The magnitude of a single extra force necessary to keep the rod in its mechanical equilibrium is 20N in the same direction with the 5.0kg blocks' weight force.
A system is in the mechanical equilibrium when it is at rest and the resultant of the force working on the system is equal to zero (0) or no movement happen within the system.
Please take a look at the drawing below.
The existance of 2 blocks attached to each of the rod's ends are enable the rod to move in a circular form. Hence, we could conclude that both of the blocks are giving centripetal forces to the rod.
Centripetal force is the force that works in a system and make a system to move within the curved path. The direction of this force is always perpendicular to the system fixed center point.
On the case, the center point is the center of gravity of rod, where the weight force (W) working in the system. Assuming that the rod is a symetrical solid rod, hence the center point is exatcly at the middle of the rod.
Both of the blocks are giving weight forces to the system with their mass. However since they are differences on their mass, their weight forces are also working differently.
Both of the blocks are working to the system and moving the system into their respective direction, which are the opposite of each other.
Hence, we could conclude that we need to give extra force into one of the block's force direction to balance all the forces working in the system.
Because we want the system to be in its mechanical equilibrium, we have to use the First Law of Newton where the force resultant of the system must be equal to zero (0).
We could write the equation for this system:
∑F = 0
Fs1 - Fs2 + Fs3= 0
(W1 × 1) - (W2 × 1) + (F3 × 1) = 0
(7×10×1) - (5×10×1) + F3 =0
70 - 50 + F3 = 0
F3 = - 20N
Hence, we need to give an extra 20N force in the same direction with the 5.0kg blocks' weight force to make system reaches its mechanical equilibrium.
Learn more about the Mechanical Equilibrium here: https://brainly.com/question/28477917?referrer=searchResults
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