# the TENSION in string will b?Two masses 5kg and 4.8kg are attached to a string passing over a pulley such that they move vertically. the tension in string will b ???

unkyd | High School Teacher | (Level 2) Adjunct Educator

Posted on

If you use the acceleration due to gravity (ag) to be 9.8 m/s2 then the masses have a weight (Fg) of:

Fg = mass * ag

= 5 kg * 9.8 m/s2

= 49 N

doing the same for the 4.8 kg give a weight of 47 N.  Since they are pulling against each other along the string that gives a net force of 2 N in favor of the 5 kg mass.  Their combined mass ( 5 kg + 4.8 kg) is  9.8 kg.

Since the entire system (masses and string) will accelerate as one the net force of 2 N will cause an acceleration of:

F = mass (m)  * acceleration (a)

a = F/m = 2 N / 9.8 kg = .204 m/s2

Now that we know what the system is doing its time to look at the masses individually.  It shouldn't matter which one we look at they should both give us the same answer for tension on the string.

Each mass is treated as an object with two forces acting on it (gravity and the tension on the string)  In one case (the 5kg mass) gravity is greater while the other case (4.8 kg mass) the tension is greater.  This is apparent by the way they accelerate  the 5 kg mass will be going down while the 4.8 kg mass will be going up.  Remember, even for independant masses the net force acting it will result in its acceleration (.204 m/s2).  But in one case (5kg) the acceleration is down so the net force is down, while the other case (4.8 kg) the acceleration is up so the net force is up.

For the 5kg mass

Fnet = m *a = 5 kg * .204 m/s2 = 1.02 N down

Fnet = Fg + Tension so

Tension = Fnet - Fg = 1.02 N down - 49 N down

= -47.98 N down  = 47.98 N up

For the 4.8 kg mass (best to check the tension should come out the same)

Fnet = m * a = 4.8 kg * .204 m/s2 = .98 N up = -.98 N down

Fnet = Fg + Tension so

Tension = Fnet - Fg = -.98 N down - 47 N down

= -47.98 N down  = 47.98 N up

Either way we get 47.98 N of tension on the string.

This is the classic atwood's maching application and equations have been derived to find tension and accleration but I find it nice to calculate this out based on laws of motion and forces first.

See the link below for these equations.

Sources: