Question

A rod of length of 5N &10N acting dawn wards find were a third force should be placed for the rod to be in equilibrium? What is the magnitide of this force?

Answer 1

Answer on Question #62362, Physics / Mechanics | Relativity

A rod of length of 5N &10N acting dawn wards find were a third force should be placed for the rod to be in equilibrium? What is the magnitude of this force?

Solution:

Schematic drawing

The third force acts as a balancing force.

The point of application of the balancing forces lies on the straight line joining the points of application of force $F_{1}$ and $F_{2}$ the right of a greater force.

With a known distance between the forces. You can find the location of the point of a third force. Then, according to the rule of moments we have:

$$F _ {1} (l + x) - F _ {2} (x)$$ $$x = \frac {F _ {1} l}{F _ {2} - F _ {1}}$$

The module balances is equal to the difference of the modules acting on the rod force:

$$F = F _ {2} - F _ {1} = 1 0 - 5 = 5 N$$

Answer:5 N

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