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Tension

When we exert a force by a rope, a chain or a string, there is a tension force. Usually this rope is massless, so we don't have to consider it as another "body" in our free-body diagram.

  • Tension force is always parallel to the rope, chain or string and is pointing away from the object ( in other words, you can only pull with a rope/string).




Wize Tip
Tension is the same at any point of a string, chain or rope. However, two ropes, chains or strings attached to an objet might have different tensions.

  • The rope/string can be used in conjunction with pulleys which are usually weightless and frictionless in the problems. These pulleys will change the direction of the forces that are applied.






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Example: A Rope Connecting Two Blocks


Two blocks are attached by a rope and are pulled upward by force F. The mass of the lower block is double the mass of the upper block. If the tension along the rope is 5N and the system is moving with an upward acceleration of 2.5m/s22.5m/s^2 , find magnitude of force F and the mass of each block.


Solution:

Since we have two objects in motion, we can write down the second law of Newton for each of them separately.

of forcesFy=m.ay                of forcesFy=2m. ay\sum_{of\ forces}^{ }\vec{F}_y=m.a_{y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\sum_{of\ forces}^{ }\vec{F}_y=2m.\ a_y

  +FTmg=m.ay              +T2mg=2m.ay\Rightarrow\ \ +F-T-mg=m.a_y\ \ \ \ \ \ \ \ \ \ \ \ \ \ +T-2mg=2m.a_y

  +F(5)m(9.81)=m.(2.5)       +52m(9.81)=2m.(2.5)\Rightarrow\ \ +F-\left(5\right)-m\left(9.81\right)=m.\left(2.5\right)\ \ \ \ \ \ \ +5-2m\left(9.81\right)=2m.\left(2.5\right)

  +F(5)m(9.81)=m.(2.5)       +52m(9.81)=2m.(2.5)\Rightarrow\ \ +F-\left(5\right)-m\left(9.81\right)=m.\left(2.5\right)\ \ \ \ \ \ \ +5-2m\left(9.81\right)=2m.\left(2.5\right)


From these two equations, we have:

 F=7.46N         m=0.2kg\Rightarrow\ F=7.46N\ \ \ \ \ \ \ \ \ m=0.2kg





A 20-kg box is hanging from two ropes as shown below. What is the tension in each rope?