What is the tension in the rope and how does it affect the man's equilibrium?

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The tension in the rope is equal to half the combined weight of the man and the seat, expressed as t = (w + s)/2. In the free body diagram for the man, three forces are acting: his weight (downward), the normal force from the seat (upward), and the tension in the rope (upward). The equilibrium condition requires that the total downward force equals the total upward force, leading to the equation w = n + t. This results in the normal force being calculated as n = (w - s)/2. Understanding these forces clarifies the relationship between tension, weight, and normal force in this scenario.
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A man sits in a seat that is suspended from a rope. The rope passes over a pulley suspended from the ceiling, and the man holds the other end of the rope in his hands. What is the tension in the rope and what force does the seat exert on the man? Draw a free body for the man.


The tension in the rope must be equal to the normal force of the seat on the man I think. For the free body I am confused. We have downward force of his weight, or gravity, and the normal force of the seat on him. But then the rope is also acting on his hand even though it is connected to the seat. Very confused.
 
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Weight of man = w
Weight of seat = s
Tension in rope = t
Normal force from seat on man = n

First look at the man + seat together. The rope is supporting the combined weight of these two things, pulling up at two points with the same force, equal to the tension:

w + s = 2t

t = \frac{w + s}{2}

Now you can draw the free body diagram for the man. There are three forces on him, his weight (acting downwards) the normal force from the seat (acting upwards) and the tension in the rope (acting upwards). The total downwards force must be equal to the total upwards force:

w = n + t

n = w - t
= w - \frac{w + s}{2}
= \frac{w - s}{2}
 
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