Question

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 kg
θ (theta) = 17°
Ø = 6°

|G| = ?
|H| = ?

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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