Three blocks, each having a mass of 1.11 kg, are connected by rigid rods of negl

Question

Three blocks, each having a mass of 1.11 kg, are connected by rigid rods of negligible mass and are supported by a frictionless surface. Forces F1and F2, of magnitude 4.90 N and 10.6 N respectively, are applied to the ends of the blocks as shown in the figure below. Find the forces acting on block B (the +x direction is to the right).

The magnitude of the force applied by the rod on the left is 5.43 N and is in the + x direction. But I can't figure out the magnitude of the force applied by the rod on the right in the - x direction. The picture is below.

Explanation / Answer

The easiest way to understand this type of problem is to imagine that you can sit on each of the blocks and look forward and backward. As you look forward, you can only see what is in front of you. As you look backward, you can only see what is behind you.

But first we need to cover all blocks with a towel and look at the entire assembly!
The total mass = 3 * 1.11 = 3.33 kg
The total force = +4.90 + 10.6 = +15.5 N to the right

Total force = total mass * acceleration of the total mass
15.5 = 3.33 * a
acceleration of the total mass = 15.5 ÷ 3.33

Since the rods are rigid, the each block is accelerating at +(15.5 ÷ 3.33) m/s^2 (to the right)

NOW you move to block B.
If you are sitting on block B, you see a +6.483 N behind you and the rigid rod on the right in front of you.


acceleration = +(15.5 ÷ 3.33) m/s^2
mass = 1.11 kg
Net Force = 1.11 * +(15.5 ÷ 3.33) = +5.166 N

rod on the left is 5.43 N given

Net force = +5.43 + Force of right rod

+5.166 = +5.43 + Force of right rod

F right rod = 5.166 – +5.43 = -0.264 N

The magnitude of the force applied by the rod on the right is = -0.264N

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