Ln order to raise a mass m a man ties it to a rope and passes the rope...
Ans.
Method to Solve :
The smaller mass is accelerated upwards which in turn lifts the the heavier mass upward at some rate as well. The heavier mass is accelerating with 'a' while the smaller mass is accelerating with 'a
m
', while 'a
r
' is the relative acceleration of smaller mass.
Here we shall establish the two equations of motion using the above diagram
T – Mg = Ma…………(b)
Now we shall calculate the value of acceleration 'a' by equating equations (a) and (b).
so,
however, we know that
so, the above relation becomes
a = [g(M-m) + m.a
r
]/(M+m)
now, here
M = 100 kg
m = 60 kg
so,
a = (-400 + 750)/160
or acceleration
a = 35/16 m/s2
= 2.1875 m/s2
now, the tension will be calculated as, using equation (b)
T = M(g+ 2.1875)
= 100 x 12.1875
i.e. T = 1218.75 N
This question is part of UPSC exam. View all JEE courses
Ln order to raise a mass m a man ties it to a rope and passes the rope...
Problem:
In order to raise a mass m, a man ties it to a rope and passes the rope over a frictionless pulley. He climbs the rope with an acceleration 3g/2 relative to the rope. If the mass of the man is m/2 and the mass of the rope is negligible, what is the tension in the rope?
Solution:
1. Identifying the forces:
To find the tension in the rope, we need to consider the forces acting on the system. The forces involved are:
- Weight of the mass m (mg): It acts vertically downwards.
- Tension in the rope (T): It acts vertically upwards on the mass m.
- Weight of the man (m/2 * g): It acts vertically downwards.
- Normal force exerted by the man on the rope (N): It acts vertically upwards.
- Force of friction between the man and the rope (f): It acts horizontally in the direction opposite to the man's acceleration.
2. Applying Newton's second law:
Using Newton's second law, we can write the following equations of motion for the system:
For the mass m:
- ΣFy = T - mg = ma₁
For the man:
- ΣFy = N - m/2 * g = m/2 * a₂
- ΣFx = f = m/2 * a
3. Solving the equations:
Simplifying the equations, we get:
- T = mg + ma₁
- N = m/2 * g + m/2 * a₂
- f = m/2 * a
Since the man is climbing with an acceleration of 3g/2 relative to the rope, we have:
- a = a₁ + a₂
- a = 3g/2
Substituting the values, we get:
- T = mg + ma₁
- T = mg + m * (3g/2)
- T = mg + 3mg/2
- T = 5mg/2
4. Final answer:
Therefore, the tension in the rope is 5mg/2.
Summary:
To find the tension in the rope, we considered the forces acting on the system, including the weight of the mass, tension in the rope, weight of the man, normal force, and force of friction. We applied Newton's second law to set up equations of motion and solved them to find the tension in the rope, which came out to be 5mg/2.
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.