Transcript Problems involving pulleys file

Problems involving pulleys
The shows two particles
joined by a light
inextensible string which
passes over a fixed pulley.
If the two particles are of
different masses then the
heavier particle will move
vertically downwards and
the lighter particle will move
vertically upwards.
However, the motions of
the two particles are not
independent because they
are connected by the string.
a ms-2
m kg
M kg
Example 1
Particles of masses 5 kg and 3 kg are attached to the ends of a light
inextensible string which passes over a smooth fixed pulley. The system
is released from rest. Find the acceleration of the system and the tension
in the string and also find distance moved by the 5 kg mass in the first 2
seconds of the motion. (Assume that neither particle reaches the pulley
or the ground.)
Solution
a ms-2
a ms-2
T – 3g = 3a
(1)
TN
5g – T = 5a
TN
(2)
(1) + (2) gives :
2g = 8a
Sub the value of a into (1)
So a = 2.45 ms-2
Gives T = 36.75 N
3gN
5gN
u=0
t=2
s = ut + ½ at2
a = 2.45
s=?
s = 0 x 2 + 0.5 x 2.45 x 22
So s = 4.9 m
Example 2
Two particles P and Q of masses 6 kg and 4 kg respectively are
connected by a light inextensible string. Particle P rests on a smooth
horizontal table. The string passes over a smooth pulley fixed at the
edge of the table and Q hangs vertically. There is a friction of 15 N
opposing the motion of the particle. The system is released from rest.
Find (a) the acceleration (b) the tension in the string and (c) the force
exerted on the pulley.
RN
P
a ms-2
TN
a ms-2
15 N
TN
6gN
Q
4g N
a ms-2
RN
Solution
P
TN
a ms-2
15 N
TN
6gN
Q
4g N
For the 4 kg mass:
4g – T = 4a
[1]
For the 6 kg mass:
T – 15 = 6a
[2]
[1] + [2] gives:
a = 2.42 ms-2
Force exerted on the pulley:
From [2]:
T = 29.52 N
29.52
Resultant force = (29.522 + 29.522) = 41.7 N
Acting at 45º to the horizontal
29.52
Example 3
A block of mass 4 kg is attached by light, inextensible strings to particles of
mass 5 kg and 10 kg. The string pass over smooth pulleys at either end of a
smooth horizontal able, as shown. Find the acceleration of the system and the
tensions in the two strings.
R
T2
4 kg
4gN
T2
5 kg
5gN
a
T1
T1
10 kg
10 g N
R
a
Solution
T2
4 kg
T1
T1
4gN
T2
10 kg
5 kg
10 g N
5gN
For the 10 kg mass:
10g – T1 = 10a
[1]
For the 4 kg mass:
T1 – T2 = 4a
[2]
For the 4 kg mass:
T2 – 5g = 5a
[3]
[1] + [2] gives:
10g - T2 = 14a
[4]
[3] + [4] gives:
5g = 19a
[1] Gives: T1 = 72.2 N
[3] Gives: T2 = 61.9 N
Gives: a = 2.58 ms-2
Example 2
Two particles P and Q of masses 8 kg and 4 kg respectively are
connected by a light inextensible string. Particle P rests on a rough
horizontal table. The string passes over a smooth pulley fixed at the
edge of the table and Q hangs vertically. The coefficient of friction
between P and the table is 0.3. The system is released from rest. Find
(a) the acceleration (b) the tension in the string and (c) the force
exerted on the pulley.
RN
P
a ms-2
TN
a ms-2
FN
TN
8g N
Q
4g N
RN
Solution
a ms-2
TN
P
a ms-2
FN
TN
(a)
8g N
Q
Resolve (P  )
R – 8g = 0
R = 8g
4g N
F = R = 0.3 x 8g = 2.4g
Resolve (P )
T – F = 8a
Resolve (Q  )
4g - T = 4a
(2)
(1) + (2)
1.6g = 12a
gives a = 1.31 ms-2
(b)
T – 2.4g = 8a
T – 2.4g = 8a
gives T = 34 N
(1)
Solution
34 N
34 N
(c)
34 N

34 N
Resultant
Resultant = (342 + 342) = 48.1 N
at 450 below horizontal
Example 3
A body of mass 5 kg is placed on a
smooth plane inclined at 300 to the
horizontal. It is connected by a light
inextensible string, which passes
over a smooth pulley at the top of
the plane, to a mass of 3 kg hanging
freely. Find the common
acceleration and the tension in the
string
T – 5gsin300 = 5a :
3g - T = 3a
(1) + (2) :
T – 2.5g = 5a
T
3g N
300
(1)
(2)
0.5g = 8a
T – 2.5g = 5a
so a = 0.6125
So T = 27.6 N