Newton's Laws: Finding the tension in a cord.

  • Thread starter Greggers
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  • #1
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Homework Statement


A 5.0-kg mass hangs at the end of a cord. Find the tensionin the cord if the acceleration of the mass is a)1.5 m/s squared up, b) 1.5 m/s squared down

Answers: a) 57 N; b)42 N
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A sample question in our physics book is "An object of mass m is supported by a cord. Find the tension in the cord if the object is a) at rest, b) moving at a constant velocity, c) accelerating upward with acceleration a = 3g/2, and d) accelerating downward at a = 0.75g

a) ay = 0: FT - mg = may = 0 or FT = mg
b) ay = 0: FT - mg = may = 0 or FT = mg
c) ay = 3g/2: FT - mg = m(3g/2) or FT = 2.5mg
d) ay = -3g/4: FT - mg = m(-3g/2)or FT = 0.25mg

Homework Equations


The relevant equations are:

See 1



The Attempt at a Solution


I know there is something I'm just not getting in this problem... It should be so simple but everytime i look at it and attempt it i just keep getting the wrong answer... Also in the sample problem, how are they getting 2.5 out of 3g/2??? It's just not clicking!!!
 

Answers and Replies

  • #2
mgb_phys
Science Advisor
Homework Helper
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Remember f = ma and forces add so you have weight f = m g, and an extra force due to accelration f = m a.
Think about wether acclerating up or down will make the tension more or less to tell you if you should add or subtract the second force.
 
  • #3
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so basically all you do is add?

(mg) + (ma) = (5 x 9.8) + (5 x 1.5) = 56.5 ~57 N
(mg) + (ma) = (5 x 9.8) + (5 x -1.5) = 41.5 ~42 N
(mg) + (ma) = (5 x 9.8) + (5 x -9.8) = 0

ahh it makes so much sense now! i was so fixed on only using one equation. i never thought of using the 2 together. i was thinking too simple now. welli suppose i can blame my teacher for telling us to think simple. thank you!
 
  • #4
mgb_phys
Science Advisor
Homework Helper
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The trick to classical physics is
1, draw a diagram
2, don't do the maths until you understand whats happening
3, it's generally simpler than you think
 

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