Q 1 :
Two masses 5 kg and 10 kg are suspended from ends of a string (mass less inelastic )
passing over a smooth pulley . Calculate the acceleration of the masses
------------------------
This is the first body acting by two force on is T other is gravity
Now we write the equation
m1g T = m1a
( I write m1g ) because T < than m1g )
Now the second body
T - m2g = m2a
( I write - m2g because T > than gravity )
Now I add both equations
m1g T = m1a
T - m2g = m2a
>
,,,m1g- m2g=m1a+m2a
a = (m1 m 2)/ ( m1 + m 2 ) X g
= - 735.75
To find T
m1g T = m1a
5 X 9.81 T = 5 (- 735.75 )
49.05 T = -3678.78
-T = -3678.78 - 49.05
T = 3727.8
Sorry I write wrong now i correct >>
In like this question required to write equation but as techer said you must beware about (- ) and ( + ) in that
First m1g – T = m1a
( I write –T ) because T less than m1g ( No m1g + T = m1a)
Also about this :
Now the second body
T - m2g = m2a
( I write - m2g because - m2g less than T)
for m1, if consider forces,
T acts upwards, and the gravitational force acts downwards so resultant force would cause the acceleration
upwards,
apply F = ma upwards
(considered a happens upwards)
if m1 accelerates upwards m2 should accelerate downwards then
since the forces act much similarly as above,
apply F=ma downwards,
now solve for T and a (and i think there's something wrong with you answer for a..)
for m1, if consider forces,
T acts upwards, and the gravitational force acts downwards so resultant force would cause the acceleration
upwards,
apply F = ma upwards
(considered a happens upwards)
if m1 accelerates upwards m2 should accelerate downwards then
since the forces act much similarly as above,
apply F=ma downwards,
now solve for T and a (and i think there's something wrong with you answer for a..)