(i)in a simple attwood machine 2 unequal masses m1 and m2 are connected by a string going over a clamped light smooth pulley.in typical arrangement m1=300g and m2=600g
*thye distance travelled by the first block in first 2 seconds is (the attachemtn is valid for first uestion only)
*the tension in the string is
*draw fbds os m1 ,m2 and pulley
(ii)look at second attachement of integer questions
(iii)2 masses m and M respectively are connected by a light string passing over a smooth pulley when set free m moves up by 1.4 m in 2 secthe ratio of m/M is
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equations of motion or dynamics for the diagram/question are:
T - m1 g = m1 a
m2 g - T = m2 a
=> a = (m2 - m1) g / (m1 + m2)
= (600 - 300) * 10 / (600 + 300) = 10/3 m/sec/sec
If u = 0 , and t = 2 sec.
s = u t + 1/2 a t^2 = 20/3 meters
============
solving the first two equations
T = 2 m1 * m2 * g / (m1 + m2)
= 2 * 600 * 300 * 10 / 900 = 4 000 newtons
=================
iii)
s = 1.4 meters
t = 2 sec.
u = 0.
s = ut + 1/2 a t^2 =>
a = 2 * 1.4 / 4 = 0.7 meters/sec/sec
a = (m2 - m1) g / ( m1 + m2)
= (m2/m1 - 1) g / (m2 / m1 + 1 )
=> m2/m1 + 1 = (m2/m1 -1 ) * g/a
m2/m1 [g/a - 1] = 1 + g/a
m2 / m1 = (a + g) / (g - a)
m2 / m1 = (10 + 0.7) / (10 - 0.7) = 10.7 / 9.3
==========================
Block B is on top of block A.
Block A has mass m. So its acceleration = 10 Newtons / m
It moves by 20 cm before block B falls.
u = 0
s = u t + 1/2 a t^2
20 cm = 0 + 1/2 * (10/m) * t^2
find t from this by substituting the value of m.
=========================
In the question Is the rod AB is resting on a surface OB ??
it is not clear what OB is.. is there friction at the points A and B ??
is it possible that n = 1 or n = 1/2 ??
T - m1 g = m1 a
m2 g - T = m2 a
=> a = (m2 - m1) g / (m1 + m2)
= (600 - 300) * 10 / (600 + 300) = 10/3 m/sec/sec
If u = 0 , and t = 2 sec.
s = u t + 1/2 a t^2 = 20/3 meters
============
solving the first two equations
T = 2 m1 * m2 * g / (m1 + m2)
= 2 * 600 * 300 * 10 / 900 = 4 000 newtons
=================
iii)
s = 1.4 meters
t = 2 sec.
u = 0.
s = ut + 1/2 a t^2 =>
a = 2 * 1.4 / 4 = 0.7 meters/sec/sec
a = (m2 - m1) g / ( m1 + m2)
= (m2/m1 - 1) g / (m2 / m1 + 1 )
=> m2/m1 + 1 = (m2/m1 -1 ) * g/a
m2/m1 [g/a - 1] = 1 + g/a
m2 / m1 = (a + g) / (g - a)
m2 / m1 = (10 + 0.7) / (10 - 0.7) = 10.7 / 9.3
==========================
Block B is on top of block A.
Block A has mass m. So its acceleration = 10 Newtons / m
It moves by 20 cm before block B falls.
u = 0
s = u t + 1/2 a t^2
20 cm = 0 + 1/2 * (10/m) * t^2
find t from this by substituting the value of m.
=========================
In the question Is the rod AB is resting on a surface OB ??
it is not clear what OB is.. is there friction at the points A and B ??
is it possible that n = 1 or n = 1/2 ??
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