Question

Two monkeys of masses 10 kg and 8 kg are moving on a vertical light rope, the first is climbing
acceleration 2 ms while the other is climbing down with acceleration 2 ms. Calculate the
Tension of the rope between the monkeys and above the first monkey. Take g= 10 m/s²​

Answer 1

Answer:

The tension will be 200N

Explanation:

According to the problem the mass of the monkeys are given , let them as m1 and m2

The acceleration of the monkey having mass m1 is 2m/s^2 (upward)

The velocity of the another monkey is 2 m/s (downward)\

Let the tension be T

Now the force applied to the first monkey is 10g so as the second one is 8g

Now the free body diagram equation will be

T - (10g+8g) = 10 x 2 + 8 x 0

=> T - 18 g = 20

=> T = 180+20[ putting g= 10 m/s^2]

T = 200

Answer 2

Answer:

Hint: In this question two monkeys are moving in the opposite directions, so, the net tension will be the addition of the forces exerted by the individual monkey on the rope. Now, one of the monkeys is moving at a constant velocity, so only gravitational force will act on it. While, the other has some acceleration, another force will be acting on it in addition to the gravitational force

Complete step-by-step answer:

Before we start solving the question that is given to us, let us take a look at all the parameters that are given in the question

Let the mass of the monkey which is climbing up the rope be m1

So, we have

m1=10kg

Also, the velocity of the monkey that is climbing up is given to us,

a = 2m/s2

So, the forces acting on the rope due to the monkey going up will be

Fup=m1a+m1g

Using the give values

We have

⇒Fup=10×2+10×10

For the ease in calculation we have taken the value of g as 10m/s2instead of 9.8m/s2

So, we have

⇒Fup=120N

Now,

Let the mass of the monkey which is coming down the rope be m2

So, we have

m2=8kg

Also, the velocity of the monkey that is coming up is given to us,

v = 2m/s

Since there is no acceleration, only the gravitational force will act on the monkey coming down

So, the forces acting on the rope due to the monkey coming down will be

FDown=m2g

Now, using the give values

We have

⇒Fup=8×10

⇒Fup=80N

Now, net tension acting on the rope due to the movement of the two monkeys will be

TNet=Fup+FDown

TNet=120N+80N

T net=200N