In a simple Atwood machine, two unequal masses m1 and m2 are connected by a string going over a clamped light smooth pulley. In a typical arrangement (figure 5−E7), m1 = 300 g and m2 = 600 g. The system is released from rest. (a) Find the distance travelled by the first block in the first two seconds; (b) find the tension in the string; (c) find the force exerted by the clamp on the pulley
Answers
Explanation:
Given In a simple Atwood machine, two unequal masses m1 and m2 are connected by a string going over a clamped light smooth pulley. In a typical arrangement (figure 5−E7), m1 = 300 g and m2 = 600 g. The system is released from rest. (a) Find the distance travelled by the first block in the first two seconds; (b) find the tension in the string; (c) find the force exerted by the clamp on the pulley
- Here given m2 > m1, so m2 is acting downwards and m1 is acting upwards. Both the masses are different and let the acceleration be a.
- Now for the mass upwards a gravitational force is applied and similarly for mass downwards a gravitational force is applied and the ropes has a force of tension in the upward direction.
- Now the force will be m1g and m2g and the tension will be T
- So we need to calculate the acceleration.
- So for m1 we have net force.
- So T – m1g = m1a
- Similarly for m2 we have
- So m2g - T = m2a
- Now adding we get
- (m2 – m1)g = (m1 + m2)a
- Or a = (m2 – m1 / m1 + m2) g
- = (600 – 300 / 600 + 300) 10
- = 300 / 900 x 10
- = 3.34 m/s^2
- Since initially it starts from rest initial velocity u is zero.
- So now distance travelled we need to find.
- So s = ut + 1/2 at^2
- So s = 1/2 at^2
- = ½ x 3.34 x (2)^2
- = 6.68 m
- Now we need to find the tension.
- So T = m1g + m1a
- = m1 (g + a)
- = 0.3 (9.8 + 3.34) (since 300 g = 0.3 kg)
- = 3.942 N
- Now we need to find the force exerted by the clamp on the pulley. So the force exerted by the clamp is same as force exerted by the pulley.
- So Force = T + T
- = 2 T
- = 2 x 3.942
- = 7.884 N
Reference link will be
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Answer:
Given,
Masses M
1
=300g,M
2
=600g
So,
T−(M
1
g+M
1
a)=0
T=M
1
g+M
1
a......1
Similarly,
T=m
2
−M
2
g.....2
By equating equation 1 and 2 then we get,
M
1
g+M
1
a+M
2
a−M
2
g=0
a(M
1
+M
2
)=g(M
2
−M
1
)
a=
(M
1
+M
2
)
g(M
2
−M
1
)
9.8(
0.6+0.3
0.6−0.3
)
=3.266m/s
2
a)
At t=2s
Initial velocity u=0
Acceleration =3.266m/s
2
So, Distance traveled by the body is:
s=ut+
2
1
at
2
=0+
2
1
3.266×2
2
=6.5m
b)
From equation 1 we get,
T=M
1
(g+a)=0.3(9.8+3.26)=3.9N
c)
The force exerted by the clamp on the pulley is given by,
F−2T=0
F=2T=2×3.9=7.8N
Thus the force exerted on the pulley is 7.8N