## Related questions with answers

To hoist himself into a tree, a 72.0-kg man ties one end of a nylon rope around his waist and throws the other end over a branch of the tree. He then pulls downward on the free end of the rope with a force of 358 N. Neglect any friction between the rope and the branch, and determine the man's upward acceleration.

Solutions

VerifiedWe're given:

$m=72.0\,\text{kg}\\ F_p=358\,\text{N}$

If we neglect the friction, we have to calculate the acceleration of the man.

The mass of a man $(m)$ is known as well as the force $(F)$ with which he acts on the free end taken, it is necessary to determine the acceleration $(a)$ which he receives. We will first analyze the free end of the rope, applying Newton's law that the force with which it acts is equal to the force of tension $(T)$. At the other end taken we have the weight $(W=m \cdot g)$ of a man as well as the force of tension $(T)$. The tightening force is equal at both ends of the rope.

$\begin{align*} m=72\ \text{kg}\\ F=358\ \text{N}\\ \end{align*}$

Analyze the free end of rope:

$\begin{align*} F=T \end{align*}$

Analyze the second end of rope and applying Newton's law:

$\begin{align*} m \cdot a=2 \cdot T-W \end{align*}$

Combine the equations:

$\begin{align*} m \cdot a=T-W\\ m \cdot a=2 \cdot F-m \cdot g\\ a&=\frac{F-m \cdot g}{m}\\ &=\frac{2 \cdot 358\ \text{N} - 72\ \text{kg} \cdot 9.81\ \frac{m}{s^2}}{72\ \text{kg}}\\ &=\frac{716\ \text{N} - 706.32\ \text{N}}{72\ \text{kg}}\\ &=\frac{9.68\ \text{N}}{72\ \text{kg}}\\ &=0.13\ \frac{m}{s^2} \end{align*}$

Therefore, the final result is:

$\boxed{a=0.13\ \frac{m}{s^2}}$

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