# Visualizing Tension in FBD -- Rope With Mass

Asked to find the tension of a point that is halfway up a rope (where the rope has mass), given the system:

The solution to the problem (FBD included) is as follows. In this diagram ($$T_m =$$tension at mid-point, and $$T_t =$$ tension at the top).

Question(s):

(1) Why is $$T_t$$ incorporated into the FBD?

(2) Why does $$T_m$$ point downward?

Also if it helps the FBD's for the individual components are given here:

• You don't identify the system that the first FDB represents. My approach to many mechanics problems is to define the system, and then draw a dotted line around it, leaving outside the line the bits that are not part of the system. Then I identify the forces on the system, and make the FBD. I can do that for a particular choice of system for the first FDB, and with that procedure I find that $T_m$ is a force on that system pulling down. $T_t$ also shows up with my choice of system. See if you can do the same. Apr 21 '14 at 21:58
• Split the rope in half and lump the mass to the blocks, making them 7 kg and 8 kg each. Then the force accelerating the bottom block up is decelerating the top block (equal and opposite forces). Apr 21 '14 at 22:58
• @ja72 TMI for a homework question IMO. Apr 22 '14 at 16:12
• Where in the FBD you have $T_m$? Apr 22 '14 at 17:07
• Tm is in the second diagram provided, not the third. Apr 22 '14 at 20:52

The actual concept here is to divide the mass string into two equal parts or inserting mass less string between the mid point of the string. Hence tension of string depends upon the the net force on the upper half so because to maintain the block with the lower half while the lower half has the overwhelming force of gravity so in order to stabilize the upper half the $T$t is necessary while for net force on upper half due upward force is downward $T$m.

You take the string and break it into two halves, then you analyse the forces on one half. In your FBD we are analysing the forces on top half. Hence, our system is the top half of the string. Mass of system is 2kg(half of 4kg) Our system is in contact with top end of the string which exerts $T_t$ and middle point which exerts $T_m$ in opposite directions(imagine pulling a rope from two ends) and we also have $mg$ the force due to gravity. We can calculate the acceleration of the entire system(i.e. of the two blocks and the string) and then we can apply Newton's second law to our system(the half string) as both will have same acceleration. Hence you get the FBD as in the question. https://drive.google.com/file/d/1icyakZJEDpY01EQr-YEg3sZH5ExV1bD4/view?usp=drivesdk

The other answers are basically right, but don't directly answer the two questions posted by OP. I will try to do so.

(1) Why is Tt incorporated into the FBD?

A free body diagram illustrates the forces acting on a particular object. In this case, the object is the midpoint of the rope. This free body diagram must include Tt, because the top part of the rope is acting on the midpoint of the rope. (what else could possibly be holding the midpoint of the rope up?)

(2) Why does Tm point downward? Since the free body diagram must include all forces acting on the object (in this case the midpoint of the rope), it must also include the tension on the bottom half of the rope. This part of the rope is pulling the midpoint downwards (because that's what tension does - it pulls in the direction of the rope).

Now more to the point - the FBD here is wrong. $$mg$$ should not be acting on the midpoint of the rope, unless we are claiming this point is somehow large enough to have it's own mass (more to the point, which mass is the mass $$m$$?).

The FBD of the 6 kg mass is correct, and the $$m$$ there is 6.0 kg.

The FBD of the 5 kg mass is correct, and the $$m$$ there is 5.0 kg.

From there, the correct way to solve this problem is to break up the mass rope into little sections, and determine the tension as a function of height $$T(x)$$, and then evaluate it at the halfway point. The answer is the same, but the concept is correct and more consistent with Newton's laws.

The figure represents the FBD OF UPPER HALF of the ROPE.

You can see that the mass used in the calculation is $$2_{{Kg}}$$, it's mass of half of the rope .It is the upper half and not lower because $$t_m$$ pulls it down .