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Physics Lesson 6.4.5 - What if the bar is heavy and as such, we cannot neglect its weight?
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Welcome to our Physics lesson on What if the bar is heavy and as such, we cannot neglect its weight?, this is the fifth lesson of our suite of physics lessons covering the topic of Moment of Force. Conditions of Equilibrium, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.
What if the bar is heavy and as such, we cannot neglect its weight?
In this case, we take the weight of the bar as an additional downward force exerted at the bar's centre of gravity. Then, we use the usual approach mentioned above. Look at the following example:
Example 5
A 4 kg uniform and homogenous bar is placed on a pivot as shown in the figure. 
What is the mass of the extra object we must place on the right end of the bar to set the equilibrium?
Solution 5
The bar exerts its weight at its centre because it is uniform and homogenous (as discussed in the tutorial "Centre of Mass. Types of Equilibrium"). Therefore, we must write a force equal at
= 4 kg × 9.81 N/kg
at middle of the bar, i.e. at
= 2m/2
= 1m
away from its left end. This means the centre of mass is 1 m - 0.6 m = 0.4 m on the right of the pivot, as shown below.
Therefore, we have dbar = 0.4 m and dobject = 60 cm = 0.6 m.
It is obvious that the bar's weight produces a anticlockwise moment while the extra object a clockwise moment because when there is equilibrium, the moments of force are equal and opposite.
Therefore, we have:
Simplifying g from both sides, we obtain for the mass of the extra object:
mobject = mbar × dbar/dobject
= 4 × 0.4/0.6
= 2.67 kg
You have reached the end of Physics lesson 6.4.5 What if the bar is heavy and as such, we cannot neglect its weight?. There are 7 lessons in this physics tutorial covering Moment of Force. Conditions of Equilibrium, you can access all the lessons from this tutorial below.
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