Acceleration with pulley

Classical Mechanics Level 2

In the above diagram, two objects with their respective masses $5$ kg and $3$ kg are hung on a pulley with no friction. Find the magnitude of acceleration $(\text{in m/s}^2 )$ of the left object with mass $5$ kg.

Gravitational acceleration is $g = 10$ m/s $^{2}$ .

2.50 2.00 5.00 1.25

2 solutions

Moulik Gaglani
Jun 19, 2014

Let m1=5 kg, m2=3kg

We know that in this pulley system Fnet=FT(tension force)+ FG(force of gravity), but since there is only one pulley it means that both masses are experiencing the same FT(supporting force FN or in this case FT) and we can reduce the FT so that we are left with Fnet=FG.

FG1= m1*g

     = 5 kg* 9.8 m/s^2     

     = 49 N

FG2= m2*g

      = 3 kg* 9.8 m/s^2

      = 29.4 N

Now we know that if one of the block is moving down, it means that the other must be moving, therefore let's make FG1 be in positive direction and FG2 be in negative direction.

Fnet= sum of all forces Fnet=FG1 + (-FG2)

Fnet= m*a

49N + (-29.4N) = (m1+m2) * a

19.6N = 8a

19.6/8 = 8a/8 , this reduces the 8 from 8a, so we divide both sides by 8

a=2.45 m/s^2

Since we used 9.8 m/s^2 for g, we get a=2.45 m/s^2 but if we used g=10 m/s^2, like it does in the question, a should equal to 2.5 m/s^2.

This is my first answering to anything, so please forgive me for any unconventional methods! Hope this solution was well explained and useful! Thanks!

Mohd Naim Mohd Amin
Jun 11, 2014

Hey yo,

As in the system, both m1 and m2 are having same tension,T and acceleration, a ,

let m1 = 5kg , m2 = 3kg,

As for m1,

W1 - T = m1a

5g - T = 5a

T = 5g - 5a = 50 - 5a(1st)

As for m2,

T - W2 = 3a

T - 3g = 3a

T = 3a + 3g = 3a + 30(2nd)

By comparing (1st) and (2nd),

3a + 30 = 50 - 5a

a = 20 / 8 = 5 / 2 = 2.5 m/s/s....

Thanks....

Acceleration with pulley

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