A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey (a) climbs up with an acceleration of 6 m s¯² (b) climbs down with an acceleration of 4 m s¯² (c) climbs up with a uniform speed of 5 m s¯¹ (d) falls down the rope nearly freely under gravity? (Ignore the mass of the rope).
Given: The mass of the monkey is 40 kg and the maximum tension in the rope is 600 N .
a)
The monkey is climbing up with an acceleration of 6 m / s 2 . As the tension in the spring is applied in the upward direction, the acceleration is also in the same direction. Therefore, the force and the weight are acting in the opposite direction with respect to tension.
The free body diagram of the rope is shown below:
The tension on the rope is given as,
T − m g = m a T = m ( g + a )
By substituting the given values in the above equation, we get
T = 40 ( 10 + 6 ) = 640 N
Thus, the tension in the rope when the monkey is moving upward with a constant acceleration is 640 N . As the tension is higher than the maximum tension, so the rope will break.
b)
The monkey is moving down with an acceleration of 4 m / s 2 . As the tension in the spring is applied in the upward direction, so the acceleration is in the opposite direction. Therefore, the force and the weight are acting in opposite direction with respect to tension.
The free body diagram of the rope is shown below:
The tension on the rope is given as,
T − m g = − m a T = m ( g − a )
By substituting the given values in the above equation, we get
T = 40 ( 10 − 4 ) = 240 N
Thus, the tension in the rope when the monkey is moving downward with a constant acceleration is 240 N . As the tension is lower than the maximum tension, so the rope will not break.
c)
As the monkey is moving with a constant velocity, the net acceleration is zero. Thus, the force applied due to the velocity is also zero. The tension in the rope is only balanced with the weight of the monkey.
The tension on the rope is given as,
T − m g = 0 T = m g
By substituting the given values in the above equation, we get
T = 40 ( 10 ) = 400 N
Thus, the tension in the rope when the monkey is moving upwards with a constant acceleration is 400 N . As the tension is lower than the maximum tension, so the rope will not break.
d)
As the monkey is free falling, the rope would be in the state of weightlessness. So the tension in the rope is zero.
Thus, the tension in the rope is 0 N , so the rope will not break.
Given: The mass of the monkey is
a)
The monkey is climbing up with an acceleration of
The free body diagram of the rope is shown below:
The tension on the rope is given as,
By substituting the given values in the above equation, we get
Thus, the tension in the rope when the monkey is moving upward with a constant acceleration is
b)
The monkey is moving down with an acceleration of
The free body diagram of the rope is shown below:
The tension on the rope is given as,
By substituting the given values in the above equation, we get
Thus, the tension in the rope when the monkey is moving downward with a constant acceleration is
c)
As the monkey is moving with a constant velocity, the net acceleration is zero. Thus, the force applied due to the velocity is also zero. The tension in the rope is only balanced with the weight of the monkey.
The tension on the rope is given as,
By substituting the given values in the above equation, we get
Thus, the tension in the rope when the monkey is moving upwards with a constant acceleration is
d)
As the monkey is free falling, the rope would be in the state of weightlessness. So the tension in the rope is zero.
Thus, the tension in the rope is
