Question

A horizontal rod with a mass of 10 kg and length 12 m is hinged to a wall at one end
and supported by a cable which makes an angle of 30º with the rod at its other end.
Calculate the tension in the cable and the force exerted by the hinge.

Answer 1

Answer on Question #79618 - Physics - Mechanics - Relativity

A horizontal rod with a mass of $10\mathrm{kg}$ and length $12\mathrm{m}$ is hinged to a wall at one end and supported by a cable which makes an angle of $30{}^{\circ}$ with the rod at its other end. Calculate the tension in the cable and the force exerted by the hinge.

Solution

First, draw a picture and agree that $\beta = 30{}^{\circ}$ :

The rod doesn't move, it means that it is in the state of equilibrium and sum of all forces in $X$ and $Y$ directions is equal to zero, or:

$$0 = - T \cos \beta + F \cos \alpha$$

for $X$ axis and

$$0 = - m g + T \sin \beta + F \sin \alpha$$

for Y,

where $F$ - the force exerted by the hinge, $\alpha$ - angle of $F$ .

Write expression for the torques equilibrium (the pivot point is in the hinge):

$$0 = T \cdot \sin \beta \cdot L - m g \cdot \frac {L}{2}.$$

Then derive $T$ :

$$T = \frac {m g}{2 \sin \beta} = \frac {1 0 \cdot 9 . 8}{2 \sin 3 0 {}^ {\circ}} = 9 8 \mathrm {N}.$$

Substitute $T$ for $\frac{mg}{2\sin\beta}$ in previous expressions and see that $\alpha = \beta$ , then

$$F = \frac {m g}{2 \sin \beta} = \frac {1 0 \cdot 9 . 8}{2 \sin 3 0 {}^ {\circ}} = 9 8 \mathrm {N}.$$

Answer

$$F = T = 9 8 \mathrm {N}.$$

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