 1: Label the diagram to illustrate the situation with all the forces and distances.
To note:
• We don’t know where on the plank the centre of mass is and as it’s non-uniform, it’s not necessarily at the centre. So, we pick a point along the plank that’s a distance m from and label it something like . We have chosen since the question has given all distances relative to .
• We don’t consider the reaction force exerted by Maria in the calculation because the plank doesn’t experience Maria’s reaction force, only her weight.
Step 2: Equate the upward/downward forces and solve for .
Because the plank is in equilibrium, the sum of the upward forces equals the sum of the downward forces.
Upward forces:
Downward forces:
We have worked out the value of to be . Now, we need to determine the distance m. To do this, we take moments about .
Step 3: Find the perpendicular distances between and the line of action of each force.
• The force acts at a perpendicular distance of m.
• The force at acts at a perpendicular distance of m.
• The force acts at a perpendicular distance of .
• The force at acts directly on the pivot, so the perpendicular distance is . Therefore, we can exclude this from our calculation.
Step 4: Determine whether each force would cause the plank to rotate clockwise or anti-clockwise about .
• The and forces would cause the plank to rotate clockwise.
• The force at would cause the plank to rotate anti-clockwise.
All of this is illustrated in the annotated diagram below.
Step 5: Calculate the clockwise and anti-clockwise moments.
Clockwise moment: Nm
Anti-clockwise moment: Nm
Step 6: Solve for .
Because the plank is in horizontal equilibrium, the net moment is , meaning that it isn’t rotating in any direction. Therefore, we equate the clockwise and anti-clockwise moments.
m
No answer provided.
Challenging Question