# Post History

$$\sum_i F_i \cdot \delta r_i$$ is virtual work when internal force is $0$. For that reason, $$\sum_i F_i \cdot \delta r_i = 0$$ Here internal force stands for what? When a object's displacement ...

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**#3: Post edited**

- $$\sum_i F_i \cdot \delta r_i$$ is virtual work when internal force is $0$. For that reason, $$\sum_i F_i \cdot \delta r_i = 0$$
- Here internal force stands for what?
~~When a object's displacement is imaginary then, that object's work is virtual work. But, how a object's displacement can be imaginary? Is that only theoretical (just to apply formulas)?~~

- $$\sum_i F_i \cdot \delta r_i$$ is virtual work when internal force is $0$. For that reason, $$\sum_i F_i \cdot \delta r_i = 0$$
- Here internal force stands for what?
- When a object's displacement is imaginary then, that object's work is virtual work. But, how a object's displacement can be imaginary? Is that only theoretical (just to apply formulas)?
**What I understood from the [video](https://www.youtube.com/watch?v=CTPxg72jbYE) that is, when we move a object in a much smaller amount of displacement than, that's called virtual displacement or imaginary displacement. For moving that, we have applied some forces hence, that force is called virtual work. Is that correct?**

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**#2: Post edited**

- $$\sum_i F_i \cdot \delta r_i$$ is virtual work when internal force is $0$. For that reason, $$\sum_i F_i \cdot \delta r_i = 0$$
~~Here internal force stands for what?~~

- Here internal force stands for what?
**When a object's displacement is imaginary then, that object's work is virtual work. But, how a object's displacement can be imaginary? Is that only theoretical (just to apply formulas)?**