# Laws of motion while climbing a Rope

• B
Gold Member

## Main Question or Discussion Point

I'm a little confused with the application of laws of motion on a man climbing a rope. Suppose a man of mass Mg is climbing a rope with an acceleration a. Rope is massless. Now if look through the frame of the piece of rope held by the man, there is a force Mg downward by man, ma downward applied by man and T upward. This balances out as the piece is at rest. This equation is correct, though my way of looking at it maybe incorrect. Now if we look through the man's frame, we have Mg downward, T upward, ma upward the reaction by rope, and as we are in an accelerated frame, ma downward. What am I doing wrong and can you explain how is the man able to climb upward i.e. how the forces are acting to give this motion.

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jedishrfu
Mentor
In these kinds of problems, you should always draw a force diagram.

In this case, the man is a weight and the rope is assumed to be attached to the ceiling. Gravity acts downward on the man. Rope tension acts upward to counter gravity assuming the rope doesn't break.

Here's a better discussion on it with a force diagram and a hand acting to hold the rope steady. (midway down in the Tension topic)

https://opentextbc.ca/physicstestbook2/chapter/normal-tension-and-other-examples-of-forces/

Gold Member
Is the tension force generated in the Rope,T equal to the force applied by the man?
In these kinds of problems, you should always draw a force diagram.

In this case, the man is a weight and the rope is assumed to be attached to the ceiling. Gravity acts downward on the man. Rope tension acts upward to counter gravity assuming the rope doesn't break.

Here's a better discussion on it with a force diagram and a hand acting to hold the rope steady. (midway down in the Tension topic)

https://opentextbc.ca/physicstestbook2/chapter/normal-tension-and-other-examples-of-forces/

jedishrfu
Mentor
Wait which man or example are you referring to?

In the second example in the link I gave you under tension the man's hand is holding the mass steady and so he is providing an upward force = the force of gravity on the mass.

In the case, a man hanging onto the rope his arms have tension forces to counteract the force of gravity.

Gold Member
Wait which man are you referring to? Man pulls the rope downwards with a force F.So,By Newton's Third Law an upward reaction force F acts on the man.
So is the tension produced in the string equal to F?

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jedishrfu
Mentor
Your forces are off. Think of the man hanging straight down.

There is upward tension in his arms due to gravity pulling down and his arms are hanging onto the rope.

Where his hands hold the rope the is a downward tension force and an upward tension force.

Where the rope is attached to the ceiling there is downward tension in the rope and is an upward force at the ceiling.

Since nothing is moving then the tension forces sum to an upward force that counters the gravitational force pulling the man downward.

sophiecentaur
Gold Member
If the climber is accelerating relative to Earth at a, the tension in the rope will be M(g + a); his acceleration will add to g. (with the appropriate sign!)

You say that the man experiences three forces: gravity down, tension up, and “ma upward the reaction from the rope”. This is incorrect. The tension is the only upward force. The idea of reaction is that the net force the man applies down on the rope is equal and opposite to the force the rope applies up on the man. This isn’t some separate force. It is a requirement on the forces.

The man has two forces acting on him: gravity downward, and the rope tension upward. The correct kinematic statement is that the SUM of these forces, that is to say, the NET force equals the mass times the acceleration

ΣF = m a

So

Ftension - Fgravity = m a

The force from gravity is m g so

Ftension - m g = m a

And thus

Ftension = m a + mg

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