So, there are a load of assumptions when we do such mechanical problems and thus we are effectively creating a model - in real-life, there would be so much more to it.
Firstly, the iron bar is uniform and thus we can assume that its weight acts through its centre. We know therefore that the distance from the wall to the centre of mass is 2 metres as this is half of 4.0 m.
Now, If we take moments about the point that the bar meets the wall, we can effectively 'ignore' any forces that are exerted on the bar by the wall and vice versa. This is because the distance from the pivot and the force is 0 and therefore the moment is 0 - thus, we can ignore the effects of those forces.
We can calculate the clockwise moment of the weight of the bar as well as the anticlockwise moment of the tension in the rope. We know that these two moments are equal as the system is in equilibrium.
Note: there is no force between the rope and the wall as the rope is not exerting a force onto the wall, unlike the bar.
To calculate the moment of the weight of the bar, it is very easy. The actual rope is more challenging as you need to apply a rule, summarised in the equation shown below:
Look at my working below, you can see how I used this above equation. Note that for the moment of the weight, the angle between the force and the perpendicular distance is 90 and sin(90) is just 1, so we can ignore the sin(90) portion.
That is the answer! I am not sure how many marks that question was, but normally it is between 3 and 4 marks - should be easy for you to collect them all up after some practice. The questions do get considerably harder than this, however.
In terms of resources, here is all you need for the time being:
@Apaul466 Yes, of course!
So, there are a load of assumptions when we do such mechanical problems and thus we are effectively creating a model - in real-life, there would be so much more to it.
Firstly, the iron bar is uniform and thus we can assume that its weight acts through its centre. We know therefore that the distance from the wall to the centre of mass is 2 metres as this is half of 4.0 m.
Now, If we take moments about the point that the bar meets the wall, we can effectively 'ignore' any forces that are exerted on the bar by the wall and vice versa. This is because the distance from the pivot and the force is 0 and therefore the moment is 0 - thus, we can ignore the effects of those forces.
We can calculate the clockwise moment of the weight of the bar as well as the anticlockwise moment of the tension in the rope. We know that these two moments are equal as the system is in equilibrium.
Note: there is no force between the rope and the wall as the rope is not exerting a force onto the wall, unlike the bar.
To calculate the moment of the weight of the bar, it is very easy. The actual rope is more challenging as you need to apply a rule, summarised in the equation shown below:
Look at my working below, you can see how I used this above equation. Note that for the moment of the weight, the angle between the force and the perpendicular distance is 90 and sin(90) is just 1, so we can ignore the sin(90) portion.
That is the answer! I am not sure how many marks that question was, but normally it is between 3 and 4 marks - should be easy for you to collect them all up after some practice. The questions do get considerably harder than this, however.
In terms of resources, here is all you need for the time being:
https://www.youtube.com/watch?v=NueuPY2Yqq4