Advertisements
Advertisements
Question
The monkey B, shown in the following figure, is holding on to the tail of monkey A that is climbing up a rope. The masses of monkeys A and B are 5 kg and 2 kg, respectively. If A can tolerate a tension of 30 N in its tail, what force should it apply on the rope in order to carry monkey B with it? Take g = 10 m/s2.

Advertisements
Solution
Let the acceleration of monkey A upwards be a, so that a maximum tension of 30 N is produced in its tail.
T − 5g − 30 − 5a = 0 ...(i)
30 − 2g − 2a = 0 ...(ii)
From equations (i) and (ii), we have:
T = 105 N (max.)
and a = 5 m/s2
So, A can apply a maximum force of 105 N on the rope to carry monkey B with it.
For minimum force, there is no acceleration of A and B.
T1 = weight of monkey B
⇒ T1 = 20 N
Rewriting equation (i) for monkey A, we get:
T − 5g − 20 = 0
⇒ T = 70 N
∴ To carry monkey B with it, monkey A should apply a force of magnitude between 70 N and 105 N.
APPEARS IN
RELATED QUESTIONS
You are travelling in a car. The driver suddenly applies the brakes and you are pushed forward. Why does this happen?
Suppose you are running fast in a field and suddenly find a snake in front of you. You stop quickly. Which force is responsible for your deceleration?
If the tension in the cable supporting an elevator is equal to the weight of the elevator, the elevator may be
(a) going up with increasing speed
(b) going down with increasing speed
(c) going up with uniform speed
(d) going down with uniform speed
car moving at 40 km/hr is to be stopped by applying brakes in the next 4 m. If the car weighs 2000 kg, what average force must be applied to stop it?
Find the reading of the spring balance shown in the following figure. The elevator is going up with an acceleration g/10, the pulley and the string are light and the pulley is smooth.

An empty plastic box of mass m is found to accelerate up at the rate of g/6 when placed deep inside water. How much sand should be put inside the box so that it may accelerate down at the rate of g/6?
In the previous problem, suppose m2 = 2.0 kg and m3 = 3.0 kg. What should be the mass m, so that it remains at rest?
Use Newton's second law of motion to explain the following instance :
A cricketer pulls his hands back while catching a fast moving cricket ball .
The correct form of Newton's second law is :
The linear momentum of a ball of mass 50 g is 0.5 kg m s-1. Find its velocity.
Calculate the magnitude of force which when applied on a body of mass 0.5 kg produces an acceleration of 5 m s-2.
Calculate the velocity of a body of mass 0.5 kg, when it has a linear momentum of 5 Ns.
A motorcycle of mass 100 kg is running at 10 ms−1. If its engine develops an extra linear momentum of 2000 Ns, calculate the new velocity of a motorcycle.
Multiple Choice Question. Select the correct option.
A force acts on a body of mass 3 kg such that its velocity changes from 4 ms−1 to 10 ms−1. The change in momentum of the body is
What causes motion in a body?
A cricket ball of mass 150 g has an initial velocity `u = (3hati + 4hatj)` m s−1 and a final velocity `v = - (3hati + 4hatj)` m s−1 after being hit. The change in momentum (final momentum-initial momentum) is (in kg m s1)
In the previous problem (5.3), the magnitude of the momentum transferred during the hit is ______.
The motion of a particle of mass m is given by x = 0 for t < 0 s, x(t) = A sin 4 pt for 0 < t < (1/4) s (A > o), and x = 0 for t > (1/4) s. Which of the following statements is true?
- The force at t = (1/8) s on the particle is – 16π2 Am.
- The particle is acted upon by on impulse of magnitude 4π2 A m at t = 0 s and t = (1/4) s.
- The particle is not acted upon by any force.
- The particle is not acted upon by a constant force.
- There is no impulse acting on the particle.
According to Newton's Second Law of Motion, what quantity is directly proportional to the applied force?
How does moving your hand backward while catching a cricket ball reduce pain?
