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Equilibrium & Moments Question

The answer for this question is A but after attempting it and checking the mark scheme I'm not sure how to get to A. I was wondering if someone could explain how to get to the answer?

A light uniform rigid bar is pivoted at its centre. Forces act on the bar at its ends and at the centre. Which diagram shows the bar in equilibrium?
moments image.png
Reply 1
Original post by Izzytee
The answer for this question is A but after attempting it and checking the mark scheme I'm not sure how to get to A. I was wondering if someone could explain how to get to the answer?

A light uniform rigid bar is pivoted at its centre. Forces act on the bar at its ends and at the centre. Which diagram shows the bar in equilibrium?
moments image.png

If you take the distance from the centre to each end to be x, then the moments produced are;
4x Nm clockwise
2x Nm anticlockwise
2x Nm anticlockwise
4x - 2x - 2x = 0, so the bar isn't rotating.
Then you have to check if it's moving. The forces acting on one side are 2N + 8N = 10N
On the other side, they are 4N + 4N + 2N = 10N
These forces are in opposite directions, and 10-10=0, so there is no resultant force, and no resultant moment, meaning the bar is in equilibrium.
B and D also have no moment, but there is a resultant force, so they are accelerating and not in equilibrium.
(edited 4 years ago)
Reply 2
Original post by Izzytee
The answer for this question is A but after attempting it and checking the mark scheme I'm not sure how to get to A. I was wondering if someone could explain how to get to the answer?

A light uniform rigid bar is pivoted at its centre. Forces act on the bar at its ends and at the centre. Which diagram shows the bar in equilibrium?
moments image.png

Calculate the net vertical forces at the three points. On the left side of the bar, the net force is 4-2 = 2N upwards. On the right side of the bar the net force is 2N upwards. In the middle the net force is 8-4 = 4N downwards.

So the total upwards force is 2+2 = 4N. Total downwards force is 4N. Therefore it is balanced. You don't have to worry about moments/distances since the problem is symmetric.
You're looking for the case where
1. the sum of forces acting upwards is equal in magnitude to the sum of forces acting downwards (otherwise it's in linear acceleration up or down)
2. the sum of clockwise moments is equal in magnitude to the sum of moments acting anti-clockwise (otherwise it's in rotational acceleration)
Reply 4
Original post by Joinedup
You're looking for the case where
1. the sum of forces acting upwards is equal in magnitude to the sum of forces acting downwards (otherwise it's in linear acceleration up or down)
2. the sum of clockwise moments is equal in magnitude to the sum of moments acting anti-clockwise (otherwise it's in rotational acceleration)


Original post by 98matt
Calculate the net vertical forces at the three points. On the left side of the bar, the net force is 4-2 = 2N upwards. On the right side of the bar the net force is 2N upwards. In the middle the net force is 8-4 = 4N downwards.

So the total upwards force is 2+2 = 4N. Total downwards force is 4N. Therefore it is balanced. You don't have to worry about moments/distances since the problem is symmetric.


Original post by Zuvio
If you take the distance from the centre to each end to be x, then the moments produced are;
4x Nm clockwise
2x Nm anticlockwise
2x Nm anticlockwise
4x - 2x - 2x = 0, so the bar isn't rotating.
Then you have to check if it's moving. The forces acting on one side are 2N + 8N = 10N
On the other side, they are 4N + 4N + 2N = 10N
These forces are in opposite directions, and 10-10=0, so there is no resultant force, and no resultant moment, meaning the bar is in equilibrium.
B and D also have no moment, but there is a resultant force, so they are accelerating and not in equilibrium.


Thanks guys! I understand the answer now
Doesn't C satisfy both of the conditions for equillibrium? Where resultant force = 0 and sum of moments = 0?
All the forces acting upwards and all the forces acting downwards added together gives you zero for resultant force.
And if you take moments about the centre, they end up being the same.

clockwise force = 2N, anticlockwise force = 2N, let distance from centre to the forces = L
2N*L = 2N*L

Am I making a mistake somewhere?
Reply 6
Original post by ifdhrewofjuhwe
Doesn't C satisfy both of the conditions for equillibrium? Where resultant force = 0 and sum of moments = 0?
All the forces acting upwards and all the forces acting downwards added together gives you zero for resultant force.
And if you take moments about the centre, they end up being the same.
clockwise force = 2N, anticlockwise force = 2N, let distance from centre to the forces = L
2N*L = 2N*L
Am I making a mistake somewhere?

Yeah, you're making a mistake.

Right hand side of the rod has 4N down and 2N up. I. E net clockwise moment around the centre.
Left hand side of the rod has 4N up and 2N down i. E. Net Clockwise moment aroundd the centre.

You need clockwise and anticlockwise to me equal for equilibrium.

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