I am confused with determining the net force on the climbing man below by the rope. My analysis is as follows. Let the tension be $T$. As both hands hold the rope then there are $T+T=2T$ upward acting on the man. Of course there is a $mg$ downward.

My question: What is the correct net tension force acting on the man?

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    $\begingroup$ More on forces and factors of two: physics.stackexchange.com/q/41291/2451 and links therein. $\endgroup$ – Qmechanic Oct 6 '14 at 19:59
  • $\begingroup$ Tension is not a type of force. The type of force in this example is static friction. The tension in the rope is a number that tells you how much force the rope will exert at its ends, not a type of force. $\endgroup$ – user4552 Oct 6 '14 at 21:24

The tension is not the same for all parts of the rope. If the tension is $T_1$ between the hands and $T_2$ between the top hand and the ceiling then.

$$ T_2 = T_1 + F_2 \\ T_1 = F_1 $$

where $F_1+F_2=W$ are the forces acted upon the arms. You arrive at this if you make two free body diagrams, one at each hand.

The result is that $T_2 = W$ and $T_1=F_1<W$. You have to establish the load distribution between the hands $$\gamma = \frac{ F_1}{F_1+F_2} = \frac{F_1}{W}$$ to show that the rope tension between the hands is $T_1 = \gamma W$, $F_1 = \gamma W$ and $F_2 = (1-\gamma) W$.

  • $\begingroup$ This is correct. If there was a flaw in the part between the arms that made the rope tear at $0.6mg$ tension, he would still survive if his weight was equally distributed to both his hands. If such a flaw was above the top hand, the rope would tear. $\endgroup$ – yo' Oct 6 '14 at 20:03
  • $\begingroup$ Sure if the left hand holds more than 60% of the weight he is safe. $\endgroup$ – ja72 Oct 6 '14 at 20:04

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