Question

You attach a meter stick to an oak tree, such that the top of the meter stick is 1.47 meters above the ground. Later, an acorn falls from somewhere higher up in the tree. If the acorn takes 0.311 seconds to pass the length of the meter stick, how high above the ground was the acorn before it fell (assuming that the acorn didn\'t run into any branches or leaves on the way down)?

Answer 1

Answer on Question #57701, Physics / Mechanics | Relativity

A rope is hung at both ends from a horizontal beam, and a weight $m$ is suspended from it. The left part of the rope exerts a force $G$ at $P$, while the right part of the rope exerts a force $H$. Find the indicated quantities from the given data.

$$m = 50 \, \text{kg}$$ $$\theta \, (\text{theta}) = 17{}^\circ$$ $$\varnothing = 6{}^\circ$$ $$|G| = ?$$ $$|H| = ?$$

Solution:

All the forces that are acting are drawn in figure:

The weight is

$$w = mg = 50 \cdot 9.8 = 490 \, \text{N}$$

The forces are resolved into their components, as shown in figure, where

$$T_{1x} = G \cos \theta$$ $$T_{1y} = G \sin \theta$$ $$T_{2x} = H \cos \phi$$ $$T_{2y} = H \sin \phi$$

The first condition of equilibrium must hold. Setting the forces in the $x$-direction to zero

$$\sum F_x = 0$$

gives

$$T_{2x} - T_{1x} = 0$$ $$H \cos \phi = G \cos \theta$$

Taking all the forces in the y-direction and setting them equal to zero,

$$\sum F_y = 0$$ $$T_{1y} + T_{2y} - w = 0$$

Using equations for the components, this becomes

$$G \sin \theta + H \sin \phi = w$$

From equation (1)

$$G = \frac{H \cos \phi}{\cos \theta}$$

If we substitute this equation for G into equation (2)

$$\frac{H \cos \phi}{\cos \theta} \sin \theta + H \sin \phi = w$$ $$H (\cos \phi \cdot \tan \theta + \sin \phi) = w$$

Hence,

$$H = \frac{w}{\cos \phi \cdot \tan \theta + \sin \phi} = \frac{490\ \mathrm{N}}{\cos 6{}^\circ \cdot \tan 17{}^\circ + \sin 6{}^\circ} = 1199\ \mathrm{N}$$ $$G = \frac{H \cos \phi}{\cos \theta} = \frac{1199 \cdot \cos 6{}^\circ}{\cos 17{}^\circ} = 1247\ \mathrm{N}$$

Answer: $|\mathrm{G}| = 1247\ \mathrm{N}$; $|\mathrm{H}| = 1199\ \mathrm{N}$.

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