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

## 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

mgb_phys
Homework Helper
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.

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!

mgb_phys